Msc Mathcomplex

1

UNIT -I

COMPLEX ANALYSIS

LESSON-1 INTRODUCTION TO THE CONCEPT OF ANALYTIC FUNCTIONS CONTENTS 1.0 Aim and Objectives 1.1 Introduction 1.2 Limits and continuity 1.3 Analytic Functions 1.4 Polynomials 1.5 Rational Functions 1.6 Let us Sum up 1.7 Lesson-end Activities. 1.8 References 1.0 AIM AND OBJECTIVES

The terms such as limits and continuity, differentiability and analyticity of

functions of complex variables at a point are introduced. After going through this lesson, you will be able to

(i) define limit and continuity of a complex function. (ii) define differentiability and analyticity of a complex function. 1.1 INTRODUCTION

A complex – valued function f(z) in a domain D is said to be analytic at a point z = a in D , if there exists a neighbourhood d<- || az , for 0>"d , at all points of which the

function is differentiable. That is f )(' z exists. If f(z) is differentiable at every point of a

domain then the function is said to be analytic in the domain D . The points at which f(z) is not differentiable are called singular points of the function.

1.2 LIMITS AND CONTINUITY

Definition: Let w = f(z) be a function defined in some region containing a point 0z

except perhaps at the point 0z . As z approaches to 0z , the value f(z) of the function is

arbitrarily close to a complex number l .Then we say that the limit of the function f(z) as z approaches to 0z is l .

Definition: A function )(zfw = is said to have the limit l as z tends to 0z , if given

,0>e there exists ed <-> |)(|0, lzfsuchthata , whenever 0< | z - 0z | <d . We

denote this by writing .)(0

lzfLimzz

=®

This watermark does not appear in the registered version - http://www.clicktoconvert.com

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2

(i.e) 0zz

Lim

® l=)(zf exists, if for every ,0,0 >$> de a such that the image of all

points }||/{ 0 d<-Î zzcz lie inside e<- || lw where ).(zfw =

The reader should remember the following points,

(1) If lzfzz

Lim=

®)(

0

exists, then it is unique.

(2) If lzfzz

Lim=

®)(

0

and ,)(0

mzgLimzz

=®

then we have following properties,

(a) ).()()]()([000

zgLimzfLimmlzgzfLimzzzzzz ®®®

+=+=+

(b) )]()([0

zgzfLimzz

-®

).()(00

zgLimzfLimmlzzzz ®®

-=-=

(c) ).().()]().([000

zgLimzfLimlmzgzfLimzzzzzz ®®®

==

(d)0zz

Lim

® )(

)(

)(

)(

0

0

zgLim

zfLim

m

l

zg

zf

zz

zz

®

®== , provided .0¹m

(e) ),()(00

zfLimcclzfLimzzzz ®®

== here c is a complex constant.

(f) lzfLimzz

=®

)(0

(g) 0zz

Lim®

Re f(z) = Re( l ).

(h) ).()(0

llmzlmfLimzz

=®

Definition: Let f (z) be a complex valued function defined in a region D of the complex plane and let .0 Dz Î We say that f(z) is continuous at ).()( 00

0

zfzfLimifzzz

=®

That is f is continuous at 0z if, given 0,0 >$> de a such that,

e<- |)()(| 0zfzf wherever .|| 0 d<- zz f is said to be continuous in D, if it is continuous

at every point of D. A continuous function is one which is continuous at all points where it is defined. If f(z) and g(z) are continuous at 0z , then ),()()(),()()( zgzfiizgzfi -+

),().()( zgzfiii ),()(|,)(|)( zcfvzfiv zviizvi Im)(,Re)( are also continuous at 0z

(viii) )(

)(

zg

zfis continuous at 0z if g( 0z ) .0¹

Definition: A function f(z) is said to be uniformly continuous in a domain D if given

0<e , >$ da 0 (depending only on e ), such that e<- |)()(| 21 zfzf , whenever

2121 ,,||0 zzwherezz d<-< being any two points of the domain D.

The function f(z) which is uniformly continuous on D is continuous on D. But the converse need not be true.



3

Definition: Let f be a complex valued function defined in a region D and let z ÎD. Then f

is said to be differentiable at z, if z

zfzzfLim

Oz D

-D+®D

)()(exists uniquely and is finite and it

is denoted by ).(' zf

Examples (1) The function 2z is differentiable everywhere. (2) The function z is no where differentiable. If f(z) is differentiable at a point z, then it is continuous at that point. But the converse is not true. If f(z) and g(z) are differentiable at a point z, then (i) f(z) + g(z), (ii) f(z)- g(z),

(iii) f(z).g(z), (iv) cf(z) are all differentiable at z , (v) )(

)(

zg

zf is differentiable at z provided

g(z) 0.

Note that 2

'

))((

)(').()(').()()()(')(')()()'.(

zg

zgzfzfzgz

g

fandzgzfzgzfzgf

-=÷÷

ø

öççè

æ+=

1.3 ANALYTIC FUNCTIONS Definition: A function f defined in a region D of the complex plane is said to be analytic at a point Dz Î if it is differentiable at every point of some unspecified neighbourhood of z = a. A function f is said to be analytic in D if it is analytic at every point of D. If the function f(z) is analytic for all finite z, then it is called an entire function.

A polynomial, ze , sinz, cosz are examples of entire function. If a function is analytic at a point ‘a’, then it is differentiable at ‘a’. But the converse is not

true. For example, f(z) = 2|| z is differentiable at z = 0, but it is not analytic there. The sum,

difference and product of two analytic functions f(z) and g(z) are also analytic. The

quotient )(

)(

zg

zf is analytic if g(z) 0.

Necessary Condition The definition of limit can be rewritten in the form

)(' zf = h

zfhzf

h

Lim )()(

0

-+

®. This is the same, irrespective of the path of approach to

zero. If we choose real values for h, then y is kept constant.

Then )(' zf = x

vi

x

u

x

f

¶

¶+

¶

¶=

¶

¶.



4

If h = ik, then )(' zf = 0®h

Lim

ik

zfikzf )()( -+ = -

y

fi

¶

¶

y

v

y

ui

¶

¶+

¶

¶-=

,,x

v

y

u

y

v

x

u

y

fi

x

f

¶

¶-=

¶

¶

¶

¶=

¶

¶Þ

¶

¶-=

¶

¶\ which are the Cauchy – Riemann equations which

are necessary for a function f(z) to be analytic.

If u possesses continuous partial derivatives and if 02

2

2

2

=¶

¶+

¶

¶

y

u

x

u(Laplace Equation), then

u is said to be harmonic. The real and imaginary parts of f(z) = u + iv are both harmonic. v is said to be harmonic conjugate of u and vice versa. Sufficient condition The function f(z) = u + iv, where u and v are functions of x and y, is analytic in a domain D if (i) u and v are differentiable in D and xyyx vuvu -== , .

(ii) The partial derivatives yx uu , , xv and yv are all continuous in D.

1.4. POLYNOMIALS Every constant is an analytic function with the derivative 0. The simplest non-constant analytic function is z whose derivative is 1.

The polynomial p(z) = nn zazazaa ++++ ...2

210 is an analytic function where 1

21 ...2)(' -+++= nn znazaazp . For n > 0, the equation p(z) = 0 has at least one root.

If )( 1ap = 0, then p(z) = )()( 11 zpz a- ; where )z(p1 is a polynomial of degree 1-n .

Thus ))...(()()( 210 nzzzazp aaa ---= .

Suppose that a is a zero of order h, then .)(),()()( opwherezpzzp hhh ¹-= aa

Lucas’s Theorem: If all zeros of a polynomial p(z) lie in a half plane, then all zeros of )(' zp lie in the same half plane .

We have .1

....11

)(

)('

21 nzzzzp

zp

aaa -++

-+

-=

Suppose the half plane H is defined by .0)(

Im <-

b

az

If ka is in the half plane H and zÏ H, then we have

.0)(

Im)(

Im)(

Im >-

--

=-

bb

z

b

z kk aaaa

But the imaginary parts of reciprocal numbers have opposite sign, therefore

0)(Im 1 <- -kzb a .

If this is true for every k, then ,0.)(

)('Im

1

<-

= å= k

m

h

k z

bI

zp

zbp

aconsequently 0)(' ¹zp .



5

1.5 RATIONAL FUNCTIONS

The rational function R(z) = P(z) / Q(z) is the quotient of two polynomials. The zeros of Q(z) are called the poles of R(z), and the order of a pole is equal to the order of the corresponding zero of Q(z).

The derivative 'R (z) = 2)(

)()(')()('

zQ

zPzQzQzP -exists only when Q(z) ¹ 0.

'R (z) has same poles as )z(R , the order of each pole is increased by one.

A rational function )z(R of order p has p zeros and p poles and every equation )z(R =a has exactly p roots.

A rational function of order 1 is a linearization .ocbda,dcz

baz)z(S ¹-

+

+=

Every rational function has a representation by partial fractions. 1.6 LET US SUM UP

1. The concept of limits, continuity, differentiability and analyticity of a function of complex variable. 2. Necessary condition for a function w = f(z) to be analytic. 1.7 LESSON-END ACTIVITIES.

1. Verify the C – R equations of the functions w = sin z, w = ze 2. Assuming the existence of continuous first order partial derivatives of u and v, prove the existence of )(' zf , for f(z) to be analytic.

1.8 REFERENCES 1. Complex analysis by L.V. Alphors, McGraw Hill, New York, 1972. 2. Foundations of Complex Analysis by Dr. S. Ponnusamy.



6

LESSON - 2 CONFORMALITY CONTENTS 2.0 Aim and Objectives 2.1 Introduction 2.2 Arcs and Closed Curves 2.3 Analytic Function in a Region 2.4 Conformal Mapping 2.5 Length and Area 2.6 Let us Sum up 2.7 Lesson end activities 2.8 References 2.0 AIM AND OBJECTIVES

In this lesson we are going to discuss · The geometric properties of functions whose derivatives exist · To prove that if an analytic function f has a non-zero derivative at a point 0z both in

magnitude and direction. After going through this lesson, you will be able to: (i) define arcs and closed curves (ii) analytic functions in a region (iii) conformal mapping (iv) length and area 2.1 INTRODUCTION The existence of the complex valued derivative 'f imposes restrictions of f. It will

be shown that infinitesimally close to 0z where 0)(' 0 ¹zf , f is a rotation by arg )(' 0zf and

a magnification by | )(' 0zf |. Further, when 0)(' 0 ¹zf , then the mapping w = f(z) preserves

the angle between two curves 1c and 2c both in magnitude and direction at 0z . To study

these geometrical properties, the necessary definitions and points are given below. 2.2 ARCS AND CLOSED CURVES

The equations of an arcg in the z-plane in the parametric form are x = O/ (t), y=

y (t) where b££a t and (t), and y (t) are continuous functions. The complex form is

z = f (t) +i y (t).

An arc is the continuous image of a closed finite interval and hence it is compact and connected.

)(')(')(' tittz yf += exists and is not equal to zero, then a tangent to g exists whose slope is

obtained from arg )(' tz .The arc is differentiable if )(' tz exists and is continuous .If

)(' tz exists, continuous and ,0¹ then g is said to be regular.

An arc is piecewise differentiable if )(' tz exists and is continuous except for a finite

number of values of t. But at these points z(t) will be continuous and left and right derivatives will be equal to left and right limits of )(' tz .



7

An arc is called simple or Jordan arc if ).t(z)t(ztt 2121 ¹Þ¹ i.e., if z is 1 – 1.

An arc is closed if the end points are coincident. The opposite arc of b££a= t),t(zz , is the

arc a-££b--= t),t(zz . Opposite arcs are denoted by g and - g . The equation of a circle

whose radius is r and centre ‘a’ is |z-a| = r is the complex form. Any point on this circle is

given by p200, ££+= oireaz .

Example: The equation given by z(t) = cos t + i sin t, where p££ 2t0 represents a unit circle C, |z| = 1, described in the anticlockwise direction which is considered as positive direction. Here z(0) = 1 = z ( p2 ). That is the end points are the same. The negative orientation of this is –C and is given by z(t) = cos ( p2 - t) + i sin ( p2 - t). This is a simple closed curve. But the curve z(t) = cost + i sint where p££ 4t0 is closed but not simple, because

÷ø

öçè

æ p=÷

ø

öçè

æ p

2

5z

2z . The curve z(t) = cost + i sint, p££ t0 is a semi circle arc above the real axis.

This is not closed since )()0( pzz ¹ .

2.3 ANALYTIC FUNCTIONS IN A REGION

If the derivative h

zfhzf

h

Limzf

)()(

0)('

-+

®= exists in an interior of some set A

in which f(z) is differentiable, then we say that f(z) is analytic or holomorphic at z. Hence we discuss about the analyticity we exclude the boundary to A; Definition: A complex-valued function f(z), defined on an open set W is said to be analytic in W ,if its derivative exists at each point of W . Therefore, in this context, we say that a function f(z) is analytic at a point 0z , if f(z) is defined and has a derivative in some

unspecified open neighbourhood of .0z

Definition: Let f be defined and continuous in a region W . Let C be a given curve by the equation ),(tzz = where bta ££ lying in W .then the function )(zf=w defines

another curve 'c in thew -plane and 'c is called the image of c under f.

2.4 CONFORMAL MAPPING Analyticity of conformal mapping Let us now prove that if w =f(z) is conformal, then f(z) is an analytic function. Let ds and dz be the line elements in z – plane and w – plane respectively. 222 dydxds += and

222 dvdud +=s But dyy

udx

x

udu

¶

¶+

¶

¶= and

dyy

vdx

x

vdv

¶

¶+

¶

¶=

222 dy

y

vdx

x

vdy

y

udx

x

ud ÷÷

ø

öççè

æ

¶

¶+

¶

¶+÷÷

ø

öççè

æ

¶

¶+

¶

¶=s\

+úú

û

ù

êê

ë

é

÷÷

ø

ö

çç

è

æ

÷÷ø

öççè

æ

¶

¶+÷÷

ø

öççè

æ

¶

¶+

úú

û

ù

êê

ë

é

÷÷

ø

ö

çç

è

æ÷ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶= 2

222

22

dyy

v

y

udx

x

v

x

udxdy

y

v

x

v

y

u

x

uúû

ùêë

é

¶

¶

¶

¶+

¶

¶

¶

¶.2

Since the transformation is conformal 2ds = 2ds is independent of direction.



8

Therefore comparing the above with 222 dydxds += , we have

011

2222

y

v

x

v

y

u

x

u

y

v

y

u

x

v

x

u

¶

¶

¶

¶+

¶

¶

¶

¶

=÷÷ø

öççè

æ

¶

¶+÷÷

ø

öççè

æ

¶

¶

=

÷ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶

)1______(

2222

÷÷ø

öççè

æ

¶

¶+÷÷

ø

öççè

æ

¶

¶=÷

ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶Þ

y

v

y

u

x

v

x

u

and 0.. =¶

¶

¶

¶+

¶

¶

¶

¶

y

v

x

v

y

u

x

u_____(2) l=

¶

¶-¶

¶

=

¶

¶¶

¶

Þ

y

ux

v

y

ux

u

(say)

y

u

x

vand

y

u

x

u

¶

¶-=

¶

¶

¶

¶=

¶

¶\ ll _____(3)

Put (3) in (1), we get

( ) 0y

v

x

u1

222 =

úú

û

ù

êê

ë

é

÷÷ø

öççè

æ

¶

¶+÷

ø

öçè

æ

¶

¶-l 012 =-l\ Hence 1±=l .

If x

v

y

uand

y

v

x

u

¶

¶--=

¶

¶

¶

¶=

¶

¶+= ,1l ________(4)

If x

v

y

uand

y

v

x

u

¶

¶=

¶

¶

¶

¶--

¶

¶-= ,1l ______(5)

Equation (4) is Cauchy – Riemann equation and hence f(z) is an analytic function. Equation (5) are reduced to (4) by writing –v in place of v, i.e., by taking as image figure obtained by the reflection in the real axis of w – plane.

Definition: Let f be a continuous function defined in a region W . Let WÎoz .Let 1c and

2c be two regular curves passing though 0z , lying in W .Let '2

'1 candc be the images of

1c and 2c is respectively under f. If the angle between 1c and 2c is equal to the angle

between '2

'1 candc both in magnitude and orientation, then f is said to be conformal at

0z . Thus a conformal mapping preserves angle both in magnitude and direction. If the

angle is preserved only in magnitude and the orientation is reserved, then the mapping is said to be isogonal.

Theorem: Let f be an analytic function defined in a region D. Let .DZo Î if 'f ( 0z ) 0,

then f is conformal at oZ .

Proof: Let G be a regular curve lying in W and passing through oZ Î W .

Let the equation of G be btawhere),t(zz ££= . Let oz = z( 0t ), for some 0t Î[a, b ].

The equation of the image curve g of G in the w plane under f is given by w (t) = f(z(t)).

Therefore ).('))((')(' tztzft =w

)('))((')(' 000 tztzft =Þ w )(')(' 00 tzz =w = )(')(' 00 tzzf= .



9

Since G is regular 0)(' 0 ¹tz and by hypothesis 0)(' 0 ¹zf .Therefore 0)(' 0 ¹tw and

=)(' 0tw arg )(' 0zf + arg ).(' 0tz ;a+l=bÞ where b = arg )(' 0tw . a = arg ).(' 0tz and l

= arg )(' 0zf .

a is the angle made by the tangent to the curve G at oz with the positive direction

of x-axis is the z-plane. b is the angle made by the tangent to the image curve g at

)(' 0zf with the positive direction of u-axis in the w -plane.

\The transformations w = f(z) rotates the tangent at oz through an angle l .Let

1G and 2G be two regular curves passing through oz Î W lying wholly in W .Let 1g and 2g be their images in the w-plane under w =f(z). Let 1a and 2a be the angle made by the

tangents to 1G and 2G at oz with the positive directions of x-axis. Let 1b and 2b be the angle

made by the tangents to 1g and 2g at f( oz ) with the positive direction of u-axis .Hence we

have

1b = l + 1a and 2b = l + 2a Þ 2b - 1b = 2a - 1a . Hence f is conformal

2.5 LENGTH AND AREA The important of consequence of conformality is discussed here. If G is an arc

whose parametric equation is z = z(t), bta ££ ,then the length of d is L( G )

= ò +b

a

dttytx 2'2' )()( = òc

a

dttz )(' .

The length of image curve g of G determined by w = w(t) = f (z(t)) is

( )( ) dttztzfLb

aò= )()( ''g \ L ( G ) = ò

g

dz and L ( g ) = dtzòg

g )('

If E is a point set in the plane whose area A(E) = dydx

Eòò and if f(z) = u (x,y) + iv (x,y) is

a bijective differentiable mapping, the area of the image 'E = f(E) is given by

( ) dydxvuvuEA xyyx

Eòò -=' . Since f(z) is conformal in an open set containing E,

using Cauchy – Riemann Equations we get ( ) ( ) dydxzfEAEòò=

2'' .

2.6 LET US SUM UP 1. Definitions of different arcs. 2. Definitions of conformality and isogonality. 3. Consequences of the derivative vanishing at a point. 4. Consequences of conformality.



10

2.7 LESSON END ACTIVITIES

1. If w = 2z find the angle of rotation and scale factor at z = 2 + i. 2. Test the conformality of the following

(a) w = nz , (n is a true integer).

(b) w = z + z

1.

(c) w = sinz.

(d) w = z

1.

(e) w = az + b (f) w = coshz.

2.8 REFERENCES 1. Complex analysis by L.V. Alphors. 2. Complex Analysis by A.R. Vashishtha.



11

LESSON - 3 LINEAR TRANSFORMATIONS CONTENTS 3.0 Aim and Objectives 3.1 Introduction 3.2 The Linear Group 3.3 The Cross-Ratio 3.4 Elementary Riemann Surfaces 3.5 Let us Sum up 3.6 Lesson end activities 3.7 References

3.0 AIM AND OBJECTIVES In this lesson we are going to discuss about a special type of conformal mapping

namely linear fractional transformation, otherwise called mobius transformation or bilinear transformation, and some of the properties of linear fractional transformations such as preservations of cross-ratio, reflection points etc. After going through this lesson, you will be able to: (i) define linear group (ii) cross ratio (iii) elementary Riemann surfaces

3.1 INTRODUCTION

A transformation of the form w = dcz

baz

+

+, with ad-bc ¹ 0 ….. (1) is called a linear

fractional transformation. It is sometimes referred to as bilinear transformation, since it is linear both in w and z. When ad – bc = 1, then (1) is said to be normalized. In the beginning of this lesson we discuss, in detail, some of the features of a linear fractional transformation. 3.2 THE LINEAR GROUP

The linear fractional transformation w = S(z) = 0, ¹-+

+bcadwhere

dcz

baz has an

inverse acw

bdwwSz

+-

-== - )(1 . If we write 21 z/zz = , 21 w/ww = , Then w = S(z) if

211 bzazw += , 212 dzczw += or .z

z

dc

ba

w

w

2

1

2

1

÷÷ø

öççè

æ÷÷ø

öççè

æ=÷÷

ø

öççè

æ Hence

÷÷ø

öççè

æ

++

++=÷÷

ø

öççè

æ÷÷ø

öççè

æ=

21212121

21212121

22

22

11

11

21ddbccdac

dbbacbaa

dc

ba

dc

baSS All linear transformations form

a group. In particular ( 21SS ) 3S = 1S ( 32SS ). The identity w = z is a linear transformation,

and the inverse of a linear transformation is linear. The simplest linear transformation

belong to matrix of the form ÷÷ø

öççè

æ÷÷ø

öççè

æ÷÷ø

öççè

æ

o

o

o

ok

o

a

1

1,

1,

1

1.The first one w = z + a is called a



12

parallel transformation. The second w = kz is rotation, if k = 1, and a homothetic

transformation if k > o. The third transformation z

1w = is called an inversion. If oc ¹ , we

can write ( )

cacdzc

adbc

dcz

baz+

+

-=

+

+

/2,this decomposition shows, that the most general

linear transformation is composed by a translation, an inversion, a rotation, and a homothetic transformation followed by another translation. 3.3 THE CROSS – RATIO

Let 2Z , 3Z , 4Z be any three distinct points. Then there exists a transformation

which map these points into 1, 0, ¥ respectively. If no point is ¥ , then the translation is

( )( )( )( )324

423)(zzzz

zzzzzS

--

--= ………….(1)

If 2z = ¥ , then the transformation is S(z) = 4

3

zz

zz

-

-


42

zz

zz

-

-


3

zz

zz

-

-

If T is another linear transformation which map 2z , 3z , 4z into 1, 0, ¥ , are invariant points

under 1ST- . But this is true only for identify transformation. Hence 1ST- =1 \S = T. Hence S is unique.

Definition: The cross-ratio of four points ( 1z , 2z , 3z , 4z ) is the image of z, under the

linear transformation which maps 2z , 3z , 4z into 1, 0, ¥ respectively.

\( 1z , 2z , 3z , 4z ) = ( )( )( )( )3241

4231

zzzz

zzzz

--

--

Theorem: The cross ratio of four points is invariant under linear transformation.

Proof: ....4,3,2,1,1 == iTz iw , )(11 iTz w-= , 1

1-¹ STz carries 432 ,, TzTzTz into ¥,0,1

Here S(z) is the linear transformation that carries 432 ,, zzz into ¥,0,1 and Tz is any

linear transformation, by the definition of cross-ratio.

( )4321 ,,, TzTzTzTz\ = )( 11 TzST - , = )z(S 1 , = ( 1z , 2z , 3z , 4z ). Hence the theorem.

Note: The linear transformation which maps the given points 1z , 2z , 3z into 321 ,, www is

given by ( )( )( )( )

( )( )( )( )213

312

213

312

zzzz

zzzz

wwww

wwww

--

--=

--

--

Theorem: The cross ratio ( 1z , 2z , 3z , 4z ) is real if and only if the four points lie on a

circle or on a straight line. Proof: Let 1z , 2z , 3z , 4z be concyclic or collinear points.

Then ( 1z , 2z , 3z , 4z )= ( )( )( )( )3241

4231

zzzz

zzzz

--

--.

Hence arg ( 1z , 2z , 3z , 4z ) = arg 42

32

41

31

zz

zzarg

zz

zz

-

--

-

-



13

If ( 1z , 2z , 3z , 4z ) lie on a circle then this difference of arguments is either 0 or p

depending on their relative positions of the four points. Similarly, if 1z , 2z , 3z , 4z are collinear, then also the difference is either 0 or p

depending on their relative positions.

In either case arg ( 1z , 2z , 3z , 4z ) = 0 or ± p . Hence ( 1z , 2z , 3z , 4z ) is real.

Conversely, let ( 1z , 2z , 3z , 4z ) be real.

We have to show that they are either concyclic or collinear. Now, zT = ( 1z , 2z , 3z , 4z ) is real on image of the real axis under 1T- and nowhere else.

\w = 1-zT for real z satisfy Tw = Tw

Hence dwc

bwa

dcw

baw

+

+=

+

+ by cross multiplying, we get

( ) ( ) ( ) ( ) 0|||| 2 =-+-+-+- bddbwadcbwbcdawacca . If acca - = 0, then this

represents a straight line. { }0,0 ¹-=-\ bcdaaccawhen . If 0¹- acca ,then the

equation becomes

acca

bcad

acca

bcda

-

-=

-

-+w and hence it represents a circle.

3.4 ELEMENTARY RIEMANN SURFACES

The Simplest Riemann surface is connected with the mapping by w = nz , where n >1, an integer. There is a one to one correspondence between each angle

n/2|kzarg)n/2()1k( p<<p- , k = 1, 2, …. n, and the whole w-plane except for the positive

real axis. The Riemann surface corresponding to w = ze maps each parallel strip p<<p- 2.kyn/2)1k( onto a sheet with a cut along the positive axis.

3.5 LET US SUM-UP

The existence of inverse of the linear fractional transformation. All linear transformations form a group. A linear transformation is composed of a translation, inversion, rotation. Cross-ratio is invariant under linear transformation. Cross-ratio is real if and only if the four points lie on a circle or on a straight line. Elementary Riemann Surfaces. 3.6 LESSON END ACTIVITIES

(1) If zT1 = 3

2

+

+

z

z,

12

+=

z

zzT , find zTT 21 , zTT 12 and zTT 2

11- .

(2) Find the linear transformation which carries 0, i, - i into 1, -1, 0.

3.7 REFERENCES 1. Complex analysis by L.V. Alphors. 2. Complex Analysis by A.R. Vashishtha.



14

UNIT-2 LESSON – 4 COMPLEX INTEGRATION

CONTENTS 4.0 Aim and Objectives 4.1 Introduction 4.2 Line Integrals 4.3 Rectifiable arcs 4.4 Line Integrals as Function of arcs 4.5 Cauchy’s Theorem for a rectangle 4.6 Cauchy’s Theorem in a circular disc 4.7 Let us Sum up 4.8 Lesson end activities 4.9 References 4.0 AIM AND OBJECTIVES

The aim of this lesson is to study line integrals, rectifiable arcs and the line integral of an analytic function over a rectifiable arc. After going through this lesson, you will be able to (i) define line integrals, rectifiable arcs, line integrals as function of arcs (ii) Cauchy’s theorem for a rectangle and a circular disk.

4.1 INTRODUCTION As in the case of real variables, the indefinite integral of a function of complex variable is the reverse process of differentiation. But in real variable, the definite integral depends only on limit points, since the only path of integration is the real axis. In the case of complex variables, the definite integrals are taken over differentiable or piecewise differentiable arcs and they are confined to analytic functions. As such, the definite integrals of complex functions generally depends on the curve joining the end points. 4.2 LINE INTEGRALS

If f (t) = )t(i)t( y+f is continuous in (a,b), then ò òò f+f=

b

a

b

a

b

a

dt)t(idt)t(dt)t(f

If b+a= ic ,

then òò =

b

a

b

a

dt)t(fcdt)t(cf = ( ) ( )òò -+-b

a

b

a

dtidt bfaybyaf

We shall now show that if f(t) is any complex function, then ( ) ( )dttfdttf

b

a

b

aòò £

Let 0iec -= ,where 0 is real. Then ( ) ( ) úû

ùêë

é=ú

û

ùêë

éòò

--b

a

ib

a

i dttfedttfe 00 ReRe

= ( )ò-

b

a

0i dttfeRe ( )ò-£

b

a

i dttfe 0 = ( )òb

a

dttf



15

{Q real part of a complex number is less than or equal to its modulus}.

( )úú

û

ù

êê

ë

é

ò-

b

a

0i dttfeRe ( )ò£

b

a

dttf ……… (1)

Let arg ( ) ,0ò =b

a

dttf then ( ) ( ) 0ib

a

b

a

edttfdttf -

òò =

(1) Þ ( )úú

û

ù

êê

ë

é

ò- 0i

b

a

0i edttfeRe ( )ò£

b

a

dttf Þ ( ) ( )òò £

b

a

b

a

dttfdttf

Suppose g is a smooth arc given by z = z(t), bta ££ and f(z) is continuous on g .

Then f(z(t)) is continuous at ‘t’ and we define ( ) ( )dttztzfdzzfb

aròò = ')()( .

If g is piecewise differentiable or if )(' tz is piecewise continuous, the interval can be

subdivided in the obvious manner. The integral is invariant under a change of parameter. A change of parameter is defined by an increasing function t = t ( t ) which maps an intervals b£t£a and bta ££ .

We assume that t( t )is piecewise differentiable. By change of variables;

( ) ( ) tttt dttztzfdttztzfb

a

b

a

)(')('))(()('))(( òò =

But ( )( ) ( )( )t=tt tzdt

d)('tt'z .

Therefore ( ) ( ) ttt

tb

a

dtzd

dtzfdttztzf

b

a

)('))(()('))(( òò = .

The integral has the same value whether g is represented by z = z(t) or by z = z(t( t )).

Definition: If g is the arc defined by z = z (t), bta ££ then g- is defined by z = z (-t),

atb -££- .

( )( )ò ò-

-

-

---=\g

a

b

dttztzfdzzf ))('()(

= ( )( )ò-

-

-

a

b

du)u('zuzf , substituting u = -t

= ò-g

dzzf )( . Further, if g = 1g + 2g +…. 'ng then ( )òò òògg gg

+++= dzzf....dz)z(fdz)z(fdz)z(f

1 2



16

Integration with respect to arc lengths:

òò =b

a

dttzzfdzzf )(')()(g

=

òb

a

dttztzf )('))(( Þ òò =

g

b

a

dt)t('z))t(z(fdzf ( )ò =£

b

a

dt)t('z)t(zf dzzfòg

|)(| Therefore if f = 1,

òògg

£ dzdz = length of g .

4.3 RECTIFIABLE ARCS

If z = z(t) is an arc defined in [a,b] and if a = bt...tt n10 =<<< ,then the least

upper bound of all sums, )()(...)()()()( 11201 --++-+- nn tztztztztztz is equal to

length of the arc. If this least upper bound is finite, then the arc is called rectifiable arc. If g is a rectifiable arc and f(z) is continuous on g , then

( )( ) ( ) ( )ò åg =

--¥®

= dztztzuzn

Lt)z(f

n

1k

1kkk , where z( ku ) is any point on the arc joining ( )1ktz -

and ( )ktz . Let us now study the class of integrals of the form òg

+ qdypdx which depend

only on end points of g . In this case if 1g and 2g have same initial and end points, then

òògg

+=+

21

qdypdxqdypdx

Theorem

The line integral òg

+ qdypdx defined in W depends only on the end points of g if

and only if there exists a function u (x,y) Î W with the partial derivatives

qy

up

x

u=

¶

¶=

¶

¶, .

Proof: Necessary part

Let the integral òg

+ qdypdx depend only on the end points of g . Choose a fixed

point ( ) WÎ00 y,x and join (x,y) by a polygon g contained in W with sides parellel to axes.

Since the integral depends on end points, let ( ) ò +=g

qdypdxyxu , .

),( 00 yx

),( yx

g



17

If we choose the segment of g horizontal we can keep y constant. \ dy = 0.

Suppose that x varies without changing the other segments. Choosing x as a parameter on the last segment, we obtain

( ) ( )ò +=

x

dxy,xpy,xu constant …………………..(1) { }0dy =Q

We do not specify the lower limit since it is immaterial for our purpose.

From (1), we get px

u=

¶

¶. Similarly if we choose the last segment vertical, then it can be

shown that qy

u=

¶

¶.

This proves the necessary part. Sufficiency Part

Suppose there exists a function u (x,y) in W such that, qy

up

x

u=

¶

¶=

¶

¶,

\If a, b are the end points of g , we have

òò ¶

¶+

¶

¶=+

gg

dyy

udx

x

uqdypdx dt

dt

dy

y

u

dt

dx

x

ub

aò

þýü

îíì

¶

¶+

¶

¶=

dt.dt

dub

aò= ( ) ( )( ) ( ){ }b

aty,txuy,xu == ))a(y),a(x(u))b(y),b(x(u -=

Thus the integral depends only on the end points, hence the theorem. Note:

Now dyy

udx

x

udu

¶

¶+

¶

¶= qdypdx += .i.e. pdx+qdy is an exact differential. Thus an

integral depends only on the end points, iff the integrand is an exact differential. Note: When is f(z)dz = f(z)dx + i f(z)dy an exact differential? By definition of an exact

differential, there must exist a function F(z) in W such that )()( zfzFx

=¶

¶and

)()( zfizFy

=¶

¶.

y

fi

x

f

¶

¶-=

¶

¶\ which is C-R equation. Further f(z) is by assumption continuous {otherwise

òg

)z(f would not be defined}. Hence f(z) is analytic. Therefore, we have the fact that the



18

integral dz)z(fòg

with continuous function f(z) depends only on the end points of g if and

only if f(z) is the derivative of an analytic function in W . Note

The integral dz)z(fòg

depends only on the end points of g if and only if the integral of f(z)

over any closed curve is zero. Definition: f(z) is analytic on the arc g if and only if it is defined and analytic in W that

contain g .

Note: By the results stated above ( )òg

=- 0dzaz n for all closed curves. curve g , since

( )naz - is the derivative of ( )

"+

- +

,1n

az 1n

integral n except n = -1. If n = -1, the above result

does not hold. If c is a circle with centre ‘a’ and radius r , then ,0ieaz r+= where

.200 p<£ Then òòp

p==-

2

0c

i20diaz

dz

Example 1: Compute dzxòg

,where g is the directed line segment from 0 to 1 + i.

Answer: The parametric representation of the line is

iyxz,ty,tx1t

y

t

x+===Þ== itt += t)i1( += , t varies from o to 1. Now dz = (1+ i) dt.

\ ( ) ( )ò+

=÷÷

ø

ö

çç

è

æ+=+

l

0

1

0

2

2

i1

2

ti1dti1t

Example 2: Compute dzzzò=

-1

1

Answer: Let C be the circle be |z| = 1. on C, z = 1e1xe 0i0i -=-Þ ( ) 0sini10cos +-=

Therefore ),0cos1(21,00 -=-= zidedz i.0d|0d||dz| == On c, 0 varies from o to p2 .

4.4 LINE INTEGRALS AS FUNCTION OF ARCS

Line integrals of the form òg

+ qdypdx are studied as functions of the arc g , where

p and q are defined and continuous in a region W and that g is free to vary in W . If

1g and 2g have the same initial point and the same end point, then òg

+

1

qdypdx =



19

òg

+

2

qdypdx .If g is a closed curve, then g and g- have the same end points, and if the

integral depends only on the end points, we obtain òòò -==- ggg

and consequently

.0ò =g

Conversely if 1g and 2g have the same end points, then

1g - 2g is a closed curve,

and if the integral over any closed curve vanishes, it follows that òò =21 gg

4.5 CAUCHY’S THEOREM FOR A RECTANGLE There are different forms of Cauchy’s Theorem which differ in their topological content. We begin with one form of Cauchy’s Theorem for which the contour of integration is the closed rectangle R defined by ,, dycbxa ££££ the order of the

vertices being (a,c), (b,c), (b,d), (a,d).

Statement: If the function f(z) is analytic on a rectangle R, then ò¶

=R

dzzf .0)(

Proof: Subdivide the rectangle R into congruent rectangles )4()3()2()1( R,R,R,R denote the

boundaries of these rectangle by )4()3()2()1( R,R,R,R ¶¶¶¶ .

Now dR = ,dRdRdRdR )4()3()2()1( +++ since the common sides each other.

\ )1.....(..........dz)z(fdz)z(fdz)z(fdz)z(fdz)z(f)4()3()2()1( RRRRR

òòòòò¶¶¶¶¶

+++=

Denote dz)z(f)k(R

ò¶

b y ))k(R(h . Clearly, for at least one )k(R (k = 1,2,3,4), we have

( )( ) )2.....(....................)(4

1RR k hh ³

For, otherwise, ( ) ( )å=

h=h4

1k

)k(RR

£ ( )å=

4

1

)(

k

kRh , (Triangle inequality)

( )Rh< . This is impossible

\(2) is true for at least one K. Let R, denote one of the rectangles )(kR for which (2) holds. Repeating the process, we obtain a sequence of nested rectangles.



20

nRRRR ÉÉÉÉ ....21 with the property that

( ) ( )14

1-³ nn RR hh ( ) ( )3322 4

1

4

1-- ³³ nn RR hh and hence

( ) ( )RRnn hh

4

1³ ……….(3)

There exists a point 0Z common to all the closed rectangles of the above sequence .

Since f(z) is analytic on R, for a given 0>e ,we can choose a 0>d such that

e<--

-Þd<- )z('f

zz

)z(f)z(fzz 0

0

00

0000 zz)z('f)zz()z(f)z(f -e<---Þ ………… (4)

Let dN )z( 0 be the set of points z for which d<- 0zz . For sufficiently large n, we have

nR ( dN )z( 0 ).

Thus (4) holds good for all zÎ nR . Note that òd

=

nR

0dz and òd

=

nR

0zdz

{Q The integrals 1 and 2 are the derivatives of the analytic functions z and 2

z2

respectively & nR is closed}.

=ò¶

dzzfnR

)( ò¶

---nR

ooo dzzfzzzfzf )](')()()([

ò ò¶ ¶

-£---£\n nR R

oooon dzzzdzzfzzzfzfR eh )(')()()()( , using (4)

Let nd denote the length of the diagonal of nR . Then n0n dzzRz £-ÞÎ …….(5)

( ) dzz'f)zz(z)z(f)z(fdz)z(f

nn R

000

Ròò

¶¶

---£

dzzz 0

Rn

-e£ òd

by (4)

dzdn

Rn

òd

e£ by (5)

nn Lde£ ……….(1) Where nL denotes arc length of nR¶ . Let d and L denote the

length of the diagonal and the perimeter of the rectangle R.

Then nd = dn-2 and nL = Ln-2 .

Now (3) Þ nnn

R

n

R

Lddzzfdzzfn

e4)(4)( ££ òò¶¶

= Ldnn

n

2

1

2

14 e , = Lde . Since e is arbitrary, ò

¶

=R

dzzf 0)(



21

Theorem: Let )(zf be analytic on the set 'R obtained from a rectangle R by omitting a

finite number of interior points ie where 0)()( =-®

zfezez

Limi

ij

.

Then ò¶

=R

dzzf 0)( .

Proof: .........

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..........................

..........................

It is enough to consider the case of a single rectangle. For R can be subdivided into

smaller rectangles so that each rectangle has almost one ie in its interior. In fact, we can

subdivide further, if necessary, and assume that je is the center of the square to which it

belongs.

Since ò¶R

dzzf )( = åfinite

ò¶ jR

dzzf )( , it is enough to consider a single square oR with

centre at e, and .0)()( 1

1

=-®

zfezez

Lt \ we get ò

¶R

dzzf )( = ò¶ nR

dzzf )( after

cancellations. Since ,0)()( 1

1

=-®

zfezez

Lt given ,0>e $ a 0>d such that

,)()( 1 e<- zfez whenever d<- 1ez choose oR so small it is contained in the disk

d<- 1ez . Therefore <)(zf1ez -

e on oRd .

Now òò¶¶

£oo RR

dzzfdzzf )()( dzez

oRò

¶-

<1

e.

If the square oR is of side 2a, then 1ez - ³ a z e .oR¶

\aez

11

1

£-

, ò¶ oR

dz = length of =¶ oR 8a,

\ ò¶ oR

dzzf )(a

1e< .8a = e8 ò

¶

=ÞoR

dzzf 0)( .



22

4.6 CAUCHY’S THEOREM IN A CIRCULAR DISK

Statement: If f(z) is analytic in an open disk D , then ò =g

,0)( dzzf for any closed curve

g in D .

Proof:

),( oo yx

Define F(z) by F(z) = òs

)1(.............)( dzzf .

Where s consists of the horizontal line segment form the centre ),(),( 000 yxtoyx and

the vertical line segment from ),( 0yx to ),( yx

i.e., F(z) = ò =OAP

dzzf )( ò +OA

dzzf )( )2(..........)(òAP

dzzf .

The parametric representation of OA is z = t + i 0y , \dz = dt, t varies from 0x to x.

The parametric representation of AP is z = x + it, therefore dz = idt, t varies from 0y to y.

\ (2) òò +++=Þy

y

x

x

dtitxfidtiytfzF00

)()()( 0 …….(3).

Since integral of F(z) along any closed rectangle is equal to zero, ò =OAPBO

dzzf 0)(

+Þ ò dzzfOAP

)( ò =PBO

dzzf 0)( òò ==ÞOBPOAP

zFdzzfdzzf ).()()(

The parametric equation of OB is Z = ox + it, therefore idtdz = , t varies from oy to y.

The parametric equation of BP is ,iytz += therefore dtdz = t varies from ox to x.

Therefore òòò +==BPOBOBP

dzzfdzzfdzzfzF )()()()(

= òò +++x

x

y

y

o

oo

dtiytfdtitxfi )()( ………..(4)

Differentiating (3) partially w.r.t. y, =¶

¶

y

F i f )( iyx + = i f (z),

Differentiating (4) partially w.r.t.x, =¶

¶

y

F f )( iyx + = f (z),

\ +¶

¶

y

Fi =

¶

¶

y

Ff (z) + i 2 f(z) = f(z) - f(z) = 0.......... (5)



23

If F = u + iv, then, ,& yyxx ivuy

Fivu

x

F+=

¶

¶+=

¶

¶

(5) 0)()( =+++Þ yyxx ivuiivu , 0)()( =++- yxyx uvivu .& xyyx vuvu -==Þ

\ The C-R equations are satisfied. Further ),(zfx

F=

¶

¶=

¶

¶

y

F i f (z) is continuous

yxyx vvuu ,,,Þ are all continuous. Hence F(z) is analytic on D . Also F(z) = f(z).

Hence by the previous theorem ,0)(ò =y

dzzf for every closed curve g in D .

Theorem: Let f(z) be analytic in the domain 'D obtained by omitting a finite no of

points ie from an open disk D . If f(z) satisfies the condition izfzez

Limii

i

"=-®

,0)()( r ,

then ò =y

dzzf 0)( for all closed g in 'D .

Proof:

),( oo yx0

P(x,y)

If ),( 00 yx itself is an exceptional point we can start with some other fixed point. Further

we cannot let s pass through the exceptional points. Suppose that iir lies on the lines

oxx = and .oyy = Then we can avoid the exceptional points by letting s consist of 3

segments. Applying previous theorem, we get that F(z) is independent of the choice of the middle segment and the last segment can be either vertical or horizontal. As before F(z) is an indefinite integral of f(z) and the theorem follows. 4.7 LET US SUM-UP

1. The necessary and sufficient condition that ò +g

qdypdx depends on the end points

only. 2. Deduction of the results (1) 3. Integral of an analytic function over a rectangle or a circle vanishes. 4.8 LESSON END ACTIVITIES

(1) Compute ò -g 12z

dz where c is |z| = 2, +ve sense.

(2) Calculate òg

xdz , where c is |z| = 1, +ve sense.



24

4.9 REFERENCES 1. Complex analysis by L.V. Alphors. 2. Complex Analysis by A.R. Vashitha. 3. Foundations of Complex Analysis by Dr. S. Ponnusamy.



25

LESSON – 5 CAUCHY’S INTEGRAL FORMULA CONTENTS 5.0 Aim and Objectives 5.1 Introduction 5.2 The Index of a point with respect to a closed curve 5.3 Cauchy’s Integral Formula 5.4 Higher Derivatives 5.5 Let us Sum up 5.6 Lesson end activities 5.7 References

5.0 AIM AND OBJECTIVES The main objective of this lesson is to know the formula for finding the value of function at any point of the domain of analyticity. From this, the formula for finding higher derivatives of an analytic function is derived. After going through this lesson you will able to: (i) define index of a point (ii) to state Cauchy’s integral formula (iii) to find Higher derivatives of an analytic function (iv) to state Morera’s theorem (v) to state Liouville’s Theorem 5.1 INTRODUCTION

Before going on to derive Cauchy’s Integral formula, let us define the notion which indicates the number of times a closed curve winds around a fixed point not lying on the curve. To explain this, consider the following lemma. 5.1 THE INDEX OF A POINT WITH RESPECT TO A CLOSED CURVE

Lemma: If the piecewise smooth closed curveg does not pass through a point ‘a’, then

the value of the integral ò -g

az

dz is a multiple of ip2 .

Proof: Letg be defined by z = z (t), .ba ££ t Consider h(t) = ò -

t

dtatz

tz

a

.)(

)('

This is defined and continuous on ],[ ba . \ atz

tzth

-=

)(

)(')(' except possibly a finite

number of points where )(' tz does not exist.

Now, { }[ ] =-- atzedt

d th )()( 0, except perhaps at a finite number of points.

Since this function is continuous, )(the- ))(( atz - = a constant c.

))(()( aze th -Þ - a = c, az -Þ )(a = { }0)( =ahe Q .

\ .)())(()( azatze th -=-- a



26

Put t = b , .)())(()( azaze h -=-- abb

\ )(bhe- = { ),()()(

)(ab

b

azz

az

az==

-

-Q since the curve is }closed ,

inheh pbb 2)(,1)( =Þ=Þ (n is an integer),

ò -

b

aaz

dttz )(' = a multiple of ip2 , ò -

gaz

dz = a multiple of ip2 .

Definition: The index of a point w.r.t g (the winding number of g w.r.t.a) is defined by

n(g ,a) = ò -g

p az

dz

i2

1.

Example: We know that iaz

dz

c

p2=-ò where c is a circle c with center a.

\ n(c,a) = 1. i.e, c winds around a only once. Remark 1 n(- g ,a) = n(g ,a). i.e, winding number of g w.r.t.a is unaltered if the

orientation is changed. Remark 2 n(g ,a) = 0 for all closed curvesg in a disc and for all points outside the disk.

Now, az -

1is analytic inside the disk {Q a lies outside the disc}

Since "=-ò ,01

gaz

, closed curveg in the disk. {by Cauchy’s theorem for a circular disk}

\ n(g ,a) = ò -g

p az

dz

i2

1= 0.

Remark 3 n(g ,a) = n(g ,b) if a and b belong to the same region determined byg and

n(g ,a) = 0 if a belongs to the unbounded region determined by g .

Proof: Join a and b by a line segment not intersecting g .

Outside the line segment log az

bz

-

- is analytic. Whose derivative

az -

1- .

1

bz -

\ ,011

=þýü

îíì

--

-òg

dzbzaz

,11

dzbz

dzaz òò -

=-

Þgg

.2

1

2

1òò -

=-

Þgg

pp bz

dz

iaz

dz

i

Remark 4 As a point set,g is closed and bounded. Its complement is open and can be

represented as a union of disjoint regions, the components of the complement.



27

5.3 CAUCHY’S INTEGRAL FORMULA

Theorem: Let f(z) be analytic in an open disc D and let g be a closed curve in D .

For any point ‘a’ not ong , n(g ,a) f(a) = .)(

2

1ò -g

p az

dzzf

i Where n(g ,a) is the index of

‘a’ w.r.t g .

Proof: Consider F(z) = az

afzf

-

- )()(. F(z) is analytic at all points except at z = a.

But ( ))()()()( afzfaz

Limzfaz

az

Lim-

®=-

® = 0 {Q f(z) is analytic}.

Hence by the previous theorem,

0)()(

=-

-ò dz

az

afzf

g

òò -=

-Þ

ggaz

dzafdz

az

zf)(

)( dz

az

zf

iaz

dz

iaf òò -

=-

Þgg

pp

)(

2

1

2

1)(

dzaz

zf

ianaf ò -

=Þg

pg

)(

2

1),()( …….. (1), hence the theorem.

Remark 1 If a DÏ , then n(g ,a) = 0, Since az -

1 is analytic in D , 0

)(=

-ò dzaz

zf

g

{by

Cauchy’s theorem for a circular disc}. \L.H.S = 0 = R.H.S of (1). Hence (1) is true for all a DÏ .

Remark 2 In the special case n(g ,a) = 1, We have f(a) = ò -g

pdz

az

zf

i

)(

2

1 and this gives

representation formula to compute f(a) as soon as the value of f(z) ong is given, together

with the fact, f(z) is analytic in D . This is called Cauchy’s Representation formula. By

change of notation, we write f(z) = ò -g

rr

r

p.

)(

2

1d

z

f

i This is referred to as Cauchy’s

Integral formula. Example:

(1) Compute dzz

e

z

z

ò=1||

Answer:

.21||

idzz

e

z

z

p=ò=

f(0) where f (z) = ze = .2 ip

(2) Compute ò= +2||

2 1z z

dz by decomposition of the integral into partial fractions.



28

Answer: We have úû

ùêë

é

+-

-=

-+=

+ iziziizizz

11

2

1

))((

1

1

12

therefore

úúû

ù

êêë

é

+-

-=

+ òòò===

dziz

dzizi

dzz zzz

11

2

1

1

1

2||2||

2

2||

= [ ])(2)(22

121 ifiifi

i-- pp ,

where 1f (z) = 1 & 2f (z) = 1, = p [1-1] = 0.

5.4 HIGHER DERIVATIVES

The Cauchy’s representation formula helps us to discuss the local properties of analytic function. In particular we can prove that an analytic function has higher derivates of all orders, which are also analytic. To do this, we require the following lemma. Lemma: Suppose that )(rf is continuous on the arcg . Then the function

ò -=

gr

rrfnn

z

dzf

)(

)()( is analytic in each of the regions determined byg and its derivative is

).()( 1 znFzf nn +=

Proof: We shall prove this lemma by the method of induction. Let gÏoz .

Step 1: We first prove that )(1 zF is continuous at gÏoz , g is compact and )(rf is

continuous ong . M<\ |||| rf for all gr Î .

Since g is closed, the complement of g is open.

\ $ a d - neighbourhood of oz (which is in the complement of g ) not intersecting g .

\ In the neighbourhood ,2

|| 0þýü

îíì

<-d

zz we find that 2

||d

r >-z for all gr Î and

.|| dr >- oz

Now, )()( 011 zFzF - = rrr

rfg

dzz o

úû

ùêë

é

--

-ò11

)( = rrr

rf

g

dzz

zz ò ---

))((

)()(

0

0

rrr

rf

g

dzz

zzzFzF ò ---<-Þ

))((

)()()()(

0

0011 rrr

rf

g

dzz

zz ò ---<

||||

)()(

0

0

ò-<g

rd

||4

)(20 dMzz }{ M<|)(| rfQ and

drdr

1

||

1<

-Þ>-

zz L

Mzz

20

2||

d-<

\ edd

dMLMLLMzFzF o ==<-

12

2|)()(|

211 where L is the length of g , d

1 =e

)(1 zF\ is continuous at 0z .



29

Step 2: ( )021 )( zFzF = . Apply step 1 to the function 0

)(

z-r

rf which is also continuous

ong . This implies that rr

rrf

g

dz

zò -

-

)(

)/()( 0 is also continuous at 0z ò --Þ

grr

rrf

))((

)(

ozz

d

is also continuous at 0z . Hence rrr

rf

g

dzzzz

zFzFò --

=-

-

))((

)()()(

00

011 tends to the limit

)(2 ozF as ozz ®

Therefore rrr

rf

g

dzzzz

Lim

zz

zFzF

zz

Limò --®

=-

-

® ))((

)()()(

000

011

0

= rr

rf

g

dzò - 2

0 )(

)(

)()( 0201 zFzF =Þ This is true for every 0z not on g ).()( 0'

20'

1 zFzF =\ Hence the

result is true for n = 1.

Step 3: Assuming ),()1()(1 zFnzF nn -=- we have to prove that ).()( 1 zFnzF nn +=

)()( 0zFzF nn - = rr

rfr

r

rf

gg

dz

dz nn òò -

-- )(

)(

)(

)(

0

rr

r

rr

rf

g

dz

z

zz n )(

)(

)()(

)( 0

01 -

-

--= ò -

rr

rf

g

dz nò -

-)(

)(

0

rrr

rf

g

dzz nò --

=- )()(

)(

01

0

rr

rd

z

zzz

þýü

îíì

-

-+- 0 rr

rf

g

dz nò -

-)(

)(

0

rrr

rf

g

dzz nò --

=- )()(

)(

01

rr

dz

zz

þýü

îíì

-

-+ 01 r

r

rf

g

dz nò -

-)(

)(

0

ò --=

-

g rr

rrf

)()(

)(

01 zz

dn

rr

rf

g

dz nò -

-)(

)(

0

+ )( 0zz - ò --g rr

rrf

)()(

)(

0zz

dn

.

Since )( 01

1 zFn- exists, )(1 zFn- is continuous at 0z . Similarly rrr

rf

g

dzz nò -- - )()(

)(

01

is

also continuous at 0z . Hence .0)(

)(

)()(

)(

001

0

=úúû

ù

êêë

é

--

--® òò -r

r

rfr

rr

rf

gg

dz

dzzzz

Limnn

Since, LM

zz

dn

zz

n 1

2)(

0

0

)()(

)(+

-

<--ò drr

rrf

g

the last term also ® 0 as ozz ® .

®-\ )()( 0zFzF nn 0 as ozz ® )(zFn\ is continuous at oz where

rr

rf

g

dz

zFnn ò -

=)(

)()(



30

Similarly ò --g rr

rrf

)()(

)(

0zz

dn

is continuous at 0z .

ò ò +-=

--®g g r

rrf

rr

rrf1

000 )(

)(

)()(

)(nn z

d

zz

d

zz

Lim = )(1 on zF +

Set ( ) ò ----

=g

rrr

rfy d

zzz

nn 100

1))((

)( ( )zn 1-y is continuous at 0z .

( ) ( )( )01

0

011

0

' zzz

zz

zz

Limn

nn-

-- =-

-

®y

yy = (n-1) ( )0zny (by induction)

= (n-1) ò +-g r

rrf1

0 )(

)(nz

d = (n-1) )( 01 zFn+

)()1()(

)(

)()(

)(101

001

000

zFnz

d

zz

d

zzzz

Limnnn ++

-=úúû

ù

êêë

é

--

---®Þ òò

gg r

rrf

rr

rrf

ò ++-

+-=-

-

®Þ

g r

rrf1

0

01

0

0

0 )(

)()()1(

)()(nn

nn

z

dzFn

zz

zFzF

zz

Lim

).()()()1()( 010101' zFnzFzFnzF nnnn +++ =+-= Since this is true for an arbitrary 0z .

).()( 1' zFnzF nn +=

Theorem: An analytic function defined in a region W has derivates of all orders and these are analytic on W . Proof: Let a WÎ and f(z) be analytic in W . Consider a d - neighbourhood D about a in

D , we take a circle c. For all z inside c, n(c,z) = n(c,a) = 1.

Hence by Cauchy’s formula, F(z) = rr

rd

z

f

mcò - )(

)(

2

1

By the above lemma, the integral on the R.H.S. is an analytic function whose derivative

is ò -c z

df2)(

)(

r

rr. Therefore by the same lemma, the integral on the R.H.S is an analytic

function. Therefore whenever f(z) is analytic in W , )(' zf is also analytic in W . Thus

)(' zf = .)(

)(

2

12

rr

r

pd

z

f

icò -

Similarly )(" zf = .)(

)(

2

!23

rr

r

pd

z

f

icò -

By induction )()( zf n = ò +-c

nz

df

i

n1)(

)(

2

!

r

rr

p are all analytic functions.



31

Morera’s Theorem

Statement: If f(z) is defined and continuous in a region W and if ò =g

0)( dzzf for all

closed curve g in W , then f(z) is analytic in W .

Proof: ò =g

0)( dzzf for all closed curveg implies f(z) dz is an exact differential which

implies f(z) is analytic in W , such that )(' zF = f(z) Þ f(z), being the derivative of an

analytic function, itself is analytic in W by the above theorem. Cauchy’s Inequality Let f(z) be analytic in W . Consider a circle c, with centre a and radius ‘g ’

contained in W . Hence, by Cauchy’s Representation Formula, ò +-=

c

na

n

az

dzzf

i

nf

1)(

)(

)(

)(

2

!,

p,

ò +-=

c

n

n

az

dzzf

i

naf

1

)(

)(

)(

2

!)(

p ò +-£

c

naz

dzzf

i

n1)(

)(

2

!

p ò+£

c

ndz

Mn12

!

gp where M =

cz Î

max |f(z)|

gpgp

22

!1+

£n

Mn =

nr

nM !.

Liouville’s Theorem Statement: A function which is analytic and bounded in the whole plane must reduced to a constant (i.e. bounded entire functions are constant). Proof: Let a be any point of the plane and C, a circle of radius r, centre ‘a’.

f(z) is bounded )(zfÞ £ m for all z, on C. \By Cauchy’s Inequality,

VV

V

pd

a

f

iaf

cò -

=2)(

)(

2

1)(' \

r

M

r

Maf ££ p

p2

2

1|)('| .

This is true for any circle with centre ‘a’. Let ¥®r , 0®r

M \ )(' af = 0, " a CÎ i.e.

)(' zf = 0. \ f(z) is a constant.

Fundamental Theorem Statement: Every polynomial in z of degree n, n ³ 1 with real or complex co-efficient, then the equation p(z) = 0 has atleast one root.

Proof: Let p(z) = .0,.....10 ¹+++ nn

n azazaa



32

If p(z) were not zero, then )(

1

zP would be analytic in the whole plane {since it is an

entire function}.

Let Q(z) = )(

1

zP. If Q(z) is not bounded, then P(z) = .0

)(

1®

zQ This is a contradiction

\ Q(z) is bounded. Hence by Lioville’s Theorem, Q(z) reduces to a constant. This means that p(z) is a constant. This is, again, a contradiction. Hence the theorem.

5.5 LET US SUM-UP (1) Statement of Cauchy’s Integral formula (2) Statement of Morera’s Theorem (3) Statement of Liouville’s Theorem (4) Statement of Fundamental Theorem. 5.6 LESSON END ACTIVITIES

(1) Find ò - 2||

||

az

dz with the condition |a| ¹ 1.

(2) Evaluate dzz

z

zò=

-2||

)1(

sin

(3) Evaluate .123||

2

2

dzzz

z

zò= +-

5.7 REFERENCES

1. Complex Analysis by L.V. Alphors.. 2. Complex Analysis by A.R. Vashistha.



33

LESSON – 6 LOCAL PROPERTIES OF ANALYTIC FUNCTIONS

CONTENTS 6.0 Aim and Objectives 6.1 Introduction 6.2 Removable Singularity 6.3 Zeros and Poles 6.4 Local Mapping Theorem 6.5 The Maximum Principle 6.6 Chains and Cycles 6.7 Let us Sum up 6.8 Lesson end activities 6.9 References

6.0 AIM AND OBJECTIVES In this lesson we are going to study the local properties of analytic functions. We are also going to study the classification of isolated singularities and their properties. We are also going see the behavior of an analytic function f(z) at (z) = ¥ . After going through this lesson you will be able to (i) define Removable Singularity, Essential Singularity, Isolated Singularity, Meromorphic function. (ii) state Taylor’s Theorem, Weierstrass’s Theorem, Local Mapping Theorem, Open Mapping Theorem, Maximum Principle. 6.1 INTRODUCTION

In the earlier lesson we have seen that derivative of an analytic function is also analytic. And that an analytic function has derivatives of all orders. Based on these results, we are going to study the local properties of an analytic function, the different types of isolated and non-isolated singularities etc. The proof of Taylor’s theorem is given using different ideas. 6.2 REMOVABLE SINGULARITY

Let f(z) be an analytic function in the region W except at z = a and ( )azaz

Lim-

®

f(z) = 0 then z = a is said to be removable singularity of f(z). That is we say that a singularity of an analytic function which can be made to disappear by redefining the function suitably is called a removable singularity. As such it is an artificial singularity of less significance Example

Prove that z = a is a removable singularity for )(

)sin()(

az

azzf

-

-=

Proof We first observe that z = a is a singularity of f(z).



34

Let us now expand f(z) about z = a.

( ) ( ) ( )úû

ùêë

é-

-+

---

-= ...

!5!3

1)(

53azaz

azaz

zF( ) ( )

¥--

+-

-= ...!5!3

142

azaz

Let us now redefine f(z) so that the singularity z = a is removed.

( )azif

azif

az

azzF ¹

ïï

î

ïï

í

ì

=

-

-= ,

,1

)(

sin)(

Hence z = a is called a Removable Singularity of f(z)

Example: If 0,sin

)(3

=-

= zz

zzzf is a Removable Singularity

Example: If 0,1

)( =-

= zz

ezf

z

is a Removable Singularity

Theorem: Suppose that f(z) is analytic in the region 'W obtained by omitting a point a from the region W . Then a necessary and sufficient condition that there exists a unique

analytic function in W which is the same as f(z) in 'W is that ( ) 0)( =-®

zfazaz

Lim. The

extended function is uniquely determined. Proof: Necessary Part: Suppose that there exists an analytic function f(z) in W satisfying the conditions of the theorem. i.e. F(z) = f(z), 'WÎ"z and F(z) is analytic. Since F(z) is analytic in W ,it is analytic at z = a and is also continuous at z = a.

i.e. az

Lim

® F(z) = F(a) exists. Equivalently, given 0>e ,there exists 0>da such that

e<- )()( aFzf .This means that )()( aFzfaz

Lim=

®

0)()()()( =®

-®

=-®

zfaz

Limaz

az

Limzfaz

az

Lim

\ z = a is a Removable singularity of f(z).

Sufficiency Part: Suppose that 0)()( =-®

zfazaz

Lim. 'WÎ" z i.e; z = a is a Removable

singularity of f(z). We now show that $ an analytic function F(z) WÎ such that F(z) = f(z) in 'W .

Since ‘a’ is a removable singularity and since WÎa and since W is an open set, we can

find a circle c about ‘a’ such that c and its interior lie in W . By Cauchy’s integral formula



35

WÎ"-

= ò zzt

dttf

izf

c

,)(

2

1)(

p and for all caz Î¹ . Since c is contained in 'W and f(t) is

analytic on c.

ò -c

zt

dt)t(f is also analytic at every point that does not lie on c. In particular it is analytic at z

= a Now, let us define.

( )

ïî

ïí

ì

=-

WÎ"

=òc

azifzt

dttf

i

zzf

zF ,)(

2

1',

)(

p

This F(z) is the require function which is analytic in W .It follows that

ò=c

dzzfi

zF )(2

1)(

p i.e. F(z) = f(z).

Let us now prove the Uniqueness Let F(z), G(z) be two functions satisfying the conditions of the theorem. \F(z) - G(z) = 0, 'z We" and f(z) = G(z) is continuous at z = a. Therefore

)()( zGaz

LimzF

az

Lim

®=

®

)a(G)a(F =Þ . Since a is arbitrary, )()( zGzF = , ZzÎ" . Hence f(z) is unique

Taylor’s Theorem

If f(z) is analytic in a region W containing ‘a’ then f(z) can be possibly written as

)()()!1(

)()()("

!2

)()('

!1

)()()(

112

zfazn

afazaf

azaf

azafzf n

nnn

-+-

-+

-+

-+=

--

where )z(fn is analytic in W and dtatzt

tf

izf

c

n ò --=

))((

)(

2

1)(

p,where c is a circle around

‘a’ whose interior lies in W .

Proof: Consider the function F(z) = ( ) ( )

az

afzf

-

- which is analytic in W except at z = a

)z(F\ is analytic in 'W = W - {a}.

Here ( ) ( ) ( )az

afzfaz

az

Limzfaz

az

Lim

-

--

®=-

®

)()(.

.)()()()( oafafafzfaz

Lim=-=-

®=

)z(F\ is analytic in 'W and ‘a’ is a Removable singularity for f(z). By a pervious theorem, we can find a unique function ),z(f)z(f1 = 'WÎ"z

( ) ( )az

afzf)z(f1

-

-=\ ……………(1)



36

Now )z(f1 is analytic as f(z) is analytic except at z = a because at z = a,

( ) ( ) .1 ozfazaz

Lim=-

®=

\z = o is removable singularity of )z(f1 . $\ an analytic function )z(f2 , WÎ"z .

Such that )z(f2

( ) ( )az

az

afzf¹"

-

-= ,11 ------ (2)

From (1), f(z) = f(a) + (z-a) )z(f1 .

From (2) ( )zf2 = ( )af2 + (z-a) ,zf3 extending this idea………………. We get

)z(f)az()a(f)z(f n1n1n -+= -- Where nf (z) is analytic in W from the above equations, we get

)3().........()()()(...)()()()()()( 11

22

1 zfazafazafazafazafzf nn

nn -+-++-+-+= -

-

Differentiating (3) with respect to z, successively.

....)()(4)()(3)()(2)()( 43

32

21' +-+-+-+= afazafazafazafzf

....)()(4.3)()(3.2)(2)( 42

32'' +-+-+= afazafazafzf

....)()(4.3.2)(3.2.1)( 43''' +-+= afazafzf …………………………………

)()(!)()!1()( 1)1( afaznafnzf nn

n -+-= --

Putting z = a, we get

)(!3

)()(

!2

)()(

!1

)()( 4321 af

afaf

afaf

afaf

¢¢¢=

¢¢=

¢= ………….

)!1(

)()(

)1(

1-

=+-

-n

afaf

n

n

Using these values in (3), we get

)!1(

)()(....)(''

!2

)(

!1

)()()()(

)1(12

-

-++

-+

¢-+=

--

n

afazaf

azafazafzf

nn

Where f(z)is analytic in W

To get the representation for nf (z)

Consider a circle around ‘a’ whose interior lies in W . By cauchy’s integral

formula for )z(fn , we have nf (z) = )1.......(dtzt

)t(f

i2

1)z(fn

c

nò -p=

( ) þýü

îíì

--

++-

¢¢+-

¢---

= --

1)1(2

)()!1(

)(...

!2

)()(

!1

)()()()(

1)( n

n

nn atn

afataf

atafaftf

attf

(1) Þ

òòþýü

îíì

--

++-

¢---

= --

c

nn

c

nn atn

afatafaftf

atizf 1

)1(

)()!1(

)(...

!1

)()()()(

)(

1

2

1)(

p

ò òòò ----

---

---=

c c

n

c

nn

c

n atzt

dtaf

iztat

dtaf

izt

dt

at

tf

at

tf

i 1))((

)(

2

1

)()(

)(

2

1

)()(

)(

)(

)(

2

1

ppp

)2(........))(()!1(

)(

2

1.....

!2

1

))((

)(

2

12 òò ---

-

--

¢¢=

-

cc

n atztn

dtaf

iatzt

dtaf

i pp

Let )(zFr = ò --c

ratzt

dt

))(( r = 1,2, …… n.



37

In particular if r = 1, )(1 zf = ò --c

atzt

dt

))((, using partial fractions

))((

1

atzt -- = ,

at

B

zt

A

-+

- we see that A = -B =

az -

1

\ )(1 zf = az -

1úû

ùêë

é

--

- òòcc

at

dt

zt

dt

= az -

1 [ ]),(2),(.2 acnizcni pp - ……. (3)

But since n (c, a) = n(c, z), )(1 aF = 0 from a Lemma, we have )(' zf n = n )(1 zFn+

\ )('1 zF = 1. ),(2 zF )("1 zF = )(2)( 3

'2 zFzF =

)('!3)(!2)( 4'

3'''

1 zFzFzF ==

……………………………..

)(!)( 1)(

1 zFrzF rr

+= ………(4)

\ !

)()(

)(1

1r

zFzF

r

r =+ . But )(1 aF = 0 \ 0!1

)()(

'1

2 ==aF

aF . Similarly )(3 aF = 0. In

general )(af r = 0, r = 1,2, ….. n,

using these results in (2),

we get ò --=

c

nnatzt

dttf

izf

))((

)(

2

1)(

p

Hence the theorem. 6.3 ZEROS AND POLES

Theorem: Suppose that f(z) is analytic in a region W containing a point ‘a’ and f(a) = 0,

)(afr = 0, r = 1,2, … then f(z) = 0 on W .

Proof Consider a circle with centre at z = a and radius R and contained in W . Let m be the maximum of |f(z)|. i.e., |f(z)| < m on c \ f(z) can have Taylor’s expansion. i.e., f(z) = f(a) +

)()()!1(

)()(.....)('

!1

11

zfazn

afazaf

azn

nnn

-+-

-++

- --

…… (1)

Where ò --=

c

nnatzt

dttf

izf

))((

)(

2

1)(

p………….(2) on c, |t-a| = R

\ t-z = t-a + a – z Þ |||||| azatzt ---³-

||

1

||

1

azRzt --£

-\



38

\ ò --=

c

nnatzt

dttf

izf

))((

)(

2

1|)(|

p ò --£

c

natzt

dttf

|)(||)(|

|||)(|

2

1

p

( ) ( )||

2

2

1)(

1 aZRR

M

azRR

RMzf

nnn--

=--

£\-

p

p………..(3)

Now f(a) and all successive derivatives vanish at z = a ie; nf (a) = 0 for n ³ o

oafafaf ===Þ )...('')()( '

\ (1) )()()(1 zfazzf nn-=Þ

nnaz

zfzf

)(

)()(

-=Þ

\nn

az

zfzf

)(

)()(

-= ……. (4)

using (3) & 4, we get ( ))(

)()(

1 azRR

Mazzf

n

n

--

-£

- ¥®® nas0 { }Raz <- ||Q ie; f(z)

is zero inside c. Let us now show that f(z) is identically zero in the whole of W . Let 1E denote the set of points where f(z) and all its derivatives vanish and ,2E the set of

points where the function or one of its derivatives is different from zero. Evidently 1E is

open. For if a 1EÎ , then there exists a circle about a of radius R such that f(z) and all its

derivatives vanish inside c. i.e. c is contained in 1E .

Hence 1E is open. Also 2E is open. For if z 2RÎ , we can find a neighborhood of z

through out which f(z) or one of its derivatives is different from zero. Hence 2E is open. From the definition of 1E and 2E , 1E È 2E = W and 1E Ç 2E = f .

Since W is connected, either 1E or 2E must be empty. Since a 1EÎ , 1E is non – empty.

\ 2E is empty. Hence 1E = W , since WÎ"z , f(z) = 0. i.e. f(z) is identically zero in W .

Note: Suppose f(z) ¹ 0 on W ie; f(z) is not identically zero. If f(a) = 0; then there exists

an analytic function nf (z) such that )(afn ¹ 0 so that f(z) = naz )( - nf (z) where f(a) =

)(' af = … = )(1 af n- = 0.

Now since )(zf n is continuous and )(afn ¹ 0 in a neighborhood of z = a, z = a is the

only zero of f(z) of order n in this neighborhood. That is, zeros of an analytic function which is not identically zero are isolated. Theorem: (i) If f(z) is analytic in a region W then its zeros are isolated (ii) If f(z), g(z) are two functions analytic in a region W and if f(z) = g(z) on a set which has an accumulation point on W , then f(z) º g(z).

Proof (i) Suppose that z = a is a zero of order n for f(z). Then f(z) = naz )( - )(zfn . In

particular )(zfn is analytic at z = a and hence is continuous at that point.



39

\0 < e < |)(| afn and $ ad > 0 such that |)()(| afzf nn - < e for all zÎNd (a)

in W . Now |)()()(||)(| afafzfzf nnnn +-= ³ .)(|)(| e-³ afzf nn

Hence |)(| zfn >0. )(zfn has no zero in Nd (a) in W .

This means that f(z) has no zero in Nd (a). \ a is an isolated zero (ie) zeros of an analytic function are isolated. (ii) consider h(z) = f(z) – g(z). Since f(z) = g(z), " zÎS, n(z) = 0, " zÎS. If a is an accumulation point of then by the continuity of h(z) at z = a, we get h(a) = 0. \ h(z) is an analytic function on W having an accumulation point aÎ W at its zero. \ h(z) º 0, " zÎ W . i.e. f(z) – g(z) º 0 Þ f(z) º g(z), " z on W . Isolated Singularities A singularity a of an analytic function f(z) is said to be an isolated singularity of f(z) if $ a d >0 such that the neighborhood 0<|z-a| <d has no other singularity of f(z).

An isolated singularity ‘a’ of f(z) is called pole of f(z) if az

Lim

®f (z) = ¥ i.e. f (a) = 0.

Note: The poles of an analytic function are isolated. Let f(z) have a pole at z = a. Then we can find a d >0 such that f(z) has no singularity in 0 < |z-a| < d and

az

Lim

®f(z) = ¥ . Since f(z) is continuous, there exists 'd £ d ,

such that f(z) ¹ 0 in 0<|z-a| < 'd .

Consider g(z) = )(

1

zf in 0<|z-a|< 1d . Then

az

Lim

® g(z) = g(a) =

az

Lim

®

)(

1

zf = 0

\z = a is a zero of g(z). Let h be the order of this zero. Then we can write g(z) = haz )( -

hg (z) where hg (z) is analytic and non-null at z = a.

i.e. )(

1

zf = haz )( - )(zgh \ f(z)

)(

)(

zg

az

h

n-.

\ f(z) has a pole of order h at z = a so that we can write f(z) = naz )( - f (z) where f (z)

is analytic and non-null at z = 0. Also f (z) does not tend to zero as z ® a.

\ There is no pole of f(z) in the neighborhood of z = a. Hence z = a is isolated pole. \ Every pole is isolated. Meromorphic Function An analytic function whose only singularities are its poles is called a Meromorphic function. That is, f(z) is called Meromorphic if it is analytic except for its poles.

Example f(z) = )3(

12 -zz

, f(z) is Meromorphic because its only singularities are its

poles z = 0,3. Note If f(z), g(z) are two Meromorphic functions defined in a region W , then



40

f(z) + g(z), f(z). g(z), )(

)(

zg

zf are also Mermorphic.

If f(z) and g(z) are analytic function in W , then )(

)(

zg

zf is a meromorphic function in W if

g(z) is not identically zero. The poles of )(

)(

zg

zf are the zeros of g(z). The common zeros

of f(z) and g(z) can be a removable singularity. Essential Singularity Let f(z) be analytic in the neighborhood of a except perhaps at z = a. Consider the following conditions.

(a) az

Lim

®|z-a| a |f(z)| = 0.

(b) az

Lim

®|z-a| a |f(z)| = ¥ , where a is real.

Now, there arise three possibilities. (1) If condition (a) holds for all a , then the function f(z) is identically zero. (2) There exists an integer h such that (a) holds for all a >h and (b) holds for all

a <h. (3) Neither (a) nor (b) holds for nay real value of a . In the third possibility z = a is

called an Essential singularity of f(z). Remark: In the case of a pole of order h at z = a,

we can write f(z) = haz )( - f(z) = )(zfn can be expanded as a Taylor’s series about z = a

as follows.

haz )( - f(z) = .))(()(...)( 111

hhhh azzazBazBB -+-++-+ -

- f

Where f (z) is analytic at z = a.

\ f(z) = az

B

az

B

az

Bh

hh

h

-+

-+

- -- 1

11

)()(+ f (z),

F(z) - f (z) = az

B

az

Bh

h

-++

-1.....

)(

\ The singular part of f(z) at the pole z = a is the function r (z) = f(z) - f (z) Þ f (z) = f(z) – p(z) is analytic at z = a.

Weierstrass’s Theorem on essential singularity Statement: An analytic function comes arbitrarily close to any given complex number in every neighborhood of an essential singularity.



41

Proof: Let z = a be an essential singularity of f(z). Let us assume that the essential were not true. There exists an analytic function which is not arbitrarily chosen to any complex value, say A. i.e. There exists ad >0 such that |f(z) – A| >d in the neighborhood of ‘a’ except for z = a.

\ For any a <0, az

Lim

®|z - a| a |f(z)-A| = ¥ .

This means that z = a is not an essential singularity of (f(z) – A).

We can find a real number B>0, such that az

Lim

®|z - a| b |f(z)-A| = 0 ……. (1)

and (1) is true for every b >0.

Assuming f(z) is not identically zero, ie; f(z) º/ 0.

We have az

Lim

®|z - a| b |A| = 0 ……. (2)

Hence |z - a| b |f(z)| = |z - a| b |f(z)-A+A| { }|||)(||| AAzfaz +--£ b

|||||)(||| AazAzfaz bb -+--£

\ |||||)(||||)(||| Aazaz

LimAzfaz

az

Limzfaz

az

Lim bbb -®

+--®

£-®

= 0 using (1) &

(2) \ |)(||| zfazaz

Lim b-®

= 0.

Hence z = a is not an essential singularity of f(z). This is a contradiction. Hence our assumption is wrong. \ An analytic function comes arbitrarily close to a given complex value in the neighborhood of an essential singularity. Definition Suppose f(z) is an analytic function and g(z) = f(1/z). If g(z) has a zero at z = 0. Then f(z) is said to have a zero at z = ¥ . If g(z) has a pole at z = 0, then f(z) is said to have a pole at z = ¥ . If g(z) has a removable singularity at z = 0, then f(z) is said to have a removable singularity at z = ¥ . If g(z) has an essential singularity at z = 0, then f(z) is said to be an essential singularity at z = ¥ . Theorem An entire function having a non-essential singularity at z = ¥ reduces to a polynomial. Proof Suppose f(z) is an entire function having a non-essential singularity at z = ¥ so that we can write

F(z) = f(z) = f(0) + ¥++++ ....)0(!

.....!2

)0("

!1

)0(' 2n

n

fn

zfzzf.

If we assume g(z) = f(1/z), then g(z) has non-essential singularity at z = 0.



42

If h is the algebraic order at z at z = 0, then )(0

zgzh

Lim h

® is neither zero nor infinity.

\ $ a constant k>0 such that |f(z) hz - A| < 1, " |z| > k

i.e. for |z| > k, |f(z) hz | = |f(z) hz -A + A| £ |f(z) hz - A| + |A| £ 1 + |A|

i.e., for |z| > k, |f(z) hz | < B, B = 1 |A|

i.e., for |z| > k, | f(z) | < B |z| h …… (1) Let R> k and c be a circle |z| = R.

Then by Cauchy’s estimate we have n

h

R

Mnf

!)0()( £ where M = Max |f(g)| on c…… (2)

But |z| = R Þ |z| > k, Þ |f(z)| < B |z| h = B hR , therefore M<B hR .

\ hnn

hh

R

Bn

R

nBRf

-=<

!!)0()( . Allowing ¥®R , we get )0()(nf £ 0 if n>h.

Hence )0(nf = 0, " n > h. Hence F(z) = f(0) + )0(!

.....)0('!1

nn

fn

zf

z++ which is a

polynomial.

Problem: Show that f(z) = aze -/1 has an essential singularity at z = a. Answer:

Let f(z) = aze -/1 , f(z) is analytic for all z except at z = a. Hence f(z) is analytic in the annulus 0 < |z-a| < R < ¥ \ f(z) can be expanded as a Laurent’s series as

aze -/1 = 1 +[ ] [ ]

¥+-

++-

+-

....1

!

1....

1

!2

1

][

1

!1

12 n

aznazaz

Here the principal part of the Laurent’s expansion has infinite number of terms. \z = a is an essential singularity.

Problem: Show that for f(z) = ze , z = ¥ is an essential singularity. Answer To know the behavior of f(z) at z = ¥ , it is enough to consider the behavior of f(1/z) at z = 0.

Now, f(1/z) = ze /1 which is analytic in the annulus 0 < |z| < R < ¥ .

\ Its Laurent’s expansion about z = 0 is ¥+++++=÷ø

öçè

æ....

1

!

1....

1

!2

11

!1

11

12 nznzzz

f .

There are infinite numbers of terms in the principle part. \ f(1/z) has an essential singularity at z = 0.

\ f(z) = ze has an essential singularity at z = ¥ . Theorem Let f(z) be an analytic function having no singularities except the poles in the extended complex plane. Then f(z) is a rational function. Proof A rational function is the quotient of two polynomials.



43

Given that f(z) is meromerphic Let f(z) have pole at z = a, b, c,…k of orders p,q,r,…. ¥ respectively.

Then ( )

( ) ( ) ( ) ( )srqpkzczbzaz

Zzf

----=

.....)(

f ----------- (1)

Where )(zf is analytic for all finite z. ( ) ( ) ( ) )(...)( zfkzbzazzsqp

---=\f

Hence by Taylor’s Theorem, )(zf can be expanded as

å¥

=0

............)( nn zazf (1) Put

uz

1=

å¥

÷ø

öçè

æ=÷

ø

öçè

æ

0

11nn

ua

uf …………..(2)

The behaviors of )(zf at z = ¥ is same as the behavior of ÷ø

öçè

æ

uf

1at z = 0.

Since u = o is a pole of )/1( uf , the principal part of )/1( uf should have only finite

number of terms \ (2) must have only a finite number of terms. Hence (1) also must have only a finite number of terms.

\ )(zf is a polynomial and therefore ( )

( ) ( )spkzaz

Zzf

--=

.....)(

fis expressed as a quotient

of two polynomials. Hence f(z) is a rational function.

6.4 LOCAL MAPPING THEOREM

Statement: Let jz be the zeros of a function f(z) which is analytic in a disc D and f(z)

does not vanish identically and each zero being counted as many times as its order indicates. Then for every closed curveg not passing through a zero,

( )[ ] ( )( )

dzzf

zf

izn j òå

¢=

gp

g2

1, where the solution has only a finite number of terms ¹ o.

Proof Let f(z) be a function which is analytic in the open circular disc D and f(z) º o. Let g be a closed curve in D .Such that f(z) ¹ o on g .

Now assume that f(z) has only a finite number of zeros in D . i.e., 1z , 2z ,…. nz where each zero is replaces as many time as its order indicates.

Since 1z ,….. nz are the zeros of f(z) and f(z) is analytic, we can write

)())....(()( 1 zgzzzzzf n--= where g(z) ¹ o and analytic

log f(z) = log (z- 1z ) +....+ log (z- nz ) + log g(z)

( )( )

( )( )zg

zg

zzzzzf

zf

n

¢+

-++

-=

¢\

1.......

1

1

( )( )

( )( )

dzzg

zg

izz

dz

izz

dz

idz

zf

zf

i nòòòò

¢+

-++

-=

¢Þ

ggggpppp 2

1

2

1.....

2

1

2

1

1

………(1)



44

Since g(z) is analytic 1g (z) is also analytic and hence ( )( )zg

zg¢ is analytic as g(z) ¹ o.

\By Cauchy’s Theorem, ( )( )

0=¢

ò dzzg

zg

g

\ (1) reduce to ( )( )

),(......),(2

11 nznzn

zf

dzzf

igg

pg

++=¢

ò

{by the definition of index of a point w.r.t a closed are g } ( )å=

=n

jjzn

1

,g

Thus we have proved the theorem if f(z) has a finite number of zeros in D . If f(z) has infinitely many zeros in D , then for any closed curve g in A, we can find a

smaller disk 1D such that AA ÌÍ 'g .

Now there are only a finite number of zeros of f(z) in 'A . Otherwise, if there are infinitely many zeros of f(z) in 'A , then closure of 'A will have an accumulation point which is impossible. That is we can find a sequence { nz } of zeros of f(z) such that nz tends to an

accumulation point 0z as n ¥® , ( ) 0)( 0 =¥®

=þýü

îíì

¥®= nn zf

n

LimZ

n

Limfzf

Hence 0z is also a zero of f(z).Thus the zero 0z of the analytic function f(z) is not

isolated. This is a contradiction. Applying first part of the proof through the disk 'D , we get

åò =j

jzndzzf

zf

i),(

)(

)('

2

1g

pg

……. (2)

Even if jz is a zero of f(z), then ),( jzn g = 0 and therefore the contribution of such zeros

to the R.H. S is zero without affecting the result (2).

Deduction (1) If w = f(z) and G is the image of g under f(z),

then òg

pdz

zf

zf

i )(

)('

2

1 = ò -

gw

w

p.

02

1 d

i

From the previous theorem, )0,(),( G=å nznj

jg

(2) If g is a circle, then ),( ¥gn = 1 or 0 according as a lies inside (or) outside D .

This means that åj

jzn ),(g gives the number of zeros of f(z) in g . Thus if g is a

circle, then n( G ,0) gives the number of zeros of f(z) in g .

(3) If a is any arbitrary complex number, applying the theorem to g(z) = f(z) – a, we get



45

òå =g

pg dz

zg

zg

iazn

jj

)(

)('

2

1)))(,((

and )(az j are zeros of f(z).

i.e., òå-

=g

pg dz

azf

zf

iazn j

)(

)('

2

1))(,(

\ ò G=-

gw

w

p),(

2

1an

a

d

i

Again if g is a circle the terms of the above summation are each equal to 1(or) 0

according as )(az j lies inside or outside g .

\ g is a circle then the number of roots of f(z) = a inside g is given by ).,( an G

(4) If a, b are in the same region determined by G , then ),( an G = ).,( bn G

i.e., number of roots of f(z) = a inside g is the same as the number of roots of f(z)

= b inside g . Thus if g is a circle, number of roots of f(z) = a is the same as the

number of roots of f(z) = b. Theorem on Local Correspondence

Statement: Suppose that f(z) is analytic at 0z , f( 0z ) and f(z) - 0w has a zero of order n at

0z . If e >0 is sufficiently small, then there exists a d >0 such that for all a with |a- 0w |

<d the equation f(z) = a has exactly n roots in the disc |z- 0z | <e .

Proof

Choose na >0, so that f(z) is defined and analytic in |z- 0z | < 0 so that 0z is the

only zero of f(z) in the disk. Let g be the circle, |z- 0z | <e and G , its image under the mapping w = f(z).

i.e. G is a closed curve. Since 0z Î g Þ f( 0z ) = 0w Ï G . Hence 0w is in the complement

of 0w Ï G .

Hence 0z belongs to an open set and so there exists a neighborhood |w- 0w | <d and for

every aÎ |w- 0w | <d all the values of a are taken same number of times insideg .

But since f(z) = 0w has exactly n coinciding roots inside g , f(z) = a also has exactly n

roots and every value of a is taken n times. Open Mapping Theorem analytic Statement: A non- constant function maps open sets onto open sets. Proof: Let f(z) be a non-constant analytic function. Let W be an open set and let

0z WÎ . Let 0w = f( 0z ) and 0w belong to the image of W we shall now prove that the

image of W is also open. i.e. there exists neighborhood of 0w contained in the image of W .



46

By the theorem on local correspondence, $ a neighborhood of 0w contained in the image

of W and $ a neighborhood || 0ww - <d of 0w for every neighborhood || 0zz - < e ,

such that each value in || 0ww - <d is taken ‘n’ times by f(z) in || 0zz - < e contained

in W . \n is the order of the zero 0z of the function f(z)- 0w .

This implies that the image of W contains the neighborhood |w- 0w | <d .

Since 0w is arbitrary it is true for every 0w which is in the image of W . This means that

the image of W is open. Corollary If f(z) is analytic at 0z with 'f ( 0z ) ¹ 0 it maps a neighborhood of 0z

conformally and topologically onto a region. Proof Taking n = 1, in the local Mapping Theorem, the function f(z) - 0w has a zero of

order 1 at 0z for every point in |w- 0w |< d , f(z) assumes the values only once in

|| 0zz - <e . By the open Mapping theorem, a non – constant analytic function maps open

sets onto open sets.

Hence there is a 1-1 correspondence between the disc |w- 0w |<d and the open disc

|| 0zz - <e .

Since open set of z- plane correspond to open set of w-plane, the inverse function 1-f is

continuous and the mapping is topological. As the inverse function is continuous, and analytic, the mapping is conformal. Conversely, if the local mapping is 1-1, the theorem on local correspondence holds only if n =1. i.e, the order of zero 0z of the function is 1. f( 0z ) = 0and 'f ( 0z ) ¹ 0. The

necessary and sufficient condition for the local mapping is that the mapping is 1-1 and

'f ( 0z ) ¹ 0.

6.5 THE MAXIMUM PRINCIPLE Statement: If f(z) is a non constant function defined and continuous on a closed bounded set E and analytic in the interior of E, then the maximum of |f(z)| on E is assumed on the boundary of E. Proof: Since E is closed and bounded, it is compact. Assume that |f(z)| has a maximum

value on E, say at 0z of E (interior point of E).

Since f(z) is non – constant and analytic, i.e |f(z) - f( 0z ) | < e if |z- 0z | <d .

i.e. |f(z) - f( 0z )| |< | f(z) – f( 0z )| < d Þ |f(z)| is continuous.

If 0z is an interior point, then | f( 0z )| is the maximum of |f(z)| in |z- 0z | <d contained in

E. But it is impossible unless f(z) is a constant in the complement of the interior of E containing 0z . By continuity, |f(z)| has its maximum on the whole boundary of that



47

complement and this boundary is non – empty and is contained in the boundary of E. Thus the maximum of |f(z)| is always only on the boundary of E.

Analytical Proof of Maximum Modulus Principle Let f(z) be a non constant analytic function defined and continuous on a region W then |f(z)| has its maximum only on the boundary of W . Proof Assume that |f(z)|, has its maximum at a point 0z Î W ( 0z is an interior point of

W ). Then we can find a neighborhood around 0z which lies entirely in W . Let g be a

circle contained in this neighborhood of radius r so that its equation is of the form (t) =

0z + itre , 0 £ t £ 2p . By Cauchy’s Integral formula.

dtrezfi

dz

f

izf it

òò +=-

=p

gp

xx

x

p

2

0

0

0

0 )(2

1)(

2

1)(

This shows that the value of an analytic function at the centre of a circle is equal to the arbitrary mean of its value on the circle only if the circle lies entirely in the region of analyticity of f(z).

Therefore |f(z)| = dtrezfi

it

ò +p

p

2

0

0 )(2

1 ||)(

2

12

0 dtrezo

it

ò +£p

p.

Since |f( 0z )| = maximum |f(z)| in W , we get )( 0itrezf + £ |f( 0z )|

If the strict inequality holds at a single point for a single t, then by continuity of f(z), it

holds on the whole of g , )( 0itrezf + < |f( 0z )| which implies |f( 0z )| < |f( 0z )| which is

absurd. Therefore |f(z)| must be a constant and is equal to |f( 0z )| for all sufficiently

small circles |z- 0z | = r and hence in the neighborhood of 0z .

\ It follows easily that f(z) must reduce to a constant. This is contrary to the hypothesis. Hence our assumption that |f(z)| has its maximum at an interior point is wrong. Hence |f( 0z )| is maximum only on the boundary of W .

Schwarz Lemma If f(z) is analytic for |z| < 1 such that |f(z)| < 1, f(0) = 0, then |f(z)| < |z| and | 'f (0)| £ 1.

If |f(z)| = |z| for some z ¹ 0 or if | 'f (0)| =1, then f(z) = cz, c is a constant and |c| = 1.

Proof: Define ïî

ïíì

=

¹=

0),0('

0,)(

)(1

zf

zz

zfzf for any z in |z| < |.

Here f(z) is analytic everywhere. If c is a circle |z| £ r < 1, then any z ,rcÎ as |z| < r is

closed and bounded, f(z) is analytic and |f(z)| £ 1for |z| £ r. Þ |f(z)| attains its maximum value on the boundary of |z| = r by the maximum modulus principle.

\||

)()(1

z

zfzf =

r

1£ As r ® 1, )(1 zf = £ 1, for z ¹ 0



48

Þ 1)(

£z

zf Þ |f(z)| £ |z| and )0(1f = )0(1f <1. If |f(z)| = |z| for some z ¹ 0 or if

)0(1f = 1, then again using maximum modulus principle, we see that there exists a

constant c such that z

zf )( = 1.

\ Letting C = z

zf )(, |c| = 1, f(z) = cz, where |z| = 1. Hence the lemma.

Corollary

If f(z) is analytic in |z| < R, |f(z)| £ 1 in |z| < R and f(0) = 0, then |f(z)| < R

z ||.

From the given conditions, we see that f(Rz) is analytic in |z| < 1 and also |f(Rz)| £ 1 and f(0) = 0 in |z| < 1.

Applying Schwarz Lemma for f(Rz), we get |f(Rz)| < |z| Changing z to R

Z,

we get |f(z)| £ R

z ||.

Corollary

If f(z) is analytic in |z| < R and |f(z)| < m, f(0) = 0, then |f(z)| £ R

zM.

Applying Schwarz lemma for ,)(

M

Rf z we get M

Rf z )( < |z| Þ |f(Rz)| £ M|z|

Changing z to z/R, we get |f(z)| < R

zM Hence the case.

6.6 CHAINS AND CYCLES

Definition: Let W be a region and nggg ,...,, 21 be a collection of arcs in W . The sum

gggg =+++ n...21 is called a chain. Here n is any arbitrary number.

We can write this chain g in the form g = ppagagag +++ ...2211 where

1g ,…, pg are the distinct arcs in W and no two of them are negative to each other and

paaa ,...,, 21 are all integers.

Definition: A chain g is said to be a cycle if the curve in the formal sum (1) are all

closed curves. That is, each 1g is a closed curve.

If g is a cycle in a region, we define the index of any point ‘a’ with respect to g

as n(g ,a) = n( 1g ,a) + n( 2g ,a) + … + n( pg ,a).



49

Definition: A cycle r in a region W is said to be homologous to zero if n(g ,a) = 0,

" a WÏ . That is, if for all a in the complement of W , n(g ,a) = 0. We denote this fact by

g � � � 0 (mod W ) and read g is homologous to zero mod W . If 21,gg are two cycles in a

region W , they are said to be equivalent. If )( 21 gg - ~ (mod W ) i.e., n( 21 gg - ,a) = 0," a WÏ i.e., n( ,1g a) = n( ,2g a),

" a WÏ . It is clear that all the results we have proved for closed curves are in fact valid for arbitrary cycles in a region. Therefore the integral of an exact differential over any cycle is zero. 6.7 LET US SUM UP

(1) Principle of maximum modulus (2) Schwarz lemma (3) Definition of chains and cycles, types of singularities, definition of removable singularity and essential singularity. (4) Behaviour of a function at z = ¥ . 6.8 LESSON END ACTIVITIES

(1) Show that |f(z)| < |f( zR )| implies the inequality 22

1

1

)(1

)('

zzf

zf

-£

-.

(2) Show that an isolated singularity of f(z) cannot be a pole of ze .

(3) Show that ze /1 has an essential singularity at z = 0. (4) Show that z = ¥ is an essential singularity of sin z and cos z.

6.9 REFERENCES

1. Complex Analysis by L.V. Alphors.. 2. Foundation of Complex Analysis by Dr. S. Ponnusamy. 3. Complex Analysis by A.R. Vasistha



49

UNIT-3 LESSON 7 THE CALCULUS OF RESIDUES

CONTENTS 7.0 Aim and Objectives 7.1 Introduction 7.2 The Residue Theorem 7.3 Argument Principle 7.4 Evaluation of definite integrals 7.5 Let us Sum up 7.6 Lesson end activities 7.7 References 7.0 AIM AND OBJECTIVES

In this lesson we define residue with respect to poles of an analytic function. This helps us to evaluate contour integrals and definite integrals. We also study about the number of zeros and poles, a meromorphic function, can have inside a closed curve. After going through this lesson you will be able to: (i) define residues and argument principle (ii) state Residue theorem and Rouche’s theorem 7.1 INTRODUCTION

The results of the preceding section show that the determination of line integrals of analytic function over closed curves can be reduced to the determination of periods. Sometimes, finding periods may be difficult. Therefore, in this lesson we suggest a method which helps us to evaluate integrals over closed curves without much calculation. This is of great value in the further development of the theory. The calculus of residues deals with this method of evaluating definite and contour integrals.

7.2 THE RESIDUE THEOREM Definition 1 Let f(z) be analytic in a region W except at ‘a’ which is an isolated singularity of f(z), we can find a d > 0 such that f(z) is analytic in the annulus region 0 < |z-a| <d . Then the residue of f(z) at z = a is a unique complex number R such that

f(z) = az

R

- is the derivative of a single valued analytic function in the said annulus.

Theorem: If f(z) is analytic in 'W = W - {a} where ‘a’ is an isolated singularity. Then

there exists a unique complex number R, such that f(z) = az

R

- is the derivative of an

analytic function. Proof: Since f(z) has an isolated singularity at z = a, we can find a d > 0 such that f(z) is analytic in the annulus 0 < |z-a| <d .

Let c be a circle, |z-a| = r where r <d write R = òc

dzzfi

)(2

1

p……. (1)



50

Now, we consider òþýü

îíì

--

c

dzaz

Rzf )( = òò -

-cc

az

Rdzdzzf )(

= ò -c

iRdzzf p2)(

òc

dzzf )( ò-c

dzzf )( az

Rzf

--Þ )( is derivative of an analytic function in the annulus

0 < |z-a| <d .

d

1c2c

To prove the uniqueness of R, it is enough to show that òò =21

)()(cc

dzzfdzzf , where 1c

and 2c are two circles 0 < |z-a| < ir , i = 1,2.

Since every closed curve G in this region is given by odzzf =òG

)( (by Cauchy’s

Theorem). Choosing the points p,q on 1c and 2c . Consider the closed curve composed of

1c from p to q in the anticlockwise direction. So the line pq, - 2c and the straight line qp .

Then =òG

dzzf )( òòòò +++- qpcpqc

dzzfdzzfdzzfdzzf )()()()(21

0 = ò1

)(c

dzzf = òò +pqc

dzzfdzzf )()(2

\ ò1

)(c

dzzf = ò2

)(c

dzzf i.e. òc

dzzf )( is independent of the radius of the circle c.

\ R is uniquely determined complex number. Cauchy’s Residue Theorem: Let f(z) be analytic except at isolated singularity in the region W . Then for and cycleg in W , which is homologous to zero and doesn’t

not pass through any of the points aj, then åò =j

ajndzzfi

),()(2

1g

pg

Residue of

f(z) at z = ja .

Proof In W , let 1a , … , an be the isolated singularities of f(z), for each aj we can find

jd such that f(z) is analytic in 0< |z- ja | <d j.

Let jc be the circle of radius < jd in the region W .



51

Let g be a cycle in W . Where 'W = W - { 1a } – { 2a } - … - { na } which is homologous to

zero with respect to 'W .

Theng W = å Wj

jcajn )(mod),(g , i.e., g =å WW )'(mod0),( cjajn g ,

i.e., g and åj

jcajn ),(g are homologous to each other, i.e., by virtue of homologous we

have òg

dzzf )( = ò dzzf )( = òj

ajn ),(g åòj

j

cj

cajndzzf ),()( g = åj

jpajn ,),(g (where

jj Rip p2= )

= åj

jiRajn pg 2),( åò =Þj

jRajndzzfi

),()(2

1g

pg

= åj

ajn ),(g Residue f(z)

at z = ja . Hence the theorem.

7.3 ARGUMENT PRINCIPLE

If f(z) is meromorphic in W with zeros ja and the poles kb , then for

every cycle g which is homologous to zero and not passing through any of the zeros or

poles, then ååò -=kj

bknajndzzf

zf

i),(),(

)(

)(

2

1gg

pg

Proof By Residue Theorem )(

)('Re),(

)(

)('

2

1

zf

zfsajndz

zf

zf

i jåò = g

pg

at z = cj.

Where cj are the singularities of )(

)('

zf

zf. Since f(z) is memomorphic, f(z) is also

meromorphic. )(

)('

zf

zf is also meromorphic and the only singularities of

)(

)('

zf

zf are zeros

and poles of f(z), every cj is either a aj or bk.

If aj is a zero of order h for f(z), then f(z) = )2()()( -- zfajz nh

where )(zfn is analytic and non-null at ajz - . Taking logarithm on both sides and

differentiating, )(

)('

zf

zf =

jaz

h

-+

)(

)('

zf

zf

h

h Þ)(

)('

zf

zf =

jaz

h

-+

)(

)('

zf

zf

h

h .

Since )(

)('

zf

zf

h

h is analytic .0)(

)('

2

1=ò

gp

dzzf

zf

i h

h

\ .0)(

)('

2

1=

ïþ

ïýü

ïî

ïíì

--ò

gp

dzaz

h

zf

zf

i jh

h

Hence by definition of residue, we have ajz

s

=

Re

)(

)('

zf

zf = h.



52

Note If G is the image of g , index w = f(z) then from the above theorem,

òòG

G== )0,(2

1

)(

)('

2

1n

w

dw

idz

zf

zf

i ppg

.

If G is in a circular disk which does not contain the origin, then n( G , 0) = 0. Rouche’s Theorem Let g be a cycle homologous to zero in the region W and n(g , z)

is either 0 or 1 for any zÏ g . Suppose that f (z), g (z) are analytic in W satisfying the

inequality |f (z) – g (z)| < |f (z)|, gÎ"z . Therefore f (z) and g (z) will have the same

number of zeros in g .

Proof: Given that |f (z) – g (z)| < |f (z)|, gÎ"z ….....(1).

Both f (z) and g (z) have no zeros on g . For, if f (z) has a zero on g , then |g (z)| < 0

from (1), which is false. Similarly if g (z) has a zero on g , then (1) Þ |f (z)| < |f (z)|

which is absurd. \ Both f (z), g (z) have no zeros ong .

From (1), we have 1)(

)(-

zf

zg < 1, gÎ"z …… (2)

Let F (z) = )(

)(

zf

zg on g \ The zeroes of f (z) are the zeros of g (z) and the poles of f (z)

are the zeros of f (z). Similarly if kb is the pole of order r for the function f(z), then we

can write f(z) = rbkz -- )( rf (z) where rf (z) is analytic and non-null at z = kb . Taking

Logarithmic differentiation, we get

)(

)(

)(

)( '

zf

zf

bz

r

zf

zf

r

r

k

+-

-=¢

)(

)(

)(

)(

zf

zf

bz

r

zf

zf

r

r

k

¢=

þýü

îíì

-

--=

¢\

Since the R.H.S is the derivative of an analytic function by Cauchy’s Theorem,

0)(

)('

2

1=

þýü

îíì

÷÷ø

öççè

æ

-

--\ ò dz

bz

r

zf

zf

i kr

r

gp

By the definition of residues, Res rzf

zf-=

¢

)(

)(……(3)

(1) Þ ( ))(

)(',

)(

)('

2

1 Re

zf

zfajndz

zf

zf

i

s

ajzj

=åò = gp

g

( ))z(f

)z(fbk,n

k

siRebkz

¢g+ å = ( ) ( )å å-+

j kj rbknhan ,, gg

Where h and r are order of zeros aj and the poles

kb ( ) ( )å åò -=j k

jj bknhandzzf

zf

i,,

)(

)('

2

1gg

pg

= åå -k

kj

j bnan ),(),( gg the summation

of zeros ja and poles kb are done according to their degree of multiplication.



53

7.4 EVALUATION OF DEFINITE INTEGRALS The evaluation of proper and improper definite is somewhat difficult. Using Cauchy’s Residue Theorem some of the definite integrals are solved in this lesson.

Integrals of types òp2

0

f(cos 0 , Sin 0 )d 0 , ò¥

¥-

f(x) dx are solved. Number of zeros of an

entire function in a given region is found using Rouche’s Theorem Unit Circle Problems

To solve problems of types òp2

0

f(cos 0 sin 0 )d 0

Evaluate 0|b|a,0cosba

0d2

0

>>+ò

p

Answer

Let l = òp

+

2

00cosba

0d Let c be the unit circle |z| = 1.

On c, z = 0ie = cos 0 + sin 0 , 1/z = 0ie- = cos 0 - i sin 0

\cos 0 = 21 ÷

ø

öçè

æ+

zz

1and sin 0 =

i2

1÷ø

öçè

æ-

zz

1

dz = 0ie . id 0 = zid 0

\zi

dz0d = on c, 0 varies from 0 to p2

ò÷ø

öçè

æ++

=

c2

1z

2

ba

zi/dzI ò ++

=

c

2 baz2bz

dz

i

2ò

++

=c

b

azz

dz

bi1

2

2

2

ixbi

p22

= x (sum of the resides)

xb

4p= (sum of the resides) …………….(1)

The poles of f(z) = 1

b

az2z

1

2 ++

are given by

01zb

a2z2 =++ i.e.

2

442

2

2

-±-

=b

a

b

a

z2

22

b

ba

b

a -±

-=

Let b

baa 22 -+-=a &

b

baa 22 ---=b

1|| =ab & =b|| 1|b|

baa 22

>---

1||

1|| <=\

ba

Hence a lies inside c and b lies outside c. Let us find residue at a

Residue at ( )( )( )ba

aa a--

-= ®zz

zLimz

122 ba2

b1

-=

b-a=



54

(1)22 ba2

bx

b

4l

-

p=Þ

22 ba

2

-

p=

Evaluate 1a,0cosa

0d

0

>+ò

p

Answer Let I = òp

+0

0cosa

0d. Let c be the unit circle |z| = 1 on c, 0iez = and 01 ie

z-=

\ cos 0 ÷ø

öçè

æ+=

zz

1

2

1 and sin ÷

ø

öçè

æ-=

zz

i

1

2

10 , dz = 0ie id 0 0zid= . Therefore d 0 =

iz

dz.

On c, 0 varies from 0 to p2

\ I ò÷ø

öçè

æ++

=c

zza

zidz

1

2

1

/ò ++

=

c

2 1az2z

zi/dz

i

2

i2xi

2p= (Sum of the residues) = p4 x (Sum of the residues)

The poles of 1az2z

1)z(f

2 ++= are given by

122 ++ azz = 0, 2

442 2 -±-=

aaz 1aa 2 -±-=

Let 1aa 2 -+-=a and 12 ---= aab

1|| =ab and =b|| 11aa 2 >-+ . Therefore 1||

1|| <=

ba

Hence a lies inside c and b lies outside C

\Residue at ( )( )( )ba

aa a--

-= ®zz

zLimz

1

12

112 -

=-

=aba

\(1) Þ I 12

14

2 -=

axp

1a

2

2 -

p=

1a

2

cosa

d

2

2

0 -

p=

q+

qòp

i.e. 1a

2

cosa

d2

2

2

0 -

p=

q+

qòp

i.e. 1acos1

d

2

2

0 -

p=

q+

qòp

.

By contour integration, show that

,e)ma1(a4)ax(

mxcos ma

3222

-¥

¥-

+p

=+ò m < 0 and a < 0.

Answer

Let f(z) = 222 )( az

e imz

+ Let G be the closed curve consisting of

(i) Positively oriented semicircular are c, sin (z) ³ 0, |z| = R. (ii) The real axis from (-R, 0) to (R, 0).



55

By Cauchy’s Residue Theorem, dz)az(

e222

imz

òG

+ = i2p x (Sum of the residues)

i2dx)az(

edz

)az(

eR

R

222

imz

c

222

imz

p=+

++ òò

-

x (Sum of the residues).

The poles of f(z) = 222

imz

)az(

e

+ are given by 222 )az( + = 0 Þ )az( 22 + = 0 Þ z = ± ai, each

is a pole of order 2. If z = ai, lies inside G . Then the Residue at ai = !1

)(' iaf

where f (z) = 2)( aiz - . 22 )()( aizaiz

e imz

+- =

2)( aiz

e imz

+

4

2

)(

)(2)()('

aiz

aizeimeaizz

imzimz

+

+-+=f

3)(

2)()('

aiz

eimeaizz

imzimz

+

-+=f

Therefore 'f (ai) ia

ame am

34

)1( +=

-

(1) Þ òò- +

++

R

R

imz

c

imz

dzax

edz

az

e222222 )()(

= ia4

)am1(exi23

am +p -

= )am1(ea2

ma

3+

p - ……. (2)

Now, 22 az + ³ |z| 2 - 2a = 22 aR - on c. 22222 )aR(

1

az

1

-£

+ ® 0 as R ® ¥ .

By a result, ¥®R

Lim0dz

)az(

e

c

222

imz

=+ò . Taking limits on both sides of (2), we get

ò¥

¥-+

+dx

)az(

xsinimxcos222

)1(2 3

amea

am += -p

Equating real parts, ò¥

¥-+

dx)ax(

mxcos222

)am1(ea2

am

3+

p= -

ò¥

¥-+

dx)ax(

mxcos222

)am1(ea4

am

3+

p= - .

Problem

Show that 2

dxx

mxsin p=ò

¥

¥-

.

Answer

g

)0,( R- )0,(R)0,( r- )0,(r



56

Let f(z) = z

eimz

. f(z) is analytic for the values of z except at z = 0.

Hence origin is a Simple Pole of f(z) . Let G be the closed curve consisting of (i) the +vely oriented Semi Circular arc c, |z| = r , im(z) > 0

(ii) the real axis from (-R, 0) to (- r , 0)

(iii) the negatively oriented Semi Circular arc g , |z| = r , im(z) > 0

(iv) the real axis from ( r , 0) to (R, 0),

Now, f(z) is analytic within and on G . Hence by Cauchy’s theorem,

0dxz

eimz

=òG

Þ òc

imz

dzz

e+ ò

r-

-R

imz

dxx

e- ò

g

dzz

eimz

+ ò =r

r

.0dxx

e imz

----- (1)

On c, |z| = R, R

1

|z|

1= ® 0 as R ® ¥ .

Hence ¥®R

Lim0dz

z

e

c

imz

=ò ….. (2)

0®r

Lim),(ikdz

z

e

c

imz

a-b=ò k = .0

zz

Lim

® z

eimz

= 1 = i ( p -0)

Taking limits on both sides of 1.

¥®z

Lim+ò

c

imz

dzz

eò

-

-

r

R

imz

dzz

e-

0®r

Limòg

dzz

eimz

+ ò =R imz

dzx

e

r

.0

0 + ò¥-

0

dzz

e imz

- i p + 00

=ò¥

dzz

e imx

pidxx

mximx=

+ò¥

¥-

sincos

Equating real and imaginary parts, 0dxx

mxcos=ò

¥

¥-

and p=ò¥

¥-

dxx

mxsin .

2dx

x

mxsin p=Þ ò

¥

¥-

Since the answer is independent of m, .2

dxx

xsin

0

p=ò

¥

-

7.5 LET US SUM UP

(1) Concept of residues (2) Cauchy’s Residue Theorem (3) Argument principle (4) Statement of Rouches Theorem (5) Methods of finding residues at poles.



57


(1) Find the residues of 22

2

az

z

+ at its poles.

(2) Find the residues of f(z) = 222

2

)az(

z

+ at its poles.

(3) Find the residue of f(z) = )3z)(2z()1z(

z4

3

--- at z = 1.

(4) Show that 3

2

cos2

d2

0

p=

q+

qòp

(5) Show that .||||,2

sin

122

2

0

baba

dba

>-

=+ò

pq

q

p

7.7 REFERENCES 1. Complex Analysis by L.V. Alphors.. 2. Foundation of Complex Analysis by Dr. S. Ponnusamy 3. Complex Analysis by A.R. Vashistha.



58

LESSON – 8 HARMONIC FUNCTIONS CONTENTS 8.0 Aim and Objectives 8.1 Introduction 8.2 Basic Properties 8.3 The mean value properties 8.4 Poisson’s Formula. 8.5 Schwarz Theorem 8.6 Let us Sum up 8.7 Lesson end activities 8.8 References 8.0 AIM AND OBJECTIVES

The objective of this lesson is to study about the harmonic functions. If f(z) = u + iv is analytic, then both u and v are harmonic functions. We are going to learn some of the basic properties of harmonic functions, such as mean value property, the maximum principle for harmonic functions, Poisson formula etc. After going through this lesson you will be able to: (i) study about basic properties of harmonic functions (ii) define homology (ii) state the mean value property (iv) state the maximum principle for harmonic functions (v) state the Poisson’s formula and Schwarz Theorem 8.1 INTRODUCTION If a real valued function u(x, y), defined and continuous together with its first and

second order partial derivatives satisfying the Laplace equation ,0y

u

x

u2

2

2

2

=¶

¶+

¶

¶ then u is

called harmonic function. The sum of two harmonic functions and a constant multiple of a harmonic function are also harmonic. The Laplace equation in Polar form is

.0u

r

ur

rr

2

2

=q¶

¶+

þýü

îíì

¶

¶

¶

¶

8.2 BASIC PROPERTIES

(1) u = log r is harmonic. (2) A harmonic function of r is of the form a log r + b.

(3) If u is harmonic in W , then f(z) = y

ui

x

u

¶

¶-

¶

¶ is analytic.

(4) If f(z) = y

ui

x

u

¶

¶-

¶

¶ = 0, then 'f (z) dz =

þýü

îíì

¶

¶-

¶

¶

y

ui

x

u (dx + idy)

= þýü

îíì

¶

¶-

¶

¶+

þýü

îíì

¶

¶+

¶

¶dx

y

udy

x

uidy

y

udx

x

u

'f (z) = du + idv … …… (1)



59

If u has a conjugate harmonic function v, then dv = dyy

vdx

x

v

¶

¶+

¶

¶ =

þýü

îíì

¶

¶+

¶

¶- dy

x

udx

y

u

(using C.R. equation) and if there is no single valued conjugate function.

Let it be du = dxy

udy

x

u

¶

¶-

¶

¶ and *du is called the conjugate differential of du.

i.e; *du = dv. Now (1) Þ f(z) dz = du + i *du. If g is a cycle, which is homologous to zero in W then by Cauchy’s theorem.

òg

= 0dz)z(f Þ òg

=+ 0du*idu .

Theorem on Homology

If 21 u,u are harmonic in a region W then ò =-g

0** 1221 duuduu , for all cycle g

which is homologous to zero in W . Proof Let 21 u,u be harmonic in W . Then 21 u,u have harmonic conjugate

i.e.; 21 du,du are harmonic conjugate differentials of 1u and 2u .

i.e.; * 11 dvdu = and * .22 dvdu = \ 12211221 ** dvudvuduuduu -=-

= )( 122121 vududvdvu -+ ……. (1)

Where )( 12vud is an exact differential and 2121 udvdvu + = Imaginary part of

))(( 2211 idvduivu ++ ……. (2)

If we put ,)(,)( 222111 ivuzFivuzF +=+= then 22'

2 )( idvdudzzF +=

\(1) & (2) Þ òòò -=-ggg

)())((Im** 1211221 vuddzzFIduuduu …(3)

Since ))()(Im( 121 ZFZF is analytic, by Cauchy’s theorem ( ) ( )( ) 0Im

1

21 =ò dzzFzFg

.

Similarly ( ) 012 =òg

vud Hence 0** 1221 =-ò duuduug

Note : If uu,1u 21 == then we get òg

= 0du* .

8.3 THE MEAN VALUE PROPERTY The Arithmetic mean of a harmonic function over concentric circles |z| = r is a

linear function of log r, a=p ò

=r|Z|

0ud2

1 log b+r and if u is harmonic in a disk, 0=a and

arithmetic mean is a constantb . In that case b = u(0).

Proof Using the previous theorem with 1u = log r, 2u = u where u is harmonic in



60

|z| < r (with origin removed). Choose W to be the punctured disc 0 <|z| < r and for g

take the cycle 21 cc - where ic , is the circle |z| = r<ir , described in the positive sense.

We know that * du || dzn

u

¶

¶= .Where

n

u

¶

¶ is the directional derivative of u in the normal

direction.

But on the circle |z| = g is replaced by |dz|n

udu*

r

u

¶

¶=

¶

¶

|dz|r

u

¶

¶= 0d

r

ur

¶

¶=

ïþ

ïý

ü

ïî

ïí

ì

=

=

=

0rd|dz|

0diredz

rez0i

0iQ

Putting u = log r, )1........(..........0d0d)r(logr

r)r(logd*du* =¶

¶==

By the previous theorem, 0duuduu 1221 =+-òl

0du*udu*u 1221

21

=-= òÏ-Ï

Putting 1u = log r , uu =2 , we get 0)(log**log||

=-= ò=

rdudurrz

( ) .00*log||

=-ò=rz

dudur

In other words it implies that the value of the integral is the same. i.e.. It is constant on different paths 21 ,cc .Let it be )( b- .

\ ò=r|Z|

(log r* du – u d 0 ) = constant = b-

Þ ÷ø

öçè

æ

p+b=

p òò=

du*2

1rlog0ud

2

1

r|Z|

rlog+b= ……(2)

If u is harmonic in a disk, then 0=a . By Cauchy’s integral formula,

)0(u0z

dz)Z(u

i2

1

r|Z|

=-p ò

=

Þ bp

p

==ò )0()(

2

10

02

0

ure

eirzu

i i

i

(say)

\ If u is harmonic with 0=a then b==p ò

p

)0(u0d)z(ui2

12

0

Note: By continuity and changing to a new origin, we have from

u(0) = òp

p

2

0

0d)z(ui2

1

( ) ,02

1)(

2

0

000 ò +=

p

pdrezuzu i because by Cauchy’s integral Formula,

ò -=

az

dzzf

iaf

)(

2

1)(

p



61

put 00),()( irezzzuzf +== and 0za = . ( )ò +=\

p2

0

000 0

2

1)( drezu

izu i is the Mean

Value Property. Maximum Principle for harmonic functions If u(z) is a non constant harmonic function in the region W then u(z) has no maximum value in W . Proof Let We0z such that |)(| 0zu has a Maximum in the neighborhood of 0z .

Since u(z) is harmonic, by the Mean value Property, we have ( )ò +=p

p

2

0

000 0

2

1)( drezu

izu i

( )ò +<Þp

p

2

0

000 |0|||

2

1|)(| drezu

izu i

Since u( 0z ) has a maximum value, )()( 00

0 zurezu i £+

Assume that the strict inequality holds for a single value of 0 .

\By continuity, it will hold on the whole arc. This means that )()(| 00

0 zurezu i <+

)()( 00 zuzu <Þ as 0r ® .This is a contradiction.

Therefore |)(| zu is consistently equal to )(| 0zu in the neighborhood of 0z . (i.e) u (z) is a

constant. Hence u(z) has no maximum value in W and maximum of |u(z)|is attained only on the boundary of W . 8.4 POISSON’S FORMULA.

Suppose that u(z) is harmonic for |z| < R and continuous, then

0)(||

||

2

1)(

||

2

22

dzuaz

aRau

Zò -

-=

p………(A)

0)(Re2

1)(

||

dzuaz

azau

Zò ÷

ø

öçè

æ

-

+=

p………(B)

If we replace firea = 0iRez = ,we get ( ) ( )

( )ò +--

-=

pf

fp

2

0

22

022

0cos2

0Re

2

1)(

rRrR

durRreu

ii ……(C)

Proof Let u(z) be harmonic for z < R. Consider the linear transformation.

taR

aRtRtsz

+

+==

)()( Which maps |t| 1£ onto |z| R£ So that

azR

azRtt

-

-=

2

)( …(1)

gives t = 0 when z = a. Since u(z) is harmonic, u(s(t)) is also harmonic in |z| |t| 1£ . From the Mean Value property



62

òp= 0d)z(u

2

1)o(U ( )( ) ( )ò

=

=Þ1||

arg2

1))((

t

tdtsuosup

{As t = o corresponds to z = a, s(t) = z Þ s(o) = a}

( )ò=

p=

R|Z|

targd)z(u2

1)a(U . ------ (2)

From (1), we see that (1) maps |z| = R to |t| = 1 and the interior point ‘a’ onto the origin, which is the centre of |t| = 1.

Putting t = fie , we have log t = fi , f= idt

dt f=Þ d

t

dti ( )td arg= …….(3)

From (1), log t = log R + log (z-a) – log ( )zaR2 - dzzaR

a

az

10

t

dt2 ÷

÷

ø

ö

çç

è

æ

-+

-=\

dzzaR

a

az

1i

t

dti

2 ÷÷

ø

ö

çç

è

æ

-+

--=-Þ …….(4)

From (3) and (4), we get d (arg t) = - dzzaR

a

az

1i

2 ÷÷

ø

ö

çç

è

æ

-+

-…….(5)

If we put z = 0ie , -idz = zd 0 …………(6)

using (6) in (5), d (arg t) = 0dzaR

za

az

z2 ÷

÷

ø

ö

çç

è

æ

-+

-

d(arg t) 0dzazz

za

az

z÷÷ø

öççè

æ

-+

-{ }2Rzz =Q

0daz

a

az

z÷÷ø

öççè

æ

-+

-=

( )( )0d

azaz

aazazazz÷÷ø

öççè

æ

--

-+-=

d(arg t) = ( )

2

22

|az|

0d|a|R

-

- using in (2)

we get ò -

-

p=

|Z|

2

22

0d)z(u|az|

|a|R

2

1)a(U …..(7). This is the Poisson’s formula

Again, we know that ÷ø

öçè

æ

-

++

-

+=÷

ø

öçè

æ

-

+

az

az

az

az

az

az

2

1Re

2

22

2

22

|az|

|a|R

|az|

|a||z|

-

-=

-

-=

Using this in (A), we get ò=

÷ø

öçè

æ

-

+

p=

R|Z|

0d)z(uaz

az

2

1)a(U ……..(B)

Now, put a = fire and z = 0iRe . Then the Poisson’s formula in polar co-ordinates become as follows.

We see that 2

22

|az|

|a|R

-

-=

( )( )azaz

rR 22

--

- ( )( )ff iiii rere

rR-- --

-=

00

22

ReRe

= 22

22

)cos(2 rRrR

rR

+--

-

fq…………..(8)

Using this in (A), we get u( fire ) = ( ) ( )

( ) 22

0222

0 0cos2

0Re

2

1

rRrR

dUrR i

+--

-ò fp

p

……..(C)



63

(A), (B), (C) are different forms of Poisson’s Formula. Note: (1) In the theorem we have assumed that u is harmonic in |z| £ R.However the result remains take under the weaker condition that u (z) is harmonic-in |z| < R and continuous in |z| £ R. Indeed if o < r < 1 then u (rz) is harmonic in |z| < R. \The above Poison formula becomes.

ò= -

-=

RZ

drzuaz

aRrau

||

2

22

0)(||

||

2

1)(

p.

Since u(z) is uniformly continuous on |z| £ R and hence u(rz) ® u(z) uniformly on |z| = R as 1®r .

We have ( )( )ò

=

÷÷ø

öççè

æ

-

+

p=

R|Z|

0d)z(uaz

azRe

2

1)a(U

This may be expressed as ( )( ) ú

ú

û

ù

êê

ë

é

÷÷ø

öççè

æ

-

+

p= ò

=R|Z|

0d)z(uaz

az

2

1Re)a(U , since t = 0ie , id 0 =

t

dt,

Changing a into z and z into t we get ( )( ) ú

ú

û

ù

êê

ë

é

÷÷ø

öççè

æ

-

+

p= ò

=R|Z|t

dt)t(u

zt

zt

2

1Re)z(u .

The expression inside the bracket is an analytic function in |t| < R and u(z) is the real part

of f(z), where ( )( )

ict

dt)t(u

zt

zt

2

1)z(f

R|t|

+÷÷ø

öççè

æ

-

+

p= ò

=

, c is an arbitrary real constant. This is known

as Schwarz Formula. Definition: If )0(U is harmonic in q£0 £ p2 and piecewise continuous in

q£0 £ p2 , then we define

uP (z) = 0)0(Re2

12

0

0

0

dUze

zei

i

ò ÷÷ø

öççè

æ

-

+p

p and this integral is called Poisson Integral of U.

Note: (i) )z(Pu is not only a function of z but also of U.

(ii) vuvu PPP +=+ )( and cP u = c uP .

8.5 SCHWARZ THEOREM

Statement: The function uP (z) is harmonic for |z| < | and )0()(0 0UzPiez

Limu =

®

only if u is continuous at oq .

Proof: From (1) we see that uP (z) is the real part of an analytic function.

)z(Pu = Re ict

dt)t(u

ze

ze

2

12

0

0i

0i

+úú

û

ù

êê

ë

é

÷÷

ø

ö

çç

è

æ

-

+

p òp

and therefore it is harmonic.



64

Let 21 c,c be two complementary arcs of the unit circle. Let us denote 1U the function

which coincides with U on 1c and vanishes on 2c , by 2U the corresponding function for

2c . Clearly 21 uuu PPP += ---- (1)

Let 2,0)( cezPez

Lim iui

Î"=®

q

q…… (2)

By continuity, suppose u( 00 ) = U, given e > 0, we can find 21 c,c such that 0ie , is an

interior point of 2c and P 2u (z) < 2/e for 0ie Î 2c .

\ 22 /)( eq <U , for all 0 .

)z(Pu2 = t

dtdU

ze

zei

i

qqp

p

ò ÷÷ø

öççè

æ

-

+2

0

20

0

)(Re2

1

But .1|ze|

|z|1

ze

zeRe

20i

2

0i

0i

<-

-÷÷

ø

ö

çç

è

æ

-

+ i.e. multiplying by the conjugate both Nr and Dr as |z| <|

( )( ) 002

1)(

2

0

22 duzPu ò<\p

p( ) 0d|0u|

2

12

0

2 òp

p< ( )|0u| 2 2/|)z(pu| 2 e<Þ ………….(3)

Since 1U is continuous and vanishes at 0ie ,

oa >$ d such that 2/|)(| 1 e<zPu for d<- || 0oiez …………..(4)

|)(||)(||)(| 21 zPzPzP uuu +£\ e=e+e< 2/2/ .

This implies that 0)(00

=®

zPuez

Limi

i.e., )0()( 000uzPu

ez

Limi

=®

. Hence the proof.

8.6 LET US SUM UP

1) Definition of Harmonic function & its basic property 2) The mean value property 3) Poisson formula in various forms 4) Schwarz’s Theorem


1) Show that the mean value formula remains valid for u = | 1+ z |, z = o,

r = 1 and use this fact to compute òp

0

log sin 0 d 0

2) If f(z) is analytic in the whole plane and if 1z- Re f(z) ® o when z ® o, show that f is a constant.

8.8 REFERENCES

1. Complex Analysis by L.V. Alphors.. 2. Foundations of Complex Analysis by Dr. S. Ponnusamy.



65

UNIT - 4 LESSON- 9 POWER SERIES EXPANSION

CONTENTS 9.0 Aim and Objectives 9.1 Introduction 9.2 Weierstrass’s theorem 9.3 Taylor’s series 9.4 Laurent’s series 9.5 Let us Sum up 9.6 Lesson end activities 9.7 References 9.0 AIM AND OBJECTIVES

In this lesson we are going to study about the existence of the limit of a uniformly convergent sequence of analytic functions which is also analytic. We are going to study about the expansion of an analytic function as an infinite series which is valid in a disk in which the function is analytic and the expansion of a function holomorphic in an annular region as an infinite series. After going through this lesson you will be able to: (i) state Weierstrass’s theorem, Hurwitz theorem and Laurent’ (ii) state Taylor’s series, Laurent’s series (iii) solve problems based on Laurent’s series

9.1 INTRODUCTION Until now the readers are vague as to how for an analytic function can be manipulated explicitly. In this lesson we are going to see that the analytic functions can take the form of infinite series, infinite products and other limiting forms. As a preliminary exercise, we establish the existence of the limit of a uniformly convergent sequence of analytic functions which is also analytic 9.2 WEIRSTRASS’S THEOREM

Statement: Suppose that )z(fn is analytic in the region W and that the sequence { )z(fn }

converges to a limit function 'f (z) uniformly on every compact subset of W .Then f(z) is

analytic in W . Moreover )z(fn converges uniformly of 'f (z) on every compact subset of

W . Proof Lt |z-a| < r be a closed disk contained in W n. The regions ¥

=W 1nn }{ form an open

covering of |z-a|<r. Since the disk is compact it has a finite sub covering. This means that it is contained in a

fixed 0nW . If g is any closed curve lying in |z-a| < r, then by Cauchy’s Theorem

( ) 0dzzfn =òg

.



66

Statement: suppose that fn(t) is analytic in the Region Wn and the sequence {fn(t)} coverage’s to a limit function f(z) in a region W, uniformly on every compact subset of W. Then f(t) is analytic in W. Moreover fn

1(z) converges uniformly to f1(t) on every compact subset of W.

Since { nf } converges uniformly on g ( ) dzzfn

Limdzzf

n

Limnn )(0 òò ¥®

==¥®

gg

= ( )dzzfòg

\ ( )dzzfòg

= ( ) 0=¥® ò dzzf

n

Limn

g

. Hence by Morera’s theorem, f(z) is analytic in |z-a| < g .

\ f(z) is analytic in the whole of W . We can prove this theorem using Cauchy’s integral

formula. By Cauchy’s integral formula, duzu

duuf

izf

c

nn ò -

=)(

2

1)(

p where c is the circle

|u-a| = r and |z-a| < r. Since }f{ n converges uniformly,

ò -¥®=

¥®c

nn

zu

duuf

in

Limzf

n

Lim )(

2

1)(

p = du

zu

uf

n

Lim

icò -¥®

)(

2

1

p f(z)

= duzu

uf

n

Lim

icò -¥®

)(

2

1

p this shows that f(z) is analytic in the disk.

By Cauchy’s formula for derivatives, we have ò -=

c

nn

zu

duuf

izf

2

'

)(

)(

2

1)(

p

).(')(

)(

2

1

)(

)(

2

1)(

22

' zfduzu

uf

izu

duuf

n

Lim

izf

n

Lim

cc

nn =

-=

-¥®=

¥® òò pp

This convergence is uniform for |z-a| < r. Since every compact subset of W can be covered by a finite number of such closed discs, the convergence is uniform on every compact subset. If a series with analytic terms f(z) = )z(f1 + )z(f2 + )z(f3 + … converges

uniformly on every compact subset of a region, then the sum f(z) is analytic in W and the series can be differentiated term by term. Hurwitz Theorem

Statement: If all functions nf (z) are analytic and not equal to zero in a region and if fn

(z) converges to f(z) uniformly on every compact subset of W then f(z) is either identically zero or never equal to zero in W . Proof Suppose f(z) is not identically zero. By Weierstrass theorem f(z) is analytic. Its zeros are isolated. For any point 0z = W , there is a real number r such that for |z- 0z | < r, f(z) ¹ 0.

|f(z)| has its minimum on the boundary of this circle. \ there exists m > 0 such that |f(z)| > m on the circle |z- 0z | = r.



67

This is because as f(z) ¹ 0, )x(f

1 is analytic and hence its maximum is attained on the

boundary. \ Minimum of f(z) is attained on the boundary, consequently )x(f

1

n

®)z(f

1

uniformly on c.

Moreover by Weierstrass theorem, )(')(' zfzfn ® uniformly on c. )(

)('

)(

)('

zf

zf

zf

zf

n

n ®Þ

uniformly on c

dzzf

zf

idz

zf

zf

in

Lim

cc n

n

òò =¥®

Þ)(

)(

2

1

)(

)(

2

1 11

pp.

But for each n, ip2

1=ò dz

zf

zf

c n

n

)(

)('The number of zeros of )(zfn enclosed by c.

\ 0dz)z(f

)z('f

i2

1

c

=p ò . Hence the number of zeros of f(z) in c is o, therefore f( 0z ) ¹ 0.

Since 0z is arbitrary, f(z) ¹ 0 in W . Hence the theorem.

9.3 TAYLOR’S SERIES

Statement: If f(z) is analytic in the region containing 0z , then the representation

f(z) = f( 0z ) + ¥+-++- ...)zz(!n

)z(f...)zz(

!1

)z('f n0

0n

00

is valid in the largest open disk of centre 0z contained in W .

Proof: As f(z) is analytic in the region containing 0z , we have

f(z) = f( 0z ) + )()!1(

)(...)('

!1

)(0

11

00

0 zfn

zzzf

zz nn

--

-

-++

-

+ 11

)(

))((!

))(( ++ -+

- non

noo

n

zzzfn

zzzf.

Where )z(f 1n+ = du)zu()zu(

)u(f

i2

1

c

1n0

ò --p +.

Hence c is any circle |z- 0z | = r such that |z- 0z | £ r is contained in W . Now the

theorem is proved if we prove that (z- 0z ) 1n+ )z(f 1n+ ® 0 as n ® ¥ .

Now, | )z(f 1n+ | £ òò --£

--++

c

n

cn

duzz

duMdu

zuzu

uf.||

)|(

||

2||

)()(

)(

2

1

1011

0rrpp

Here M is such that |f(u)| < M on c

| )z(f 1n+ 1n0 )zz( +- | £ M

2

1

ppr

--rr

-+

+

2x|)zz|(

zz

01n

1n0



68

= |)zz|(

zzM

01n

1n0

--rr

-+

+

£ 1n

0

0

|zz|

|zz|1

M+

÷÷ø

öççè

æ

r

-

÷÷ø

öççè

æ

r

--

® 0 as n ® ¥ .

Problems (1) Expand f(z) = ze , stating the region of validity as a power series of z. Answer Let |z| = R be a circle c. Here R is arbitrary. Now ze is analytic within and on c. Hence by Taylor’s theorem,

ze = f(0) + )0('!1

fz

+ )0(''f!2

z2

+ …. + )0(!

)(nn

fn

z+ …. + ¥ . Hence f (0) = 0e = 1.

)0(f n = ze , " n \ )0(f n = 1, " n \ ze = 1+ !1

z +

!2

z2

+ !3

3z …. +

!n

zn

+ …. ¥ . This is valid in

the entire z-plane. (2) Expand f(z) = log (1-z) stating the region of validity Answer f(z) is analytic for all z except z = 1. Let c be |z| = r < 1. Now f(z) is analytic for all z within and on c. Hence by Taylor’s theorem

Log (1-z) = f(0) + )0('!1

fz

+ )0(!2

)(2

nfz

+ … + )0(!

)(nn

fn

z + ... + ¥ .

f(z) = log (1-z) f(0) = log 1 = 0,

)z(f n = n

nn

z

n

)1(

)!1()1()1( 1

-

--- -

)0(f n = - (n-1)!

¥+---=\ ....3

z

2

zz0)z(f

32

This is valid in |z| < |

9.4 LAURENT’S SERIES

A series of the form +++++nn

221

0z

b....

z

b

z

bb ……. (1)

can be considered as an ordinary power series in the variable 1/z. It will therefore converge outside some circle |z| = e except in the extreme case R. The convergence is uniform in every region |z| > r > R. Hence the series represents an

analytic function in the region |z| > R. If the series (1) is combined with an ordinary power series we get more general series of

the form å¥

-¥=n

nn za ------- (2).

It will be convergent if the parts consisting of non-negative powers are separately



69

convergent. Since the first part is convergent in a disc |z| < 2R and the second series is

convergent in a region | 1z | > 1R , there is a common region of convergent only if 1R < 2R

and (2) represents an analytic function in the annulus 1R < |z| < 2R .

Laurent’s Theorem Let f(z) be analytic in a region W containing an annulus 1R < |z| < 2R or more

generally 1R < | z-a | < 2R .

Then f(z) = å¥

-¥=

-n

nn )az(A where du

zu

uf

iA

rau

nn ò=-

+-=

||

1)(

)(

2

1

p where 1R < r < 2R

Proof

z

1c

2c

1R2R

Let z be any point in the annulus. We see that n( 12 cc - , z) = n( 2c , z) – n( 1c , z) = 1

-0 = 1. By Cauchy’s integral formula

n( 12 cc - , z) f(z) = duzu

)u(f

i2

1

12 ccò-

-p

i.e., f(z) = duzu

)u(f

i2

1

12 ccò-

-p = du

zu

)u(f

i2

1

2cò -p

- duzu

)u(f

i2

1

1cò -p

= )z(f1 + )z(f2

Where )z(f1 = duzu

)u(f

i2

1

2cò -p

and )z(f2 = duzu

uf

icò -

-

2

)(

2

1

p.

Now )z(f1 is an analytic function in |z-a|< 2R .

Hence it has a power series expansion in |z-a| < 2R as

)z(f1 = å¥

=

-0n

nn )az(A , in |z-a| < r < 2R .

Where duau

uf

iA

rau

nn ò=-

+-=

||

1)(

)(

2

1

p …….. (1)

Now, let us get the development of )z(f2 .

In )z(f2 , we make the transformation

u = a + 'u

1, z = a +

'

1

z



70

This transformation caries |u-a| = r into

| 'u | = r

1 with negative orientation

\ ''

1'

'

1

'

1

'

1

2

1

'

12

1'||

2 duu

u

zu

uaf

izaf

ru

÷ø

öçè

æ -

÷ø

öçè

æ-

÷ø

öçè

æ+

=÷ø

öçè

æ+ ò

=p

= 'du)'u'z(

'u

1af

'u

'z

i2

1

r

1|'u|

ò=

-

÷ø

öçè

æ+

p.

)z(f2 considered as a function of 'z has a power series expansion given by

÷ø

öçè

æ+

!2

1af = å

¥

=1k

ikkzB

Where kB = 'du'u

'u

1af

i2

1

r

1|'u|

1kò=

+

÷ø

öçè

æ+

p =

( )du

au

uf

iru

kò=

+--'||

1)(2

1

p

Putting k = -n nB = ( )

n

r|'u|

1nAdu

)au(

uf

i2

1=

-p ò=

+

Hence å¥

=1k

kk zB becomes )z(f2 = å

¥

=

-

1n

nn 'zB

\ f(z) = å¥

=

-1n

nn )az(A + å

¥

-=

-1n

nn )az(A

= å¥

=

-1n

nn )az(A .

Here all the co-efficient An are determined by (1). Hence the theorem Problems

Prove that the Laurent’s development is unique. Answer

Let f(z) = å¥

-¥=

-n

nn )az(A …… (1)

Let us now prove that (1) is identical with the Laurent’s expansion

f(z) = å¥

-¥=

-n

nn )az(a ……. (2)

Where na = ( )

ò +-p0c

1n)au(

uf

i2

1

The result will be proved if we prove that nn aA = .



71

The equation of the circle 0c is |u-a| = r where 1R < r < 2R . \ u – a = qire .

Now, na = ò å¥

-¥=+-

-0

1)()(

2

1

c mn

mm

au

duauA

ip

= ò å¥

-¥=

---p

0c m

1nmm du)au(A

i2

1

= òå¥

----¥

-¥= 0

)1(1

2

1q

pqq deirerA

iinminm

mm

= ò òå --¥

-¥=0

2

0

)(

2

1

c

nminmm

m

derAi

qp

pq

If m ¹ n, qòp

q- de

2

0

)nm(i = p

q-

÷÷

ø

ö

çç

è

æ

-

2

0

)nm(i

)nm(i

e = 0 .

If m = n, qòp

q- de

2

0

)nm(i = p=qòp

2de

2

0

0

\ na = nnn

n A2rA2

1=p

p- . Hence Laurent’s Development is unique.

(2) If 0 < |z-1| < 2, then express, F(z) = )3z)(1z(

z

-- in a series of positive and negative

powers of (z-1). Answer

f(z) = )3z)(1z(

z

-- =

1z

A

- + ,

3z

B

- using partial fractions we get A =

2

1-, B =

2

3

\f(z) = 2

1

3z

1

2

3

1z

1

-+

- =

2

1

2)1z(

1

2

3

1z

1

--+

-

= 2

1

2

11

1

2

1

2

3

1

1

--

÷ø

öçè

æ

-+

- zz =

2

1-1

2

)1z(1

4

3

1z

1-

÷ø

öçè

æ ---

-

= 2

1-

÷÷

ø

ö

çç

è

æ¥+÷

ø

öçè

æ -+÷

ø

öçè

æ -+

-+-

-....

2

1z

2

1z

2

1z1

4

3

1z

132

is the Laurent’s Series.

(3) Expand 2z3z

12 +-

in 1 < | z | < 2.

Answer

When 1 < | z | < 2, 12

,1||

1<<

z

z

2z3z

12 +-

= )2z)(1z(

1

-- =

1z

1

2z

1

--

- =

2

1-

÷ø

öçè

æ-

-

-2

11z

1

2

z1

1



72

= 2

1-11

2

11

z

1

2

z1

--

÷ø

öçè

æ--÷

ø

öçè

æ- =

2

1-úû

ùêë

é++++ ....

2221

2

2

n

nzzz -

z

1úû

ùêë

é++++ ....

2

1

2

1

2

11

32 is the

Laurent’s series. 9.5 LET US SUM UP

(1) The Statement of Weierstrass theorem (2) The statement of Hurwitz theorem (3) Technique of Expansion of analytic function in a simply connected region and in annular region .


(1) Expand f(z) = ze about z = 1 (2) Expand f(z) = log (1 + z) and state region of convergence (3) Expand sin z as a Taylor’s series about z = 0.

9.7 REFERENCES 1. Complex Analysis by L.V. Alphors.. 2. Foundations of Complex Analysis by Dr. S. Ponnusamy.



73

LESSON – 10 PARTIAL FRACTIONS AND FACTORIZATION CONTENTS 10.0 Aim and Objectives 10.1 Introduction 10.2 Partial Fractions 10.3 Infinite Products 10.4 Canonical Products 10.5 Gamma Functions 10.6 Stirling’s Formula 10.7 Let us Sum up 10.8 Lesson-end Activities. 10.9 References

10.0 AIM AND OBJECTVES The main objective of this lesson is to learn the method of representing

meromorphic functions with an infinite sequence { }¥

=1nnb with ¥=¥® nLtn b .We also study

about infinite products, and representation of entire function as an infinite product. We also gain knowledge about Gamma function which is meromorphic and non-vanishing After going through this lesson you will be able to: (i) define canonical product, canonical product – Genus, partial fractions and infinite

products (ii) Gamma functions and its properties and Stirling’s formula 10.1 INTRODUCTION

A complex function is meromorphic if it is analytic expcept for its poles. Corresponding to each pole bv,the function f(z) contains the negative power of z – bv in its Laurent development. If a function has finite number of poles, it is a rational function. The problem is somewhat difficult if the number of poles is infinite,. However it is true such functions can always find limit point. We also discuss about the finite products

Õ¥

1nP in this lesson.

10.2 PARTIAL FRACTIONS

Mittag – Leffler’s Theorem: Let { }nb be a sequence of complex number with

¥=¥® nLimn b and )(zpn y be a polynomial with out constant term. Then there are functions

which are meromorphic in the whole plane with its poles at the points nb and the

corresponding singular parts np ÷÷ø

öççè

æ

- nbz

1Moreover, the general meromorphic function of

this kind can be written in the form )()(1

)( zgzpbz

pzfn

n

n

n +÷÷

ø

ö

çç

è

æ-÷÷

ø

öççè

æ

-= å , where np (z)

are suitably chosen polynomial and g(z) is analytic in the whole plane.



74

Proof: Since ¥=¥® nLimn b , obn ¹ after a certain stage \We may assume that no bn is

zero.

The function ÷÷ø

öççè

æ

- n

nbz

p1

= s

ss zaå

¥

=0

11

1210 1

... ++

- +++++=-

n

n

n

n

n

n

mm

mm

mm zkzazazaa

Let np (z) = s

ss zaå

¥

=0

.The difference nP - np can be expanded.

Let nM = Max þýü

îíì

=÷÷ø

öççè

æ

-||

2

1||:

1n

n

n bzbz

p .

Let |)(||| 1

1+

+£- nmnnn zzfpP

( )ò=

+

+ -

÷÷ø

öççè

æ

-=

||||

1

1

21

.

1

2

1

n

n

n

b

m

m

n zdz

bpn

ix

xxx

x

p|||

2

1|||

2

12

11

1

zbb

zM

in

m

n

mn

n

n

-÷ø

öçè

æ£

+

+

p ò= ||||

21

||nb

dzx

||4

1||

2

1|||

2

12

11

1

nn

m

n

mn

bbb

zM

i n

n

-÷ø

öçè

æ£

+

+

p þýü

îíì

£ ||4

1|| nbzSince

= 2

1

||

||2

1

11

C+

++

nn

mm

nmb

zM

nn

1

||22

+

£nm

n

nb

zM

1

||

|4

|22

+

ïþ

ïý

ü

ïî

ïí

ì

£

nm

n

n

nb

zb

Mnn Mn

Mn MM

22

21

==+

------ (1)

If we choose nM so large that n

m

n

n

M2

2£ ,we have n

nn pP -£- 2||

As the absolute values of the terms of the series in (1) are less than the terms of a

geometric series of common ratio 21 , the series converges absolutely, on |z| <

4

|| nb

except at poles. As ¥®|| nb , the series converges uniformly in that disk provided, we

omit the terms with Rbn £|| .

By Weierstrass theorem, the remaining series is analytic in Rz £|| . If follows that the

full series is meromorphic in the whole plane with singular parts ÷÷ø

öççè

æ

- n

nbz

P1

.

Let h (z) = þýü

îíì

-÷÷ø

öççè

æ

-å

¥

=

)(1

1

zpbz

p n

n

nn

. If f(z) has the same poles and principal part as h

(z), then the function defined by g (z) = f (z)- h (z) is analytic in the whole complex plane ie; f (z)= h (z) + g (z) where g (z) is an entire function.



75

Problem

Using Mittag –Leffler’s Theorem, show that( )22

2 1

sin nzz n -= å

¥

-¥=p

p

Deduce that p cot zp = å¥

= -+

122

21

n nz

z

z

Proof

Let f(z) = zp

p2

2

sin, f(z) is meromorphic in the whole plane with poles of order z

at each integer n.

By Laurent’s expansion, the principal part at z = 0 is 2

1

z.

Since sin 2 p (z-n) = sin 2 p z, the principal part at z = n is 2)(

1

nz -.

Now å¥

-¥= -n nz 2)(

1 is convergent for every z ¹ n. By limit comparison test, compare it

with å¥

12

1

n moreover if K is any compact set then Kc {z/|z|} £ R

|z – n| = |n – 2| ³ n - |2| > n - R 22 )(

1

||

1

Rnnz -£

-.

\ The series å¥

-12)(

1

nz converges uniformly except at finite number of points.

Hence å¥

¥- - 2)(

1

nz converges uniformly in K after omitting the terms which becomes

infinity on the set.

Thus å¥

¥- - 2)(

1

nz is an entire meromorphic function with poles at z = + n and the

principal part at z = n is 2)(

1

nz -.

Hence by Mittag – Leffler’s Theorem, f(z) = zp

p2

2

sin = å

¥

¥- - 2)(

1

nz + g(z) where g(z) is

an entire function. Now we show that g(z) = 0.

The function zp

p2

2

sin and å

¥

¥- - 2)(

1

nz are both periodic function with period 1.

The function g(z) is also periodic with period 1.

Now, 22 sinsin zz pp = =

2)(sin yix +p =

2sincoscoshsin yhxiyx pppp +

= yhxyhx pppp 2222 sincoscossin +

= )1(coscoscos)cos1( 2222 -+- yhxyhxh pppp



76

= xyxyxy pppppp 222222 coscoshcoscoshcoshcosh -+- = xyh pp 22 coscos -

\zp

p2

2

sin0® uniformly as |y| ¥® , since yp2cosh ¥® .

Hence g(z) = f(z) – h(z) 0® uniformly as |y| ¥® Þ g(z) is bounded in a period strip 0 £ x £ 1. Since g(z) is periodic function with period 1, it is bounded in the whole plane. \ By Liouville’s Theorem, g(z) is a constant.

Since ¥®|| y

Ltg(x + i y) = 0, g (z) = 0.

Hence zp

p2

2

sin = å

¥

¥- - 2)(

1

nz. Now,

zp

p2

2

sin = å

¥

¥- - 2)(

1

nz…… (2)

Where the prime to the summation indicates that we have omitted the term corresponding

to n = 0 we see that the L.H.S is the derivative of zz

pp cos1

- .

Let f(z) = zz

pp cos1

- . This has a removable singularity at z = 0.

We define f (o) = 0. Since the R.H.S. of (2) is uniformly convergent on every compact subset, term by term integration of the series is allowed. Integrating along the path from o to w.

F(w) – F(o) = å¹

÷ø

öçè

æ-

-

-

0

11

n nnz

Replacing w by z, zz

pp cos1

- = å¹

÷ø

öçè

æ-

--

0

11

n nnz

\ zcosp = +z

1å

¹ -0

1

n nz

þýü

îíì

=å¹

01

0n nQ

= +z

1å

¥

÷ø

öçè

æ

+-

-1

11

nznz = å

¥

-+

122

21

nz

z

z .

10.3 INFINITE PRODUCTS

An infinite product å¥

=

=1

21 ......n

nn pppp is said to converge, if the sequence }{ np

of partial product nn pppp ...... 21= converges. We say that the infinite product converges

to p = npn

Lim

¥®, if this limit exists and is different from zero.

We omit the value zero since any infinite product with one factor zero would converge. We follow this convention that the infinite product is said to converge if and only if atmost finite number of the factors are zero and if the partial products formed by the non vanishing factors tend to finite limit which is different from zero.



77

In a convergent series we have npn

Lim

¥® =

1-¥® n

n

p

p

n

Lim = 1, the zero factors being

omitted. Usually we write an infinite product in the form Õ¥

=

+1

)1(n

na so that na ® 0 is a

necessary condition for the convergence. Theorem 1

The infinite product Õ¥

=

+1

)1(n

na with 1 + na ¹ 0 converges simultaneously with

the series å¥

+1

)1log( na whose terms represent the values of the principal branch of the

logarithm. Proof Let us first prove the sufficiency part Let nP = )1( 1a+ )1( 2a+ … )1( na+ , nS = log )1( 1a+ + log )1( 2a+ + … + log )1( na+

= log )1( 1a+ )1( 2a+ … )1( na+ = log nP , \ nP = nse .

Assuming that å log )1( na+ is convergent, we have ¥®n

Lim nS = S, a finite number

\ npn

Lim

¥® = nse

n

Lim

¥® = se = a finite non – zero number.

Hence Õ + )1( na is convergent. Let us now prove the necessary part. Suppose

Õ + )1( na is convergent and converges to a non – zero finite number P.

i.e., npn

Lim

¥® = p ¹ 0, then nP = )1( 1a+ )1( 2a+ … )1( na+

log nP = log )1( 1a+ + log )1( 2a+ + … + log )1( na+ + nihp2

i.e. log nP = nS + nihp2 , where nh is an integer for large n. We show that nh is a

constant.

Let na be the imaginary part of log )1( na+ and nb be the imaginary part of

log nP respectively. Equating imaginary parts on both sides. nb = 1a + 2a + … + na +

nhp2 .

Also 1+nb = 1a + 2a + … + na + 1+na + 12 +nhp ( 1+nb - nb ) = 1+na + )(2 1 nn hh -+p .

Since p (1 + an) is convergent, ¥®n

Lim )1( na+ = 1

i.e. ¥®n

Lim log )1( na+ = log 1 = 0 Þ

¥®n

Lim imaginary part of log )1( na+ = 0

\¥®n

Lim na = 0 =

¥®n

Lim 1+na .



78

Let ¥®n

Lim1+nb = b =

¥®n

Limnb = p2

¥®n

Lim )( 1 nn hh -+ .

i.e. b - b = p2¥®n

Lim )( 1 nn hh -+ = 0.

As nh is an integer, nh = h = a constant for large n.

\¥®n

LimnS =

¥®n

Lim log nP - p2 ih Þ å

¥

+1

)1( na = log P - p2 ih, which is finiteÞ

å¥

+1

)1log( na is convergent.

Note: If an are non – negative real numbers, then Õ + )1( na converges if and only if an

converges. Theorem 2 A necessary and sufficient condition for the absolute convergence of the product

å¥

+1

1 na is the convergence of the series å¥

1na .

Proof

na+1 £ 1 + na £ an

e

na+1 21 a+ … na+1 £ naaae

+++ .....21

Þ .||...|||||)1log(|1

21 nnn

n

k ASaaaa £Þ+++£+å Also

( ) ( )( ) ( ) nSnnn eAaaaaa £Þ+++£++ ||1....||1||1||...|| 211

\ nsnn eAS ££ . Thus Lim nA and Lim nS are finite or infinite together.

Problem

Prove that for |z| <z-

=++++1

1)........21)(21)(21)(21(,1 842

Answer

nP = ( )1242 1)........1)(1)(1(-

++++n

zzzz

= 12...4212 ...1

-+++++++n

zzz = 122 ...1

-

++++n

zzz = z-1

1 as n ® ¥ .

Problem

Prove that Õ¥ -

÷ø

öçè

æ+

1

1 n

z

en

z converge absolutely and uniformly on every compact set.



79

Answer

n

z

en

z-

÷ø

öçè

æ+1 is an entire function

n

z

en

z-

÷ø

öçè

æ+1 = ÷÷

ø

öççè

æ+-+-÷

ø

öçè

æ+ ...

6211

3

3

2

2

n

z

n

z

n

z

n

z

= 1 - 3

3

2

2

32 n

z

n

z+ +….. = 1-

2

2

n

z÷ø

öçè

æ+- .....

32

1

n

z

= 1 - 2

2

n

z g(z) = 1+ na (z) where g(z) is an entire function. Let |z| £ R and

M = Rz £||

max|g(z)|, and | na (z)| £

2

2

n

MR

We see that na (z) converges absolutely Þ (1 + na (z)) converges absolutely in |z| £ R.

10.4 CANONICAL PRODUCTS

In this section we construct an entire function with zeros of prescribed orders at prescribed points. A function which is analytic in the whole plane is said to be entire or

integral. The simplest entire functions which are not polynomials are ze , sinz and cosz.

If g(z) – is an entire function, then f(z) = )( zge is also entire and not equal to zero. Conversely, if f(z) is an entire function which is never zero, then f(z) must be of the form

)( zge . To prove this, we note that the function

)(

)('

zf

zfis analytic in the whole plane since f(z) ¹ 0.

\)(

)('

zf

zfis the derivative of an entire function F(z).

\ =)(zfdz

d

)(

)('

zf

zf

Integrating between the limits 0z to z, we get =- )()( 0zFzF log f(z) – log f )( 0z

log f(z) = F (z) – F( 0z ) + log f )( 0z

f(z) = )()(log)()( 00 zgzfzFzF ee =+- . Since f(z) entire and F( 0z ) and log f( 0z ) are constants, it

follows that F(z) – F( 0z ) + log f( 0z ) = g(z) is also entire. \ we have f(z) = )( zge , g(z) is

entire. Convergence producing factors Multiplication of infinite products by certain factors will help the product to converge Such factors are called convergence producing factors. Theorem on Canonical Product

There exists an entire function with arbitrarily prescribed zeros na provided that,

in the case of infinitely many zeros, na ¥® . Every entire function with these and no

other zeros can be written in the form.



80

( )

nm

nnnn a

z

ma

z

a

z

n n

zgm ea

zezzf

÷÷ø

öççè

æ++÷

÷ø

öççè

æ+¥

=Õ ÷÷

ø

öççè

æ-=

1...

2

1

1

2

1)(

Where the product is taken overall na ¹ o, the nm are certain integers and g(z) is an entire

function. Proof Assume that f(z) is an entire function having m zeros at the origin (m may be

zero) and denote the other zeros at 1a , 2a , 3a ,…. na , multiple zeros being repeated. Then

we can write

Õ ÷÷ø

öççè

æ-=

N

n

zgm

a

zezzf

1

)( 1)(

If there are infinitely many zeros we obtain a similar representation

Õ¥

÷÷ø

öççè

æ-=

1

)( 1)(n

zgm

a

zezzf …….(1), provided that the infinite product converges

uniformly on every compact set. If this is convergent the product represents an entire function with zero at the same points and with the same multiplies as f(z). It follows that the quotient can be written in the

form )( zgmez .The infinite products Õ¥

÷÷ø

öççè

æ-

1

1na

zconverages absolutely if and only if

å¥

1 || na

z converges and in this case the convergence is also uniform in every closed disk

|z| £ R. In this case the representation (1) holds, In the general case convergence producing factors must be introduced

We consider an arbitrary sequence of complex numbers na ¹ o with Ltn ¥® na ¥= and

prove the existence of the polynomials np (z) such that Õ¥

÷÷ø

öççè

æ-

1

1na

z )( zPne converges to an

entire function. The product converges if and only if the series

)(zrn = log ÷÷ø

öççè

æ-

na

z1 + nP (z) converges. Here the branch of the logarithm shall be chosen

so that the imaginary part of nr (z) lies between - p andp .

For a given R, we consider only terms with | na | > R. In the disk |z| £ R, the principal

branch of log ÷÷ø

öççè

æ-

na

z1 can be developed in a Taylor’s series.

log ÷÷ø

öççè

æ-

na

z1 = ....

3

1

2

132

÷÷ø

öççè

æ-÷÷

ø

öççè

æ-

-

nnn a

z

a

z

a

z

we choose np (z) as



81

np (z) = nm

nnnnn a

z

ma

z

a

z

a

z÷÷ø

öççè

æ+÷÷

ø

öççè

æ+÷÷

ø

öççè

æ+

1...

3

1

2

132

…

Then,

nr (z) = -÷÷ø

öççè

æ

+-÷÷

ø

öççè

æ

+

-++ 21

2

1

1

1nn m

nn

m

nn a

z

ma

z

m…

\ | nr (z)| £ +÷÷

ø

ö

çç

è

æ

++÷

÷

ø

ö

çç

è

æ

+

++ 21

2

1

1

1nn m

nn

m

nn a

z

ma

z

m…

£ïþ

ïýü

ïî

ïíì

+++÷÷

ø

ö

çç

è

æ

+

+

...||||

1||

1

12

21

nn

m

nn a

z

a

z

a

z

m

n

= 2

11

1

111

<ïþ

ïýü

ïî

ïíì

-÷÷

ø

ö

çç

è

æ

+

-+

nn

m

nn a

Ras

a

R

a

R

m

n

, as na ® ¥ .

Now the series

1

1 1

1+

¥

=å

úúû

ù

êêë

é

+

nm

n nn a

R

m can be made convergent by choosing

nM = N. For, then as na ® ¥ 2

1<

na

R can be made to hold good for are large n so that

1

1

1+

úúû

ù

êêë

é

+

n

na

R

n<

1

2

1

1

1+

úû

ùêë

é

+

n

n. Since the geometric series å +12

1n

is convergent.

Thus the thn term converges and hence nr (z) ® 0 and hence nr (z) has an imaginary part

between - p and p and as soon as n is sufficiently large.

Be Weierstrass’s M – Test å )(zrn converges absolutely and uniformly on |z| £ R.

Thus m )(

1

1 zp

n

nea

zÕ

¥

÷÷ø

öççè

æ- is convergent and represents an analytic function in |z| £ R.

It is for the sake of this reasoning we have excluded the values na £ R.

But the uniform convergence of f(z) is not affected when the corresponding factors are again taken in to account. If f(z) is an arbitrary entire function with prescribed zeros then

)(

1

1

)(

zp

n

m nea

zz

zf

Õ¥

÷÷ø

öççè

æ-

= g(z) an entire function.

Thus we obtain f(z) = mz )( zge

nm

nnnn a

z

ma

z

a

z

n

ea

z ÷÷ø

öççè

æ++÷÷

ø

öççè

æ+¥

Õ ÷÷ø

öççè

æ-

1...

2

1

1

2

1 .



82

Corollary Every function which is meromorphic in the whole plane is the quotient of two entire functions. Proof Suppose F(z) is an entire meromorphic function. If f(z) has only a finite number of poles naaa ....., 21 (counted with multiplication), then

F(z) = )(

)(

zQ

zp+ g(z), by subtracting the principal part at ( naaa ....., 21 ), where p and q are

polynomials and g(z) is an entire function.

Þ f(z) = )(

)()()(

zQ

zQzgzp + quotient of 2 entire functions.

If F(z) has an infinite number of poles naaa ....., 21 , then by Weierstrass’s theorem, there

exists an entire function g(z) whose zeros are the poles of F(z) and hence F(z) g(z) is an

entire function F(z). This implies that F(z) = )(

)(

zg

zf.

Canonical Product – Genus In the representation of an entire function as

f(z) =

nm

nnnn a

z

ma

z

a

z

n

zgm ea

zez

÷÷ø

öççè

æ++÷÷

ø

öççè

æ+¥

Õ ÷÷ø

öççè

æ-

1...

2

1

1

)(

2

1 we put nm = h " n.

The product

h

nnn a

z

ha

z

a

z

n

ea

z ÷÷ø

öççè

æ++÷

÷ø

öççè

æ+¥

Õ ÷÷ø

öççè

æ-

1...

2

1

1

2

1 converges and represents an entire function

provided that the series å¥

=

+

+

÷÷

ø

ö

çç

è

æ

1

1

1n

h

n

h

a

R

converges for all R. i.e. provided ¥<å¥

=+

11

1

nh

na. If

h is the smallest integer for which the series converges then the expansion

å¥

=

÷÷ø

öççè

æ++÷÷

ø

öççè

æ+

÷÷ø

öççè

æ-

1

1...

2

12

1n

a

z

ma

z

a

z

n

nm

nnnnea

z is called the canonical product associated with the

sequence { na } and h is the genus of the canonical product.



83

Genus of an entire function f(z)

If in the representation, f(z) =

h

nnn a

z

ha

z

a

z

n n

mzg ea

zze

÷÷ø

öççè

æ++÷

÷ø

öççè

æ+¥

=Õ ÷÷

ø

öççè

æ-

1...

2

1

1

)(

2

1.

g(z) reduces to a polynomial, the function f(z) is said to be finite genus and the genus of the function f(z) is equal to the degree of this polynomial or the genus of the canonical product which ever is larger.

For example, an entire function of genus 0 is of the form Õ¥

÷÷ø

öççè

æ-

1

1n

m

a

zcz with £

na

1 < ¥ .

The canonical representation of an entire function of genus 1 is either of the form

Õ¥

÷÷ø

öççè

æ-

1

1 na

z

n

zm ea

zeCz a with å 2

1

na < ¥ ,

å 2

1

na = ¥ or of the form Õ

¥

÷÷ø

öççè

æ-

1

1n

zm

a

zeCz a with å

na

1 < ¥ , a ¹ 0.

Canonical Representation of Sin z

The zeros of sin z are the integers of z = ± n since ån

1 diverges and å 2

1

n

converges, we take h = 1.

Hence we obtain a representation of the form. Sin p z = z na

z

zg en

z

ne ÷

ø

öçè

æ-

¹

Õ1

0)( .

Let us now find g(z). Taking logarithms and differentiating,

å¹

÷ø

öçè

æ+

-++=

0

11)('

1

sin

cos

n nnzzg

zz

z

p

pp

Õ cot zp = å¹

÷ø

öçè

æ+

-++

0

11)('

1

n nnzzg

z

But we have the result p cot zp = å¹

÷ø

öçè

æ+

-+

0

111

n nnzz.

Hence we get )(' zg = 0. \ g(z) is a constant. Also we get 0®z

Lim

z

zpsin =

0®z

Lim )( zge =

0¹

Õ

h÷ø

öçè

æ-

n

z1 n

z

e , p = )( zge , \ )( zge = p .

Thus sin zp = zp0¹

Õ

h÷ø

öçè

æ-

n

z1 n

z

e , combining the factors ÷ø

öçè

æ-

n

z1 n

z

e and ÷ø

öçè

æ+

n

z1 n

z

e-

,

we get sin zp = zp Õ¥

=÷÷ø

öççè

æ-

12

2

1n n

z.

Remark We see that sin z is an entire function of genus 1.



84

10.5 GAMMA FUNCTIONS We have seen that the function sin z has all the integers as its zeros. The simplest function having the negative integers as its zeros is the canonical product G(z) =

Õ¥ -

÷÷ø

öççè

æ+

1

1 n

z

n

ea

z. We see that G(-z) has the +ve integers for zeros.

Now, G(z) G(-z) = Õ¥

÷ø

öçè

æ-÷

ø

öçè

æ+

1

11n

z

n

z = Õ

¥

÷÷ø

öççè

æ-

12

2

1n

z =

z

z

p

psin Þ z G(z) G(-z) =

p

p zsin.

The function G(z) = Õ¥

÷ø

öçè

æ+

1

1n

z n

z

e-

has for its zeros the negative integers together with

zero i.e. G(z-1) has for its zeros the zeros of G(z) and the origin. Hence we can write G(z-

1) = z )( zen G(z), where )(zn is an entire function.

Let us now find )(zn . We have Õ¥

÷ø

öçè

æ+

1

1n

z n

z

e-

= z )( zen Õ¥

÷ø

öçè

æ+

1

1n

z n

z

e-

.

Taking logarithmic differentiation, åå¥

=

¥

=úû

ùêë

é-

+++=úû

ùêë

é-

-+ 11

11)('

11

1

1

nn nnzz

znnzn .

But L.H.S = å¥

=úû

ùêë

é-

-+1

1

1

1

n nnz = å

¥

=úû

ùêë

é

+-

++÷

ø

öçè

æ-

1 1

111

1

n nnzz

= å¥

=úû

ùêë

é-

++-

1

111

1

n nnzz + å

¥

=úû

ùêë

é

+-

1 1

11

n nn

= å¥

=úû

ùêë

é-

++

1

111

n nnzz

i.e. )(' zn = - 1 + å¥

=úû

ùêë

é

+-

1 1

11

n nn = 1 + 1 = 0. i.e. )(zn is a constant which we denote

byn .

Hence G(z – 1) = )( zen G(z) and H (z – 1) = )1( -zen G(z-1)

The function G (z) defined by G (z) = )(

1

zzH is called Euler’s Gamma function.

Where H(z) = )( zen G(z). To find the Value of r (Euler’s Constant)

Taking z = 1 in G(z – 1) = )( zen G(z)

G (0) = ne G(1).

1 = ne Õ¥

÷ø

öçè

æ+

1

11

nhe1-

= n-e = Õ¥

÷ø

öçè

æ+

1

11

nhe1-

.



85

Then thn partial product is Õ¥

=-+

1

11

ne

n

n (n + 1)

úû

ùêë

é+++-

ne

1...

2

11

We get, n = ¥®n

Limúû

ùêë

é-÷

ø

öçè

æ++++ )log(

1...

3

1

2

11 n

n

= ¥®n

Limúû

ùêë

é-÷

ø

öçè

æ++++ n

nlog

1...

3

1

2

11 .

The constant n is called Euler’s constant. Its approximate value is 0.57722. Some Properties of Gamma Function

G (z) = )(

1

zzH, G (z - 1) =

)1()1(

1

-- zHz =

)()1(

12 zHz -

= )1(

)(

-

G

z

z

\ (z – 1) G ( z – 1) = G (z) Þ G (z) = )(

1

zzH =

)(

11

zGez zn

=

Õ¥

-

-

÷ø

öçè

æ+

1

/2

)(

1

1

n

z

en

zz

e n

= z

e z )(n-

Õ¥

-÷ø

öçè

æ+

1

/21 nen

z.

We observe that n (z) is a meromorphic function with poles at z = 0, -1, -2, … but without zeros.

We have zG (z) G (-z) = p

p zsin =

)()(

1

zGzzG - =

zp

p

sin

)()(

1

zGezGze zz -gg =

zp

p

sin

)()(2

1

zHzH - =

zp

p

sin

G (z) (-z) G (-z) = zp

p

sin

G (z) G (-z) = zp

p

sin.

We have G (z) = z

e zg-

Õ¥ -

÷ø

öçè

æ+

1

/

1

1 nzen

z

G (z) = 1

ze g-

Õ¥ -

÷ø

öçè

æ+

1

/1

1

1 nen

z

= úúû

ù

êêë

é÷ø

öçè

æ+Õ

¥-

-

1

/1

11

1 nen ú

úû

ù

êêë

é÷ø

öçè

æ+Õ

¥ -

1

/1

11

1 nen

\ G (2) = 1 G (1) = 1 G (3) = 2 G (2) = 2.1 = 2.1=2. Similarly G (4) = 3! …… G (n) = (n – 1).



86

In G (z) G (1 – z) = zp

p

sin, if we put z = ½, we get

{ G 2)}2/1( = zp

p

sin = p

\ G (1/2) = p . Problem

Prove that p G (2z) = 22.1 G (2) G ÷ø

öçè

æ+

2

12

Proof

G (z) = z

e zg-

Õ¥ -

÷ø

öçè

æ+

1

/

1

1 nzen

z.

Taking logarithmic differentiation.

)(

)('

z

z

G

G = å

¥

=

÷ø

öçè

æ-

+---

1

11

2

1

n nzng .

Again differentiating with respect to z,

)(

)('

z

z

dz

d

G

G = å

¥

= ++-

022 )(

11

n znz = å

¥

= +02)(

1

n zn

÷ø

öçè

æ+G

÷ø

öçè

æ+G

2

1

2

1'

z

z

dz

d = å

¥

=

úúúú

û

ù

êêêê

ë

é

++0

2

2

1

1

n zn

i.e. )(

)('

z

z

dz

d

G

G +

÷ø

öçè

æ+G

÷ø

öçè

æ+G

2

1

2

1'

z

z

dz

d = å

¥

= +02)(

1

n zn + 4 å

¥

= ++02)12(

1

n zzn

= 4 å¥

=úû

ùêë

é

++0

2

1

n zn+ 4å

¥

= ++02)12(

1

n zzn = 4å

¥

= +02)(

1

n mzz = 4

)2(

)2('

z

z

dz

d

G

G.

Integrating,

)(

)('

z

z

G

G +

÷ø

öçè

æ+G

÷ø

öçè

æ+G

2

1

2

1'

z

z

= 2)2(

)2('

z

z

G

G + a.

Again integrating, log G (z) + log G (z + 1/2) = log G (2z) + az + b log G (z) (z + ½) = log G (2z) + az + b

Þ G (z) G (z + ½) = log G (2z) aze + b

Z = 1 Þ G (1) G (3/2) = G (2) bae +

2

1G ÷

ø

öçè

æ

2

1 = bae +



87

bae + = 2

p ……….. (1)

Z = 2

1 Þ G ÷

ø

öçè

æ

2

1G (1) = G (1)

ba

e+

2

p = bae +2/ ……….. (2) (1) Þ a + b = log 2 + ½ log p (2) Þ a/2 + b = ½ log p a/2 = - log 2 \ a = - 2 log 2 and b = log 2 + ½ log p .

Hence (z) G (z + 1/2 ) = )log2/12log)2log2( p++- ze (2z)

= ze 22log - p2loge G (2z) = z22

)(2 pGG (2z)

G p G (2z) = 122 -z G (z) G ÷ø

öçè

æ+

2

1z .

10.6 STIRLING’S FORMULA

Consider the second derivative of log G (z)

i.e. dz

d÷÷ø

öççè

æ

G

G

)(

)('

z

z = å

¥

= +02)(

1

n nz =

2

1

z +

2)1(

1

+z+ … +

2)(

1

nz +.

Introduce the function )(xf = 2)(

cot

x

pxp

+z, with residues

2)(

1

n+z at the integral points v.

Here x is the variable and z = x + iy; x > 0, a parameter. We apply the residue formula to

the rectangle whose vertical sides lie on x = 0 and x = n + ½ and with horizontal sides

h = ± y. Denote this contour by K, passes through the pole at 0.

We obtain pr.n ip2

1 ò

k

dxxf )( = å= +

+- n

v vzz 022 )(

1

2

1. Since the factor 2)/(1 x+z ® 0.

The corresponding integrals have the limit zero, on each line x = n + ½, cot px is

bounded and because of the periodicity the bound is independent of n. \ ò+=

+2

12||

nz

d

xx

h

can be evaluated for on the line of integration x = 2n + 1 - x .

We obtain by using residues

0212

2

)12)((

1

||

12

®++

=+-++

=+ òò xnznz

d

iz

d

i

p

xx

x

x

x as ¥®n .

Writing cot hph = 1 + 1

22 -phe

We obtain hh

h

p

h

h

h ph

phde

z

z

e

d

z òò¥¥

--+

-=

-+ 0

2

222

22

0222

)11log()(

1

1

2.

Integrate with respect to z, and obtain



88

log G (z) = hhp ph

dez

zzzczc ò

¥

÷ø

öçè

æ

-++÷

ø

öçè

æ-++

0222 1

1log

1log

2

1'

We can find the constants c and 'c .

Denote J(z) = hhp ph

dez

zò¥

-+0222 1

1log

1

J(z) ® 0 as z ¥® provided that ‘z’ keeps away from the imaginary axis. One can see

that c = -1 and 1c = 2

1log p2 .

Thus log G (z) = 2

1log p2 - z + (z – ½ ) log z + J(z) or G (z) = p2 2/1-zz ze- )( zJe is

the required Stirling’s formula.

10.7 LET US SUM UP 1. Techniques of expanding general meromorphic function using Mittag Lefflers theorem 2. Necessary and sufficient condition for the convergence of infinite products 3. Weierstars theorem on canonical products 4. Canonical product of an entire function and its genus 5. Gamma function and its properties 6. Stirling’s formula

10.8 LESSON END ACTIVITES

1. Prove that sin app

aap +

-¥

¥-Õ ÷

ø

öçè

æ

++=+ nzz e

z

zez

2cot 1)(

2. Show that 2

3

132

6

1 21

21

÷ø

öçè

æG÷

ø

öçè

æ=÷

ø

öçè

æG

-

p.

10.9 REFERENCES

1. Complex Analysis by L.V. Alphors..



89

UNIT - 5 LESSON 11 RIEMANN MAPPING THEOREM

CONTENTS 11.0 Aim and Objectives 11.1 Introduction 11.2 Riemann Mapping Theorem 11.3 Boundary Behaviour 11.4 Use of the Reflection Principle 11.5 Analytic Arcs 11.6 Let us Sum up 11.7 Lesson end activities 11.8 References

11.0 AIM AND OBJECTIVES

The main objective of this lesson is to study Riemann mapping theorem which states how a simply connected region W is mapped onto a unit disk |w| < 1. We also study the use of the reflection principle. After going through this lesson you will be able to: (i) define connectedness and univalent function. (ii) Explain Boundary behaviour (iii) Use of the reflection theory

11.1 INTRODUCTION We know that in the theory of Analytic function the concept of conformal

mapping with the geometrical aspect plays a significant and vital role. Analytic functions defined with geometric properties of the mapping function and in particular Riemann mapping. Theorem deals with the mapping of simply connected regions based on the theory of normal families and related geometric properties. We wish to define an equivalence relation between regions in D. It will be shown that all proper simply connected regions in C are equivalent to one another by showing that each simply connected region is equivalent to the open disk { }1|z:|zD <=

11.2 RIEMANN MAPPING THEOREM

Definition: A region W is said to be connected if its complement relative to the extended complex plane is connected Definition: A one – to – one analytic function g(z) defined in a region W is called a univalent function if )( 1zg = )( 2zg 21 zz =Þ .

Example: ( )z

zzg

-=

1)( is univalent.



90

Definition: A subset of a metric space is connected if it cannot be represented as the union of two disjoint relatively open sets none of which is empty. Riemann Mapping Theorem Statement: Let W be a simply connected region which is not the whole plane and let W . There exists a unique analytic function f on W having the following properties. a) 0)z(f 0 = and 0)zo('f <

b) f is one – one c) f( W ) = {w : |wk|} Proof:

Let us first prove the uniqueness of f. let 1f , 2f be two functions such that

1f ( 0z ) = 0, 0)(,0)(,0)( '20

'102 >>=W- ozfzfzf ,

so that each of 1f , 2f maps W onto |w| < 1.

This implies that the function )W(ffs 21

1= maps the unit disk onto itself.

Because, sW" , in the unit disk |W| < 1, 12-f ( sW ) 1

21, -WÎ ff ( sW ) belong to the unit disk

|w| < 1.

Also, s (0) = 1f . 12-f (0) = )f(f 2

11

- (0) = 1f ( 0z ) = 0

[ ] 0')0(0)0(' 21

1 >= -ffs as both '1f ( 0z ) and 1f ( 0z )

are greater than zero. we also know that a function which maps a circle onto a circle is a

linear fractional transformation and in particular if we suppose s (z) =1jz

zei

-

a-l it gives the

family of transformations which transform the unit disk onto the unit disk.

Since s (0) = 0, lie 0=a

\ a = 0 Putting 0=a we have s(z) = l- ie z………..(2) 'sÞ (z) = l- ie as 's (z) > 0. The least possible value of l is p . Therefore s(z) = z.

Hence s is the identity transformation i.e. '1f

12-f = l.

Hence 1f = 2f i.e. the analytic function f is unique.

Let us now prove the existence of such function f. Let 1f be the family formed by the analytic function g(z) satisfying the following

characteristics, i) g(z) is univalent in W ii) | g(z) | £ 1, WÎ"z and

iii) g ( 0z ) = 0, g’ ( 0z ) > 0

we shall complete the proof by showing (a) The family 1$ is non – empty

(b) $ a function f 1$Î with the maximal derivative i.e. 'f ( 0z ) is maximal

(c) The function f is analytic and univalent and f( 0z ) = 0, 'f ( 0z )> 0

(a) To prove the family 1$ is non - empty

Since by hypothesis W is not the whole complex plane, $ at least one point a ¥¹ which is not in W )a( WÎ¥¹ .



91

Since W is simply connected, it is possible to define a mono valued branch of az - is W .As WÎ¥¹a , z-a is a non vanishing analytic function in W .Which is simply connected \Z – a has an analytic square root, denoted

By h (z) = az - . Consider )z(h|z|h 21 = azaz 21 -=-Þ 21 zz -Û

Then let 1211 )(,)( wzhwzh == i.e. )z(h)z(h 21 -= azaz 21 -=-\ 21 zz -Û

i.e., H(z) does not take the same value twice nor does it take opposite values as h(z) is single valued. This implies that h(z) is a non-constant function by open mapping theorem we know that a non-analytic function maps open sets.

\ The images of W under h, i.e. h( W ) is an open set since ,0 WÎz )z(h 0 )(zhÎ .

Since h( W )is open $ a real number f > 0 such that |)z(hw| 0- < r contained in h( W ).

Since opposite values are not taken by h in W ,there is no points Wez for which h(z) takes opposite values )z(h 0 - )z(h 0 .

This implies that the disk |)z(hw| 0- ³ r does not meet the disk |)z(hw| 0+ ³ r (i.e.) to say

h(z) Îh( W ),we have the condition )1.(..........)()(( r³+ zohzh

In particular, | r³+ |)()( zohzh 2/)z(h 0 r<Þ …..(2)

Now consider the function defined by úû

ùêë

é

+

-l=

)z(h)z(h

)z(h)z(h)z(g

0

0 where l is given by

)z('h

)z(h

|)z(h|

|)z('h|

4 0

02

0

0r=l ……….(3) 0gÞ (z) is a liner transformation of h(z)

We will verify that og (z) Î 1$ so the family 1$ is non-empty.

Since h(z) is a single valued analytic function and h(z) ¹ -h(z), WÎ"z an 0g (z) is

defined using h(z), 0g )z( 0 is analytic WÎ"z . Since h(z) is univalent so is 0g (z) and

0g )z( 0 = 0.

To show 10g )z( 0 > 0, consider

20

0100

))z(h)z(h(

)z(h)z('h2)z(g)z(g

dz

d

+

l==

)(2

)(')'(

0

0

zh

zhzgo

l=\

Now using (3) )z(g 01

0 0|)z(h|

|)z('h|

8|)z(h|

|)z('h|

2

1.

4 20

02

0

0 >r

=r

=

We shall prove | 0g (z)| < 1 in W )z(h)z(h

)z(h2)z(h)z(h

)z(h)z(h

)z(h)z(h

||

|)z(g|

0

00

0

00

+

-+=

+

-=

l

)z(h)z(h

2

)z(h

1)z(h

||

|)z(g|

000

0

+-£

l………..(4)

Since r

£r

³2

|)z(h|

1m

2|)z(h|

00 ………… (5)

r

2

)()(

1

0

£+

Þzhzh r

2

)()(

2

0

£+

Þzhzh

….. (6)



92

Using the results in (3), we have, )z(h)z(h

2

)z(h

1|)z(h|

||

|)z(g|

000

0

++£

l

£ ÷÷ø

öççè

æ

r+

r

22|)z(h| 0

rl

4|)(|||)(| 00 zhzg <

r

r<

4)z(h

)z('h

)z(h

)z(h

)z('h

40

0

0

20

0£ 1 in W .

i.e., 0g )z( 0 has all the properties of an univalent function. \ 0G )z( 0 1$ is non-empty.

(b)To show that there exist a function f 1$Î and 'f ( 0z ) is maximum

Now, 1$Î"g and )(' 0zg > 0, it follows that )(' 0zg has a least upper bound say

B, where B is finite or infinite, $ a sequence }g{ n Î 1$ such that

WÎ"=¥® 000 )()( zzfzg nLimn …….(7)

Since 1)z(g £ 1$Î"g WÎ" z , it follows that 1$ is a normal family.

Hence there exists a sub – sequence }g{ nk of the sequence }g{ n such that )}({ zg nk

converges to an analytic function f(z) uniformly on every compact subset of W .

\ )(')(' 00 zfzg nLimn =¥® .But from (7), we have BzgLim

n =¥® )( 0 .

Hence )(' 0zf = B. This implies that B is finite. It is true that 1)z(f £ , We"z , )z(f 0 = 0 and

)z(f 0 )B(B ¥<< .We will prove that f is univalent. We first observe that f is not a constant

as )z('f 0 0B >= .

Choosing a point We1z and defining )z(g)z(g)z(g 11 -= 1g $e" .We see that )}z(g{ 1 are all non-

zero functions in }z{ 11 -W=W

\By Hurwitz Theorem we know that the limit function of any sequence is either identically zero or never zero. But )z(f)z(f 1- is a limit function of any sequence in the

above family and it is not identically zero. \ 11 zz)z(f)z(f)z(f ¹"¹ , since 1z is arbitrary, 11 zz)z(f)z(f ¹"¹ .This implies, that f(z) is a

univalent function on W .Hence f(z) has the property of maximal derivative i.e. 'f ( 0z ) is

maximal. \G (z) is a member of $ and 'G ( 0z ) > 'f ( 0z )

But we know 1f ( 0z ) is only maximum. )z(f)z(G 01

01 > is a contradiction to the definition

of f(z). Hence the unique univalent function f(z) assumes all values and maps W onto |w| <1 and

since 0)( 0 =zg , $Î"g .

We get in particular 0=kng , " k = 1,2,3…..and 0)()( 00 =-¥® zfzg

knLimk .

Corollary Between any two simply connected regions there is always a (1,1) onto analytic function



93

Proof: Let 1W , 2W be two simply connected regions and neither of them is the plane.

Consider 1W , 2z e 2W .

By Riemann mapping theorem, $ a(1,1) analytic function 1W D® and g: 2W D® such

that )z(f 1 = 0, 1f ( 1z ) > 0.

)z(g 2 = 0. g’ ( 0z ) > 0. This implies that 1g- .f is a mapping from 1W onto 2W 1g- of ( 1z ) =

1z and so 1g- is the required (1.1). Analytic function existing between 1W and 2W .

(c) To show that the univalent function f maps 0W onto |w| < 1.

If possible assume that $ a point ow with | ow | < 1 and f(z) ¹ ow . We"z . Since W is

simply connected, it is possible to define a single valued branch )z(fw1

w)z(f)z(f

o

o

-

-= , which

never vanishes in W . Since f(z) is univalent, naturally f(z) is defined. Using f(z) is also univalent on W and |f(z)| £ 1 normalize f(z), let us define a new functions.

÷÷ø

öççè

æ

-

-=

)()(1

)()(

)('

)(')(

0

0

0

0

zfzf

zFzfx

zF

zFzG

Naturally 0)z('G,0)z(G 00 >= .

In particular ( )( )00

200

0wqw2

w1)z('f)z('G

-

-=

( )

0

00

w2

w1)z('f +as 1

4

1w 0 <=

B)z('f)z('G 00 =<Þ .

Thus G(z) is analytic and univalent in W , and 1)z(g £ . G(z) = o and 1)z('G 0 >

11.3 BOUNDARY BEHAVIOUR Let W be a simply connected region. Consider a sequence |z| n of points in W (or)

an arc )(tz for 0 £ t<1. We say that the sequence }z{ n or the arc z(t) in W is said to tend to

the boundary of W if the points nz (or) the arc z(t)will ultimately stay away from any

point of the region. That is, a sequence }z{ n or an arc z(t) in W is said to tend to boundary

of W if to each Wez , $ an 0>e and an integer on or a real number ot Î{0,1} such that

³"e³- n,zzn on for all t < ot , such that e³- z)t(z .Therefore, the disk with centre z and

radius e form an open covering of W . Hence any compact subset of W can be covered by a finite number of such disks. Theorem A sequence of points or an arc in a simply connected region W tends to the boundary of the region if and only if every compact set ÍK W , there is a tail end of the sequence or the arc which does not meet K. Proof Consider the sequence }z{ n of points on the arc z(t) in W such that it tends to the

boundary of W . Let K be any compact subset of W . Then for every zÎK and 0>e depending on z and a natural number on = on (z), or a real number ot = ot (z).



94

Such that for all n > on , zzn - < e and for all t < ot , e³- |))(| ztz .This implies that the

neighborhoods of the points of K will form an open covering for the compact set .WÍK .

If N = Man { |)z(N)......z(N),z(N 2o2o1o or )}z(t),z()....,z(t),z(t{MaT n012010=

We say that for n > oN , 0n zz e³- or Tt ³ 0zz)t(z e³- . Where 0e = Man

{ 0e ( 1z ), 0e ( 2z )…. 0e ( nz )} i.e. n > N or t > T. }{ nz ÏK or z (t) ÏK. Thus we see that the

condition is necessary. Conversely suppose that }z{ n is a sequence or z(t) for 1t0 ££ is an arc in W which has

tailed every form of compact set 0z Î W .

Let 0z Î W , then $ a 0>d . Such that |z- 0z | < d lies entirely in W .

If f< d then the disk |z- 0z | < r lies entirely in W . Now the tail end of the sequence }z{ 0 or

the arc z(t) must be away from the compact set (z- 0z ) < r

i.e. $ a natural n > no, such that | 1z -z| > r or for t > ot £ r<- 0z)t(z .That is }z{ n of z(t)

tends to the boundary of the region W . Note: If W is a simply connected region having a line segment g , a straight line as a part

of its boundary, by rotating the region if necessary we can assume that g is a segment of

the real axis. Definition: Let W be a simply connected region having a segment g of the real axis as a

part of its boundary. g is said to be a free boundary if to each point of g , $ a

neighborhood whose intersection with the boundary 2S of W is the same as its

intersection with g . i.e. g is a free boundary to each on Î W there exists a neighborhood

a of on such that gÇ-D W ASn real diameter of the disk a along real axis.

Theorem: Let f(z) be the topological mapping of the region W onto a region 1W .If the sequence of points }f{ n or an arc z(t) in W tends to the boundary of W then the sequence

)}({ nzf or f(z(t)) will tend to the boundary of 1W .

Proof: It is enough to show that the sequence )}({ nzf of the disk {f(z(t)} stays away

from every compact subset K of W .Let K be a compact subset of W . 'f (K) is compact in W .

Since f is (1,1), 1f - is continuous and as the sequence }z{ n or the arc z(t) tends to the

boundary, we get that the tail end of the sequence or of the arc does not meet 1f - (K). i.e.

$ an integer on or a real number ot such that }{0 oznn <" is not contained in 1f - (K) or

for all t > ot , z(t) Ï 1f - (K) \ )z(f n ÏK if 0nn < or f(z(t)) ÏK for ott > .This implies that

the set )}z(f{ n or the arc )}z(f{ n does not meet K ultimately. Let for t > ot e³- |z)t(z| .

')()())(( e=³- zfzftzf

Þ )}z(f{ n or f(z(t)) tends to 'W WÎ" z .



95

Definition: A free boundary arc g of a region W is said to be an one sided free boundary

arc if to each WÎ0g , there is a neighborhood D which has one half disk lying entirely in

W . If either of half disks are on W then the arc g is said to be a two sided boundary arc.

11.4 USE OF THE REFLECTION PRINCIPLE

The principle of reflection is based on the observation. That is if u(z) is harmonic then u(z) is also likewise harmonic where z is the reflection of z on the real axis and if f(z) is an analytic function then )z(f is also analytic ware precisely if u(z) is harmonic and f(z) is analytic, then u ( )z is harmonic and )E(f is analytic in the region W obtained by

reflection of W on the real axis. i.e. WÎz if and only if WÎz . Theorem: Suppose that the boundary of a simply connected region W contains a line segment g as a one –sided free boundary arc. Then the function f(z) which maps W onto

the unit disk can be extended to be function which is analytic and (1,1) on gUW . The

image of g is an arc 'g on the unit circle.

Proof: Since W is a simply connected region which is not the whole plane, by Riemann mapping theorem. We can find a unique univalent function D®W:f such that )z(f 0 = o and )z(f 0 > W for some

0z Î W .

Since g is a free boundary arc, every point of g has a neighborhood whose intersection

with the whole boundary of W containing a segment g of a straight line is the same as its

intersection with g .

Consider the disk D around Îon g which is so small that the half disk in the region W

will not contain the point

0z with )z(f 0 = o.

That is, for every z in this disk f(z) ¹ o this implies that log f(z) has a single valued branch in the said half disk D and its real part log f(z) ® o as z approaches the diameter. Because as z tends to the boundary of g in gUW ,f(z) will tend to the boundary of D so

that {f(z)} ® 1. This implies log |f(z)| ® 0 as z ® diameter. Hence by the reflection principle log f(z) can be extended analytically to the whole disk as the function g(z) and g(z) is analytic in gWU and g(z) = log f(z), We"z where we can

have G(z) = )z(ge and is the analytic extension of f(z) in gUW and as f(z) is univalent and

(1,1). Moreover, since f(z) is analytic at 0x , 'f (z) ¹ 0 on 'g , if at all 'f ( 0x ) = o. f( 0x )

were a multiple value so that the two squares of g meeting at gÎ0x would be mapped on

as arcs forming an angle 2n,n

³p

which is impossible.

Therefore this is a contradiction. 0x 'f (z) ¹ 0 is only true.

Hence considering the upper half discs lying in W we have g<¶

¶-

¶

¶on0)z(flog

x)z(flog

y.



96

Hence by the reflection principle, f(z) moves constantly in the same direction. i.e. the analytic continuation mapping is (1,1) on g and hence on gWU .Since the univalent

function f(z) maps W onto the unit disk, clearly the images of g under this analytic

extension is an arc 1g contained in the unit disk. Hence the theorem.

11.5 ANALYTIC ARCS A real or complex function j (t) of a real variable t, defined on an interval

a < t < b, is said to be real analytic, y for every ot in are interval, the Taylor development

200000 )tt)(t('

2

1)tt)(t(')t()t( -j+-j+j=j +….. converges in some interval ),( 00 rr +- tt ,

r > 0. The function )t(j can be defined as an analytic function in a region D , symmetric

to the real axis, which contains the segments (a,b). We say that )t(j determines an

analytic arc. It is regular if )t('j ¹ 0, and it is a simple arc if )t( 1j = )t( 2j , only when 1t = 2t .

11.6 LET US SUM UP

1) Definition of simply connected region. 2) Definition of univalent function. 3) Riemann Mapping Theorem. 4) Use of the reflection principle in the following mapping theorem. 11.7 LESSON END ACTIVITIES

1) If oz is real and W is symmetric with respect to the real axis, prove by the

uniqueness that f satisfies the symmetry relation )()( zfzf = .

11.8 REFERENCES

1. Complex Analysis by L.V. Alphors..



97

LESSON – 12 CONFORMAL MAPPING OF POLYGONS CONTENTS 12.0 Aim and Objectives 12.1 Introduction 12.2 The Behaviour at an angle 12.3 The Schwarz – Christoffel formula 12.4 Mapping on a Rectangle 12.5 Let us Sum up 12.6 Lesson end activities 12.7 References

12.0 AIM AND OBJECTIVES Our aim in this lesson is to study Schwarz - Christoffel formula and learn the mapping on a rectangle. We shall come to know that when W is a polygon the mapping problem has an solution. We shall learn that the mapping function can be given through a formula with parametric values that depend on the shape of the polygon. After going through this lesson you will be able to: (i) study about Schwarz – Christoffel formula (ii) know about the mapping rectangle

12.1 INTRODUCTION In this lesson we consider W as a bounded simply connected region which is bounded by a closed polygon. If 1z , …….. , nz are the consecutive vertices in the positive

cyclic order with 1+nz = 1z , the angle at kz is given by arg kk

kk

zz

zz

-

-

+

-

1

1 which lies between 0

and 2p . We shall denote the interior angle at kz by ,pa k where 0 < ka 2< . The outer

angles 1b + ….. + nb = 2. Since, sum of the interior angles of a polygon of n sides is

åå -=-=n

k

n

k nn11

)2(,)2( appa . With these basic information’s let us move on to

study some important results such as Schwarz – Christoffel formula. 12.2 THE BEHAVIOUR AT AN ANGLE

Introduce the outer angles kb p = ,)1( pa k- -1< kb <1. Observe that 1b + 2b +

… + nb = 2. The polygon is convex if and only if all kb > 0. Consider a circular sector

kS which is the intersection of W with a sufficiently small disk about kz . A single –

valued branch of V = k

kzz a/1)( - maps kS onto a half disk 'kS . A suitable branch of



98

kz + kaV has its values in W and we may consider the function g(V ) = )( k

kzf aV+ in 'kS . It follows that | g(V )| ® 1 as V approaches the diameter. The reflection principle

applies and we concluded that g(V ) has an analytic continuation to the whole disk.

12.3 THE SCHWARZ – CHRISTOFFEL FORMULA

Statement: The function z = F(w) which maps |w| < 1 conformally onto polygons with

angle pa k (k = 1,2,….n) are of the form F(w) = c òÕ=

- +-w n

kk cdwww k

0 1

')( b

where kb = 1 - ka and the kw are points on the unit circle and c, 'c are complex

constants. Proof: Let W be a bounded simply connected region whose boundary is a closed polygonal line without self intersections. Let 1z , 2z , … , nz be the n consecutive vertices

of the n added polygon.

The angle subtended at kz , by the sides joining ( kz , 1-kz and kz , 1+kz ) is given by any

÷÷ø

öççè

æ

-

-

+

-

kk

kk

zz

zz

1

1 and we know that, in any polygon, the interior angles are always less than

p2 . If we let, paR = arg ÷÷ø

öççè

æ

-

-

+

-

kk

kk

zz

zz

1

1 Þ 0 < ka < 2

By the hypothesis, kb = )1( ka- .

\ If the outer angles are ab k , then pkb = (1- ka )p Þ å = pb p 2k å = 2kbp .

Since W is the bounded simply connected region, by Riemann Mapping Theorem, we can find a unique univalent function f(z) and by the reflection principle, f(z) can be analytically extended to any side of the polygon and that each side is mapped in an one-to-one way into the arc of the unit circle. Let kS be a small circular sector. Which is also the intersection of W with a small disc

about kz so that the single valued Brach t = kakzz /1)( - maps kS onto a half disk '

kS .

Hence a suitable branch of z = ktzka+ has its values in W so that the function g(t) =

f( ktzka+ ) is such that |g(t)| ® 1 and log |g| ® 0 by the reflection principle.

\ By the same reflection principle, g(t) has an analytic extension to the whole disk. Since g(t) is analytic, g(t) can be represented in series using Taylor’s development given

by g(t) = f( ktzka+ ) = kw + å

¥

=1m

mmta …… (1).

Where 1a ¹ 0 because 'kS is not contained in the unit disk. The above series

representation can be inverted by assuming w = f( kz + kt a ) …. (2).



99

\ From (1) & (2) w = kw + å¥

=1m

mmta , t = å

¥

=

-1

)(m

mkm wwb (say) b ¹ 0…. (3)

and this expression is valid in the neighborhood kw .

Since from (2), )(1 wf - = ktzka+ and putting )(1 wf - = F (w),

we have, F (w) – kz = kt a . Using (3) we get,

F (w) – kz = k

m

mkm wwb

a

÷ø

öçè

æ-å

¥

=1

)( = )()( wGww kkka- (say) …. (4).

Where kG is analytic and ¹ 0 in the neighborhood of kw , differentiating, we get

)()()()(0)(' '1 wGwwwGwwwF k

kkkkkk

aaa -+-=- -

= k

k

kk

kk

kk

ww

wGww

ww

wGaa

a--

-+

-- 1

'

)(

)()(

1)(

)(, since kb = 1- ka .

\ )(' wF k

kww b)( - = )( kww - )(' wGk + ka )(wGk is analytic in the neighbourhood of

kw and not equal to zero.

Now, put H(w) = )(' wF Õ=

-n

kk

kww1

)( b ….. (5)

This implies that H(w) is also analytic and not equal to zero in the closed unit disk |w| £ 1. We shall complete the proof by showing that H(w) is a constant. Let us consider the argument of H(w) on |w| £ 1.

When w = qie lies on the unit circle |w| £ 1, argument kw = kie q and 1+kw = 1+kie q

Since 'F = qd

df

arg 'F = arg df – arg dw …….. (6)

Where arg df denotes the angle of the tangent to the unit circle at w = qie and Arg qd

indicates the angle of the tangent to its image f(w) = F( qie ).

Since F is only straight line arg df is a constant and we can have arg dw = q + 2

p.

Hence kww - = qie - kie q .

kww - = 2

)( ki

eqq +

÷÷ø

öççè

æ-

-+

2

)(

2

)( kk ii

eeqqqq

= 2i 2

)( ki

eqq +

sin ÷ø

öçè

æ -

2

)( kqq.

This implies that arg (w-wk) is 2

q +a constant

\ Adding the arguments of all factors on the R.H.S of (5), we find

arg h(w) = arg 1F + å=

-n

kkwwBr

1

)arg( using (6)

= arg df – arg dw + z + ÷ø

öçè

æå

=

n

k

Bk1

+2

q some constant

arg H(w) = constant. Since H(w) is continuous, arg H(w) is a constant in the whole unit circle.



100

\ By the Maximum modulus principle, arg H(w) = mJ log H(w) = constant inside the

unit disk. Þ itself is a constant (say c)

(5) Þ )(' wf = cÕ=

--n

kk

kww1

)( b . Integrating we get

F(w) = c òÕ=

- +-w n

kk cdww k

0 1

')( qb Hence the theorem.

12.4 MAPPING ON A RECTANGLE We know that from Schwarz – Christoffel formula,

z = F(w) c òÕ=

- +-w n

kk cdqww k

0 1

')( b ….. (1)

and F(w) maps the unit disk |w| £ 1 conformal onto polygons with angles pa k (interior)

and kb = 1- ka (interior) and kw are points on the unit circle and c, 'c are complex

constants and W is a bounded simply connected region whose boundary is the above said closed polygonal lines.

If w becomes a rectangle, then .22 432

4

11 =+++=å

=

bbbpbpbk

k

Since 2/4321 ppbpbpbpb ==== in a rectangle,

2/4321 pbbbpb ====

Choosing the three vertices, as 1w = 0, 2w = 1, 3w = r > 1, the above mapping function

from (1) gives for c = 0 and 'c = 0.

f(w) = ò---

---w

dwww0

2

1

2

1

2

1

)()1()0( qr = ò --

w

www

dw

0 ))(1( r …… (2)

Where (2) is an elliptic integral. To avoid ambiguity, let w , 1-w , r-w lie in the

1st quadrant. Consider the mapping f(w) as w traces the real axis, since w is real each square root is either positive or purely imaginary with positive imaginary part. As 0 < w < 1, there are one real and two imaginary square roots. This means f(w) decreases from 0 to a value –k where it can be proved that

K = ò--

1

0 ))(1( ttt

dt

r …… (3)

If it is assumed 1 < w < r , then there is only one square root and the integral

ò--

w

www

dw

1 ))(1( r is purely.

Imaginary with a negative imaginary part. Thus f(w) follows a vertical segment move from –k to –k- i 'k where 'k is given by



101

'K = ò --

r

r1 ))(1( ttt

dt \ For w > r

the integrand is +ve and f(w) traces a horizontal segment in the positive direction and it terminate at - i 'k and the length of the segment is given by the integral

ò¥

--r r ))(1( ttt

dt. Now let t =

u

u

-

-

1

r Þ u =

t

t

-

-

1

r and t = 1 +

u

u

-

-

1

r

dt = duu

ut2)1(

)(

-

-r and if t ® r , u ® 0, if t ® ¥ , u ® 1.

\ ò¥

--r r))(1( ttt

dt = ò

-

-÷ø

öçè

æ

-

-÷ø

öçè

æ

-

-

--1

0

2

)1(

)1(

1

1

1

)1()1(

u

u

uu

u

u

du

rrr

r

= ò--

1

0 )1)(( uuu

du

r = ò

¥

--r r ))(1( ttt

dt = K by ….. (3)

By Cauchy’s Theorem, as ))(1(

1

r-- ttt

is analytic within a semi circle with radius R and as R ® ¥

ò- --

R

R ttt

dt

))(1( r = .0

))(1(®

--ò¥

¥- rttt

dt

This implies that k ® 0. i.e., the real part becomes zero so that we can claim that the horizontal segments are equal. Similarly when the imaginary part is zero, - ¥ < w < 0 is mapped onto the segment i 'k to 0 (as already k = 0).

\ -k means (-k, 0), -k -i 'k $ (-k, - 'k ) and - i 'k means Þ (0, - 'k ) \ these parts form the rectangle OABD. Note:

If we consider the vertices as ± 1, ± 1/k, 0 < k < 1, F(w) = ò--

w

wkw

dw

0222 )1)(1(

are +ve, 21 w- , 221 kw- have the +ve real parts so that the vertices of the rectangle

become, -k/2, k/2, k/2 + i 1k , -k/2, + i 1k , where k = ò- --

1

1222 )1)(1( tkt

dt and

'K = ò- --

1

1222 )1)(1( ktt

dtk and the corresponding rectangle is given by ABCD.



102

12.5 LET US SUM UP

(1) The sum of the interior angle of a polygon of n sides = p)2( -n .

(2) The sum of the exterior angle of a polygon of n sides = .2p (3) Note that the sum of the exterior angles is independent of the number of sides of a polygon. (4) Schwarz – Christoffel formula. (5) Mapping on a rectangle. 12.6 LESSON END ACTIVITIES

(1) Show that f(w) = ò-

-w

nnw0

2

)1( dw maps |w| < 1 onto the interior of a regular

polygon with n sides. (2) Determine a conformal mapping of the upper half plans on the region W = { z = x + iy; x > 0, y > 0, min (x,y) < 1}. (3) Show that the mappings of a disk onto a parallel strip or onto a half strip with two

right angles, can be obtained as special caps of the Schwarz – Christoffel formula.

12.7 REFERENCES 1. Complex Analysis by L.V. Alphors.



103

LESSON 13 ELLIPTIC FUNCTIONS CONTENTS 13.0 Aim and Objectives 13.1 Introduction 13.2 General Properties 13.3 Let us Sum up 13.4 Lesson end activities 13.5 References 13.0 AIM AND OBJECTIVES Our main aim of this lesson is to learn about elliptic functions, period modulus, canonical basis of the period module. We also study about some general properties of elliptic functions. After going through this lesson you will be able to: (i) know about general properties of an elliptic function. (ii) theorem an elliptic function.

13.1 INTRODUCTION We say that 1z is congruent to 2z , 1z º 2z (modm) if 1z - 2z Îm. i.e. if 2z - 1z

= 1n , 1w + 2n , 2w . The function f(z) takes identical values at congruent points. Therefore

f(z) may be regarded as a function on the congruent classes. Let aP denote the

parallogram with vertices at a, a + 1w , a + 2w , a + 1w + 2w . We can represent each

congruence class by exactly one point in aP . The point ‘a’ can be any point provided f(z)

has no poles on the boundary of aP .

13.2 GENERAL PROPERTIES Theorem: An elliptic function without poles is a constant. Proof: Since f(z) has no poles, it is an entire function. Since f(z) is entire, it is bounded

on the bounded region aP , including, its boundary. Hence it is bounded in the entire

plane. Therefore by Liouville’s theorem, f(z) reduces to a constant. Theorem: The sum of the residues of an elliptic function is zero. Proof: Let us choose the point ‘a’ such that none of the poles of f(z) lies on the boundary

of aP . By Cauchy’s Theorem,

ò¶

=aP

idzzf p2)( x (sum of the residues) ….. (1)



104

But ò¶

=aP

dzzf )( ò+ 1

)(wa

a

dzzf + ò++

+

+21

1

)(wwa

wa

dzzf ò+

++

2

21

)(wa

wwa

dzzf + ò+

a

wa

dzzf2

)( …….. (2)

In 2I , put z = u + 1w , dz = du

When z = a + 1w , u = a, 2I = a + 1w , u = a + 2w

\ 2I = ò+

+2

)( 1

wa

a

duwuf = ò+ 2

)(wa

a

duuf = ò+ 2

)(wa

a

dzzf {Q 1w is a period}

Similarly 3I = ò+

a

wa

dzzf2

)( .

Now (2) Þ òa

Ipa

dzzf )( = ò+ 1

)(wa

a

dzzf + ò+ 2

)(wa

a

dzzf + ò+

a

wa

dzzf1

)( + ò+

a

wa

dzzf2

)( = 0 ……. (3)

Using (3) in (1) we get 0 = ip2 x (sum of the residues).

\ Sum of the residues at poles is ap¶ = 0. Hence the theorem.

Note: From the above theorem it is clear that an elliptic function with a simple pole does not exist. Theorem: A non constant elliptic function has equally many poles as it has zeros.

Proof: We know that the zeros and poles of f(z) are simple poles of )(

)('

zf

zf, itself being

an elliptic function.

The multiplicities of zeros and poles of f(z) are the residues at simple poles of )(

)('

zf

zf.

The multiplicities are counted positive for zeros and negative for poles.

Since the sum of residues at poles of an elliptic function is zero and )(

)('

zf

zf is an elliptic

function ò¶

=ap

dzzf

zf

i0

)(

)('

2

1

p Þ N – P = 0 Þ N = P

Þ Number of zeros of f(z) = Number of poles of f(z).

Theorem: The zeros 1a , …… na and poles 1b , 2b , …… nb of an elliptic function

satisfy 1a + …… + na º 1b + …… + nb (modm).

Proof: We know that when g(z) is analytic in W , then

ò =g

pdz

zf

zfzg

i )(

)(')(

2

1 = å

jjj agayn )(),( - å

kkk bgbyn )(),( ……. (1)

Choose ‘a’ such that none of these zeros and poles lies on the boundary of aP .



105

\ ò¶

=ap

dzzf

zzf

i )(

)('

2

1

p( 1a + …… + na ) – ( 1b + …… + nb ) ……. (2)

{from (1), g(z) = z . \ g( ia ) = ia and g( kb ) = kb }

But, ò¶ ap

dzzf

zzf

i )(

)('

2

1

p

= êêë

é++ òò

++

+

+ 21

11

1

)(

)('

)(

)('

2

12

wwa

wa

wa

a

dzzf

zzfwdz

zf

zzf

ip +

úúû

ù+ òò

+

+

++

a

wa

wa

wwa

dzzf

zzfdz

zf

zzf

21

2

21)(

)('

)(

)('

Now 1I + 3I = êê

ë

é

úúû

ù+ òò

+

++

+ 2

21

1

)(

)('

)(

)('

2

1wa

wwa

wa

a

dzzf

zzfdz

zf

zzf

ip

= êêë

é

úúû

ù

+

++- òò

++ 11

)(

)(')(

)(

)('

2

1

2

22

wa

a

wa

a

duwuf

wufwudz

zf

zzf

ip

= êêë

é

úúû

ù- òò

++ 11

)(

)('

)(

)('

2

1wa

a

wa

a

dzzf

zzfdz

zf

zzf

ip

= êê

ë

é

úúû

ù-- òòò

+++ 11

1

1

)(

)('

)(

)(

)(

)('

2

12

wa

a

wa

a

wa

a

dzzf

zzfdz

zf

zfwdz

zf

zzf

ip {Q 2w is a period}.

= ò+

- 1

.)(

)('

22

wa

a

dzzf

zf

i

w

p

But ò+ 1

)(

)('

2

1wa

a

dzzf

zf

ip represents the winding number around the origin of the closed

curve described by f(z) when z varies from a to a + 1w and therefore it is an integer.

The same result applies for 2I + 4I .

\ L.H.S of (2) is of the form 11wn + 22wn .

(2) Þ ( 1a + …… + na ) – ( 1b + …… + nb ) = 11wn + 22wn e M.

Hence 1a + …… + na º ( 1b + …… + nb ) (mod m). Hence the theorem.

13.3 LET US SUM UP (1) Period Module (2) Unimodular representation (3) Canonical Basis (4) Fundamental region (5) Properties of elliptic functions.



106

13.4 LESSON END ACTIVITIES (1) Show that any elliptic function with periods 21, ww can be written as

Õ-

-

)(

)(

k

k

bz

azc

s

s where c is a constant, using )2/(

111)()( wzezwz +-=+ hss and

)2/(2

22)()( wzezwz +-=+ hss in the above theorem.

13.5 REFERENCES

1. Complex Analysis by L.V. Alphors.. 2. Theory of Functions by Konrad Knopp – Part - II.



107

LESSON 14 THE WEIERSTRASS THEORY CONTENTS 14.0 Aim and Objectives 14.1 Introduction

14.2 The Weierstrass P -function

14.3 The function x and s associated with P .

14.4 The differential equation for P (z) 14.5 Let us Sum up 14.6 Lesson end activities 14.7 References 14.0 AIM AND OBJECTIVES

In this lesson we learn about Weierstrass P - function, zeta function and sigma function. we also obtain Lagender’s relation and differential equation of Weierstass function. After going through this lesson you will be able to:

(i) know about Weierstrass P -function.

(ii) the association between P -function and x and s functions.

(ii) the differential equation for P (z). 14.1 INTRODUCTION The simplest elliptic functions are of order 2.Thesefunctions have either a double pole with residue zero or two simple poles with residues which are equal in value but opposite in sign. We now construct a function with one double pole. This double pole can be assumed to be at the origin. We can assume that the singular part at the origin is

2z

1 .There is no z

1term in the singular part as the sum of residues at poles of an elliptic

function is zero.

12.4THE WEIERSTRASS P -FUNCTION Now f(z)- f(-z) will have no singular part but will be elliptic with the same

periods. So it should be a constant and by putting 2

wz 1= , we conclude that the constant

should be zero. This is an even function. Since f is any constant is again elliptic and has the same periods we can assume that the Laurent’s expansion for f near the origin is of the form

........1 4

22

12+++ zaza

z{an even function cannot have odd power of z in its expansion}

Having this form in mind, we assert that



108

P( )å

¹

úû

ùêë

é-

-+=

0222

111)(

w wwzzz , 2211 wnwnw += ………(1)

possesses the desired properties.

First, if 2

|w||z| < ,

( )( )( )2222

211

wzw

zwz

wwz -

-=-

- 3||

||10

w

z£ .

Hence the series on the R.H.S. converges uniformly on every compact subset

provided å¹

¥<0

3||

1

w w.

To verify note that 1

2

w

w is not real so that there exists k > 0 with

22211 |wnwn| + )|n||n(|k 2

12

1 +³ for all real pairs )n,n( 11 .

If 0 = arg ÷÷ø

öççè

æ

1

2

w

w 2

2211 |wnwn| + |n|w|n 22

21

21 += |w,w|n,n2|w 2121

22 + cos 0 .

But 1

2

w

wbeing in the upper half plane 0 < 0 < n. So that

cos 0 = 1o, <m£m± Hence 2

2211 |wnwn| + |w,w|n,n2|w|n|w|n| 21212

22

22

12

1 m±+= 2

2211

2

22

112 ||||)(||)1( wnwnnwn ±++-= mm

)|w|n|w|n)(1( 22

22

211

2 +m-=

)nn(k)nn(a)1( 222

12

22

12 +=+m-³

If a = min |)w||,w(| 21 .Thus ( )åå

¹¹ ++£

0n22

21

21

0w3

22

23

nn

1n

k

1

|w|

1 and the latter double series

converges.

Thus the series representing P (z) converges uniformly on every compact subsets of

Ë }wnw,n{ 2211 + , znn Î21 and therefore represents a meromorphic function with poles at

znn,wnnwnw 2122111 e+= .

We next verify that P (z) is periodic with period 2211 wnwnw += .

i.e. P (z) is a doubly period i.e. function with a double pole at origin are at congruent points. We have thus proved the existence of such a function and also that it can be represented by the series (1)

14.3 THE FUNCTION x and s ASSOCIATED WITH P

P (z) is a meromorphic function with zero residue at its poles. So it is the derivative of an analytic function except at the poles. By uniform convergence over compact sets we can obtain this meromorphic function denoted by - x (z) by integrating from o to z along any path not passing through the poles.

We have then



109

å¹

÷ø

öçè

æ++

-+=

02

1111)(

w wwwzzzx

This function x is different from the Riemann – zeta function.

From the form of x (z), it is clear that 11 )()( hxx +=+ zwz and 22 )()( hxx +=+ zwz .

Where 1h , and 2h do not depend on z but only on 21, ww .

In any parallelogram aP only one w lies, so that ò¶

=aP

dzzi

1)(2

1x

p

But ò¶

=aP

dzzi

)(2

1x

p dzz

i

a

wa

wa

wwa

wwa

wa

wa

a

)(2

1

2

2

21

21

1

1

xp ú

úû

ù

êêë

é

òòòò+

+

++

++

+

+

= òòòò+

+

++

+ a

wa

wa

wa

a

wa

wa

a

dzzdzzdzzdzzi

2

2

11

1

)())(())(()(2

112 xhxhxx

p

= ip2

1[ 1n 2w - 2n 1w ]

\ 1n 2w - 2n 1w = ip2 .

The above relation is known as Legendre’s relation. Integrating once again )(zx as can

be done because of uniform convergence and exponentiating to get rid of function s

defined by s (z) = z Õ¹

÷ø

öçè

æ+

÷ø

öçè

æ-

0

2

12

1w

w

z

w

z

ew

z

The definition of s shows that from 11

1

1 )()()(

)('hxx

s

s+=+=

+

+zwz

wz

wz = 1

)(

)('h

s

s+

z

z.

Integrating and exponentiating, we get

,)()(' ,11

zezcwz hss =+ where 1c is a constant.

By its form )(zs is an odd function. Putting z = 2

1w-, we get

2/11

1 11

22we

wc

w hss -÷ø

öçè

æ-= = - 2/1

111

2we

wc hs -

÷ø

öçè

æ. So that 1c = - 2

11

w

em-

.

Hence )( 1wz +s = - ÷ø

öçè

æ+

21

1

)(

wz

ezh

s and analogously )( 2wz +s = - ÷ø

öçè

æ+

22

2

)(

wz

ezh

s .

14.4 THE DIFFERENTIAL EQUATION FOR P (Z)

We have )(zV = å¹

÷ø

öçè

æ++

-+

02

111

w w

z

wwzz

The Laurent’s expansion of )(zV around the origin can be obtained.

We have 2

11

w

z

wwz++

- =

2

11

11

w

z

ww

z

w++÷

ø

öçè

æ--

-



110

= úû

ùêë

é++++ ...1

13

3

2

2

w

z

w

z

w

z

w +

2

1

w

z

w+ = ¥-+

-...

4

3

3

2

w

z

w

z.

Since )(zz is an odd function by the definition of )(zz , we have

)(zV = å¥

=

--2

121

k

kk zG

z where kG = å

¥

¹02

wkw

q.

Since P (z) = )(' zV- , we obtain P (z) = å¥

=

--+2

22

2)12(

1

k

kk zGk

z.

Omitting terms of higher degree, P (z) = 43

222

531

zGzGz

++ +….

P )(' z = 332

3 206/2 zGzGz ++- + ….

P 2)(' z = 32

26 80/024/4 GzGz - + ….

4P 3)(z = 32

26 60/36/4 GzGz ++ + ….

3G 60 2G P (z) = 2

260

z

G + …. Using multiplication power series.

\P 4)(' 2 -z P 23 60)( Gz + P 3140)( Gz -= + …..

But the L.H.S is a doubly periodic function which therefore does not have a pole because of the expansion on the R.H.S. We therefore conclude that the function should be a constant. So, we have

P 4)(' 2 -z P 23 60)( Gz + P 3140)( Gz -=

P 4)(' 2 =z P 23 60)( Gz - P 3140)( Gz -

P 4)(' 2 =z P 23)( Gz - P 3)( Gz -

2G = 60 2G and 3G = 140 3G is the Differential Equation satisfied by P (z).

This is a first order differential equation for w = P (z). It can be solved explicitly by the formula,

ò +--

=

3234 gwgw

dwz constant

Which shows that P (z) is the inverse of an elliptic integral . More accurately, this connection is expressed by the identity.

z- 0z = ò--

)(

)( 322

04

zp

zp gwgw

dw

where the path of the integration is the image of a path from oz to z, that avoids the zeros

and poles of P )(' z , and where the sign of the square root must be chosen so that it

actually equals P )(' z .



111

14.5 LET US SUM UP

(1) Construction of Weierstrass P -function

(2) The functions s (z) and P (z) (3) Legender’s Relation

(4) Differential Equation of P (z). 14.6 LESSON END ACTIVITIES

(1) Show that every even elliptic function with periods 21, ww can be expressed in the

form.

Õ= -

-n

k k

k

bPzP

aPzPc

1 )()(

)()( provided that o is neither a zero nor a pole.

(2) Show that any even elliptic function with periods 21, ww can be written as

cbz

azc

n

k k

kÕ= -

-

1 )(

)(

s

s c is constant.

14.7 REFERENCES 1. Complex Analysis by L.V. Alphors.. 2. Theory of Functions Part - II. by Konrad Knopp.



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