UNIT - 2 BILINEAR TRANSFORMATION

Bilinear Transformation

( 41 )

2.0 Objective

In this unit in addition to Bilinear Transformations, Cross ratio, normal

form of Bilinear Transformations will be discussed. Special emphasis will be

given on conformal mapping, special bilinear transformation, the mapping,

2,nW z W z and the inverse mapping 1/ 2W z . At the end extended complex plane

will be discussed.

2.1 Introduction

Before going to discuss Bilinear Transformation, we shall define

UNIT - 2 BILINEAR TRANSFORMATION

LESSON STRUCTURE

2.0. Objective

2.1 Introduction

2.2 Bilinear Transformation

2.3 Resultant or Product of two Bilinear Transformations

2.4 Cross-ratio

2.5 Preservation of Cross Ratio

2.6 Fixed point and normal form of a Bilinear Transforma-tions

2.7 Some special cases of Bilinear Transformations

2.8 Conformal mapping

2.9 Necessary and Sufficient Condition for a mapping to beConformal

2.10 The mappings, 2,nW z W z and the inverse mapping

1/ 2W z

2.11 Summary

2.12 Solved Examples

2.13 Model Questions

2.14 References


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Transformation or mapping, conformal transformation and linear transformation.

The equation

,

,

u u x y

v v x y

...... (1) define a transformation or mapping by means

of which a correspondence between uv plane and xy -plane can be established. If

corresponding to each point of xy -plane, there is a unique point of uv -plane and

conversely, then such a transformation is called one-one transformation. The

corresponding points in the two planes are called images of each other. Hence,

by means of equations (1), a region or curve of xy - plane is said to transform or

mapped on or represented by the corresponding curve of the uv -plane.

The linear transformation : A transformation of the form w az b , is

called a linear transformation, where a and b are complex constants.

2.2 Bilinear Transformation or Mobius transformation :

A transformation of the form az b

wcz d

...... (1) is called a Bilinear transformation

of linear fractional transformation, where a,b,c,d are complex constants.

Such type of transformation was first studied by Mobius and hence it is

sometimes called mobius transformation.

Equation (1) may be derived as 0zCW Wd az b which is clearly a linear

both in W and Z. That is why, it is called Bilinear. Here we assume that

0ad bc which is called the determinant of the transformation. The

transformation (1) is called normalized if 1ad bc / if the determinant vanishes,

then w is merely a constant as shown below.

Let 1 2W and W be two values of W corrosponding to 1Z and 2Z in (1), then

W Waz b

cz d

az b

Cz d2 1

2

2

1

1

– –

= acz z bcz az d bd acz z bcz az d bd

Cz d Cz d1 2 1 2 1 2 2 1

1 2

– – – –

( )( )

= z ad bc z ad bc

Cz d Cz d2 1

1 2

( – ) – ( – )

( )( )


( 43 )

= ( – )( – )

( )( )

z z ad bc

Cz d Cz d2 1

1 2

0

if ad - bc = 0

2 1 0W W if ad - bc = 0 This shows that W is a constant

2.2.1 Critical Points : Let

az bw

Cz d

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

zb wd

wc a

–

–........... (2)

From (1) 2

dw ad bc

dz cz d

This means /dw

if z d cdz

and 0dw

if zdz

The points 'dz z

c are called critical points where the conformal

property does not hold good.

Also from (2) it is clear that for each dzc

we have a value of W and for

each aWc

there corresponds a value of Z and the correspondence between W

and Z is one-one. The execptional points dZc

and aWc

are mapped into

the points W abd Z respectively.

2.2.2 Extended complex plane : These points will not remain exceptional if we

adjoin a new point called point at infinity denoted by to the complex plane

and the complex plane in this case is called extended complex plane. Thus the

critical point Z corresponds to the point aWc

.

Hence we can now say that the bilinear sets up a one-one correspendence

between the points of entire extended complex z-plane upon the entire extended

w-plane.


( 44 )

2.3 Resultants or Product of two bilinear transformations

Consider two bilinear transformation

az bW

cz d

such that 0ad bc ............. (1)

and 1 1

1 1

a w d

c w d

such that 1 1 1 1 0a d b c .............. (2)

Putting the value of W from (1) in equation (2), we get

1 11 1 1 1

1 1 1 11 1

az ba b

aa b c z b d a bcz d

az b c a d c z d d c bc d

cz d

Writing 1 1 1 1 1 1 1 1, , ,A aa b c B b d a b C c a d c D d d c b

We get AZ B

CZ D

Here AD-BC 1 1 1 1 1 1 1 1aa b c d d c b b d a b c a d c

1 1 1 1 1 1 1 1aa d d bb cc b c ad a d bc

= (a1d

1 – b

1c

1) (ad – bc) 0 by (1) & (2)

Thus AZ B

CZ D

such that 0AD BC

This is a bilinear transformation. This bilinear transformation is called the

resultant or product of the transformation resultant (1) and (2). Thus we may say

bilinear transformation forms a group.

2.3.1 Geometrical Inversion : Every bilinear transformation is the resultant of

bilinear transformation with simple geometrical imports.

Consider the bilinear transformation

az bw

cz d

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

Where 0, 0ad bc c


( 45 )

from (1)

. 1b b dza aa a cw

ddc c zz cc

2

1.

a bc adw

dc c zc

Putting Z z dc

Zl

zZ

bc ad

cZ1 2

1

3 2 2 , ,–

.

We obtain 3

aw Z

c which is similar to 1

dZ zc

.

The above three auxiliary transformations namely 1 2 3, ,Z Z Z are of the form

1, ,W Z W W z

z

This proves that every general bilinear transformation is the resultant of

the bilinear transformations

1, ,W Z W W z

z

Where (i) W Z represents translation

(ii)1

Wz

represents inversion

and (iii) W z represents dilation.

2.4 Cross Ratio

If 1 2 3 4, , ,Z Z Z Z are distinct points, then the ratio

4 1 2 3

2 1 4 3

Z Z Z Z

Z Z Z Z

is called

the cross ratio of 1 2 3 4, , ,Z Z Z Z . This ratio is invariant under the bilinear

transformation.

Number of four distinct cross-ratio :

From four points 1 2 3, ,Z Z Z and 4Z lying in the Z-plane, we can obtain different

cross ratios according to the order in which the points are taken. Since the four

points can permute themselves in be only six-distinct cross ratios. This is so


( 46 )

because if we inter change any points and then interchange remaining

two, the cross ratios of the points in this new order will be the same. In

fact, we have

1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1, , , , , , , , , , , ,Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z ......... (1)




1 4 2 3 2 3 1 4 3 2 4 1 4 1 3 2, , , , , , , , , , , ,Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z .......... (5)

1 4 3 2 2 3 4 1 3 2 1 4 4 1 2 3, , , , , , , , , , , ,Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z .......... (6)

2.4.1 The cross ratio 1 2 3 4, , ,Z Z Z Z is real iff the four points 1 2 3 4, , ,Z Z Z Z lie on a

circle or on a straight line.

Proof : By definition of cross ratio

( , , , )( – )( – )

( – )( – )

( – )( – )

( – )( – )

Z Z Z ZZ Z Z Z

Z Z Z Z

Z ZZ Z

Z ZZ Z

1 2 3 44 1 2 3

4 3 2 1

4 1

4 3

2 1

2 3

4 1 2 11 2 3 4

1 3 2 3

arg , , arg argz z z z

z z z zz z z z

0 or if points lie on a circle or straight line

2.5 Preservance of Cross Ratio :

Prove that cross ratio remains invariant under a bilinear transformation.

Proof : Let 1 2 3 4, ,w w w w be the images of four distinct points 1 2 3 4, ,z z z z

respectively under a bilinear transformation.

az bw

cz d

where 0ad bc ............. (1)

If we show that


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4 1 2 3 4 1 2 3

2 1 4 3 2 1 4 3

w w w w z z z z

w w w w z z z z

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

The result will follow from (1)

W Waz b

cz d

az b

cz dr s

r

r

s

s

– –

W Wad bc z z

cz d cz dr s

r s

r s

–( – )( – )

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

This 4 1 2 3w w w w

( )( )( )( )

( )( )( )( )

4 1 2 3

1 2 3

.ad bc z z ad bc z z

cz d cz d cz d cz d

- - - -=

+ + + +4

2

4 1 2 34 1 2 3

1 2 3 4

ad bc z z z zw w w w

cz d cz d cz d cz d

Similarly again from (3) prove that

2

2 1 4 32 1 4 3

1 2 3 4

ad bc z z z zw w w w

cz d cz d cz d cz d

Dividing the last two results we get the required equation (2).

2.6 Fixed Point of a Bilinear Transformation :

To prove that in general there are two values of Z (invariant points) for

which w=z, but there is only one if

2

4 0a d bc

Show that if there are distinct invariant points p and q, the transformation

may be put in the form (normal form)

k z pw p

w q z q

and that if, there is only one invariant points p, the

transformation may be put in the form (normal form) 1 1

kw p z p

.


( 48 )

Proof : Let us consider the bilinear transformation waz b

cz d

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

(i) The invariant points are given by

this gives 2 0cz a d z b ........ (2)

Solving

24

2

a d a d bcZ

c

2

2

4

2

4

2

a d a d bcTaking p

c

a d a d bcand q

c

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

This shows that p, q are distinct points and are invariant points for the

transformation (1).

If 2

4 0a d bc , then (3) deduce that

2

a dp q

c

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

Hence there is only one invariant point namely 2

a dp

c

if

24 0a d bc

(ii) Suppose p and q are distinct invariant points the transformation (1), so

that they satisfy (2). As a result of which

2 0Cp a d p b

2 0Cq a d q b

for which

2

2

Cp ap b pd

Cq aq b qd

.............. (5)

. .

w z

az bi e z

cz d


( 49 )

az bp

az b pcz pdw p cz d

az bw q az b cqz qdq

cz d

2

2

a pc z b pd a pc z cp ap

a cq z b qd a cq z cq aq

using (5)

a cp z p

a cq z q

Taking a cp

ka cq

, we get

w p z pk

w q z q

.............. (6) which is required.

(iii) Let there be only one invariant point p, so that

2,2

a dp cp ap b pd

c

by (4) and (5)

Now 1 1 cz d

az bw p a cp z b pdpcz d

2

2cz cp a cz cp a cp

a cp z cp ap a cp z p

c z p a cp

a cp z p a cp z p

1c

a cp z p

Taking c

ka cp

, we get 1 1

kw p z p

which is required

This completes the proof.


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(iv) Every bilinear transformation maps circles or straight lines into circles or

straight lines.

or

In a bilinear transformation, a circle transforms into a circle and inverse

points transform into inverse points. In the particular case in which the circle

becomes a straight line, inverse points become symmetrical about the line.

or

Find the condition that the transformation waz b

cz d

transforms the unit

circle in w-plane into straight line of z-plane.

Proof : The equation of a circle with p and q as inverse points in z-plane is

given by

, 1z p

k kz q

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

(In case of a striaght line k=1)

The bilinear transformation az b

wcz d

So that dw b

zcw a

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

transforms the equation (1) into the form

dw bp

cw a kdw b

qcw a

or

w d cp ap bk

w d cq aq b

or'

.'

w p d cqk

w q d cp

Where 'z p

ap b az bp

cp d cz d


( 51 )

'z q

aq b az bq

cq d cz d

Thus '

''

w pk

w q

evidently ' 1k even if k = 1

Clearly the transformed equation (3) represents a cirele in w-plane with

'p and 'q as inverse points. In particular case when k=1, (1) represents a straight

line which bisects perpendicularly the line joining p to q. The bilinear

transformation transforms the line (1) of z-plane into a circle in w-plane and the

points p and q symmetrical about the line becomes inverse point in w-plane.

Note : The above theorem may be read as Obtain an expression which can

be used to transform one given circle into another given circle or straight line.

2.7 Some specieal Cases of Bilinear Transformations.

2.7.1 Find all the Mobius (linear) transformation which transforms the half

plane ,0I z into circles w 1.

Proof : Suppose that the bilinear transformation az b

wcz d

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

transforms the upper half plane I (z) = y 0, into the unit circle | w | 1.

(1) is expressible as

.bza aw

dc zc

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

This 0c , otherwise the point at infinity in the two planes would

correspond.

If w = 0, then from (2) bza

Figure - I

u

y

x

0


( 52 )

If w then (2) dzc

Also 0w , w are the inverse points of 1w we have seen that the

transformation (2) transforms a straight line of z-plane into a circle and points

symmetrical about the line transform into inverse points of the circle w-plane.

Hence the points , zz symmetrical about the real axis ie.e 0zI will correspond

to 0w , w respectively. Hence we may take

,b da c

Then (2) becomes a z

c zw

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

The point z = 0 on the boundary of the half plane I (z) , 0 must correspond

to a point on the boundary of the circle 1w so that

01

0

a aw ascc

This a

ec

where is real

i zw e

z

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

evidently z gives w = 0. But w = 0 is an interior point of the circle 1w .

Hence z must be a point of the upper half plane, i.e. 0I . With

this condition (4) is the required transformation.

Verification wwz

ze

z

zei i1 1

–

–.

–

–––

1z z

z z


( 53 )

or

1

zz z z zz z z

z zw w

2 2

z z z z

z z

2 2

2 .2 4 .i I i I z I I z

z z

finally, 2

2

4 .1

I I zw

z

.................... (5)

Also we have proved that 0I

Now from (5), it is clear that

(i) I (z) = 0 corresponds to 2

1 0w i.e. 1w

(ii) 0zI corresponds to 2

1 0w i.e. 1w

finally, ,0I z corresponds to 1w by virtue of transformation (4)

2.7.2. Find the Bilinear Transformation which transforms 0R z into the

unit circle 1w .

Figure - 2


( 54 )

Proof : Suppose the bilinear transformation

az bw

cz d

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

transforms the half plane R (z) 0 into the circle 1w .

(1) is expressible as

bzaaw

c dzc

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

This 0c , otherwise the points at infinity in the two planes would

correspond.

We have seen that the transformation (2) transforms a straight line of z-

plane into a circle of w-plane and points symmetrical about the line transform

into the inverse points w.r.t. 1w .

Here the points z and z symmetrical about the imaginary axis 0x R z

will correspond to w = 0 and w , the inverse points of the circle 1w .

Thus we may write ,b da e

on putting above in eqn (2) , we get

0w , w .

Then (2) takes the form a z

wc z

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

The point z = 0 on the boundary of the half plane R (z) 0 must corrrespond

to a point on the boundary of 1w so that

01 .

0

a awcc

so that

ia ec

,

Where is real.

Hence i z

w ez

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


( 55 )

evidently z gives w = 0. But w = 0 is an interior point of circle 1w .

Hence z must be a point of the right half plane i.e. R () > 0. With this

condition (4) is the required transformation.

Verification :

1 1i iz zw w e e

z z

= zz z z zz z z

z z

– – – ( )

( )( )

2

z z

z

2 2

2 .2z z R R z

z z

or 2

2

4 .1

R R zw

z

............... (5)

Also we have proved that 0R now from (5) it is clear that

(i) 0R z corresponds to 2

1 0 1w or w

(ii) R (z) > 0 corresponds to 2

1 0 1w or w

finally R (z) >, 0 corresponds to 0w , by virtue of transformation (4).

2.7.3 Find all the bilinear (Mobius) transformation which transforms the unit

circle 1z into the unit circle 1w .

or

Find the general homographic transformation which leaves the unit circle

invariant.


( 56 )

Proof : Suppose the transformation

az bw

cz d

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

transforms the unit circle 1z into the circle 1w ,

bza awdc z

c

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

Cleraly 0c , otherwise the point at in the two plane will correspond.

We have seen that the transformation (2) maps a circle of z-plane into a circle of

w-plane and inverse points transform into inverse points.

If w =0, then c 2) bza

If w , then (2) z = –dc

Thus 0,w , the inverse points of 1w are the transform of bza

and

dzc

respectively.

Hence we may write ba

, –dc

1

Such that 1

[For , and 1 are the inverse points of circle 1z ]

Hence from (2) 1

a zw

c z

or 1

zaw

c z

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


( 57 )

The point z = 1 on the bondary of 1z must correspond to a point on the

boundary of 1w

So that 1

1 .1

aw

c

ia a

or ec c

Hence (3) becomes 1

i zw e

z

with 1 ................(4)

This is the required transformation.

Verification

1 11 1

i iz ze e

z z

11 1

z z

z z

2

1

1

zz z z zz z z

z

2

1 1

1

zz

z

or w2

1 (| | –1)( –| | )

| – |

z

z

2 2

2

1

1

.................. (5)

Also 1 so that 2

1 0

Now from (5) it is clear that

(i) 1 corresponds to

( | –1)( –| | )

| – |

z

z

2 2

2

1

10

or

21 1z or z


( 58 )

(ii)2

1 1 0or corresponds to

2 2

2

1 10 1

1

zor z

z

finally, 1z is transformed to 1 by virtue to transformation (4)

2.7.4 The general linear transformation of the circle z into the circle

'e is

2' i z

ee ez

Such that


az b

cz d

............... (1) transforms z

onto the circle '

from (1)

bzaa

c dzc

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

Clearly 0c , otherwise the point at infinity in the two planes would

correspond.

We have seen that transformation (2) maps a circle of z-plane into a circle

of -plane and inverse points transform into inverse points,

If 0 then (2) bza

If then (2) dzc

Thus 0 , , the inverse points of 'e are the transforms of

,b dz za c

respectively. Hence we may write – ,–ba

dc

2

such that

.


( 59 )

[For and 2

are the inverse point of circle z ]

Hence, by (2),

2 2

a z a zor

c c zz

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

(3) Satisfies the condition z and ' .

Hence for z , we must have ' so that (3) becomes

21'

a z a zas zz

c z zz c z z

.

But z z

Hence ' ' ia aee and ee e

c c

Putting this in (3),

'2

ziee e such thatz e

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

Where is real, this is the required transformation.

Verification

22 2

z zi ie ee e e e e e ez z

22 2 2

22

ez z z z

z

2 2

2 2 222

zz

z


( 60 )

or

2 22 22 2 2'

22

ze

z

................ (5)

Also we have proved that 22 0or

from (5) 2 2 0

2 2 0z z

Again 2 22 2' - ' 0 ' - ' 0w Þ w z

z

Thus corresponds to z

2.7.5 To show that the region z R is mapped conformally on 1

by the transformation.

2

iR z c e

R z a c a

where is real and z c is the point which is transformed into the

origin.


Az B

Cz D

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

transforms z a R onto 1 such that z c transformed into the

origin.

BzA A.DC z

C

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

Evidently 0C , otherwise points at infinites in the two planes would

correspond.


( 61 )

We have seen that the transformation (2) transforms a circle of z plane

into a circle of -plane and inverse points transforms into inverse-points.

If 0 , then (2) BzA

If , then (2) DzC

Thus 0 , , the inverse point of 1 are the transforms of

BzA

, DzC

respectively.

But it is given that z c corresponds to 0 .

The inverse of z c relative to z a R is given by

or

2Rz a

c a

Now, we may write 2

,B D R

c aA C c a

Now (2) is reduced to 2

A z c

C Rz ac a

or

2

A c a z c

C z a c a R

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

ww w corresspondsto z a z a R 2 21 ( )( )

Then (3) gives 1=|w|

A

C

c a z c

z a c a z a z a

.

( ) ( )b g b g or A

C

c a z c

z a c z

.1

2z a z a R


( 62 )

But z c z c and z a R Hence the

last gives, A c a

Rc

or

R

A c a iec

where is real.

By (3),

R

2

ie z c

z a c a R

or

R

2

ie z c

R z a c a

which is the result.

For – .e e e ei i i i

where is real.

Verification

Let z a R so that

2z a z a R

. .1

2 2

iR z c e R z c

R z a c a R z a c a

.

R z c

z a z a z a c a

1R z c R

z z z a z a

or 1

finally, z a R corresponds to 1

Thus the interiors correspond.

2.8 Conformal Mappings

Suppose the transformation u=u(x,y)

v=v(x,y)


( 63 )

maps the two curves C1, C

2 [intersecting at the point P(z

o)] of z-plane on the

curves ! !1 2C C [intersecting at P !(wo)] of w plane

I f t h e an gle bet w een C1 and C

2 and at Z

0 is equal to the angle between and c

1

and c2 at

0, then the transformation is called isogonal. If the sense of rotation

as well as the magnitude of the angle is preserved, the transformation is called

Conformal. In figure (A), the transformation is conformal.

While in figure (B) the transformation is isogonal. = z is isogonal transformation.

2.9. The necessary and sufficient condition for the transformation = f(z) to

be conformal is that f(z) is analytic.

First suppose that = f(z) be an analytic function in a domain D on the

z-plane and z0 be an interior of D. Also c

1 and c

2 be the curves (intersecting at

0) in -plane. Corresponding to the two curves c

1 and c

2 (intersecting at z

o) of

z-plane. Let 1 and

2 be points on c

1 and c

2 respectively corresponding to the


( 64 )

points z1, z

2 on c

1 and c

2 respectively such that Distance between z

1 and z

0 =

distance between z2 and z

0 = r (suppose).

So we can write z1 – z

0 = rei1, z

2 – z

0 = r ei2 let the tangents at z

0 to the curves

c1+ and c

2+ make angles

1,

2 with the real axis so that

1

1,

2

2 as r 0.

Also let the tangents at 0 to the curves c

1 and c

2 make angle

1 and

2 with

the real axis and let 1 –

0 =

1 ei1,

2 –

0 =

2 ei2 where

1

1 as

1 0 and

2

2 as

2 0.

f(z0) = lim

z z 0

f z f z

z z

( ) – ( )

–0

0

= limz z1 0

1 0

1 0

–

–z z = lim

z z1 0

1

1

1

e

r e

i

i

or, f (z0) = lim

z z1 0 1 1 1

rei ( – )

By assumption f(z0) 0. Write f(z

0) = R ei

z – Plane w – Plane

Then Rei = lim –( – )

z z

i

re

1 0

1 111 1

LNM

OQP

equating the modulus and argument,

R = limz z1 0

e

r1FHGIKJ, = lim (

1 –

1) =

1 –

1 or,

1 = +

1


( 65 )

Similarly, we can show that 2 = +

2

from which 1 –

2 =

1 –

2.

This proves that the angles between c1 and c

2 at

0 is equal in magnitude

and as well as in sign to the angle between the curves c1 and c

2. Therefore the

transformation is conformal.

Again let us suppose that the mapping = f(z) is conformal then here we shall

see that f(z) is an analytic function of z.

Proof : Let us suppose = u (x, y) + iv (x, y) = f(z).

Also suppose u = u(x, y), v = v(x, y) are equations defining conformal

transformation from z-plane to -plane. Let ds and d be the line of elements in

z-plane and -plane respectively so that ds2 = dx2 + dy2, d2 = du2 + dv2.

u

u

xdx

u

ydy

v =

v

xdx

v

ydy

Squaring and adding d2 = du2 + dv2

=

FHGIKJ

FHGIKJ

LNMM

OQPP

FHGIKJ

FHGIKJ

LNMM

OQPP

u

x

v

xdx

u

y

v

yy

2 2

2

2 2

2

LNM

OQP

2u

x

u

y

v

x

v

ydx dy.

Since transformation is conformal and hence the ratio 2 : s2 is independent

of direction. Comparing the last equation with ds2 = dx2 + dy2

FHGIKJ

FHGIKJ

FHGIKJ

FHGIKJ

u

x

v

x

u

y

v

y

2 2 2 2

1 1

=

u

x

u

y

u

x

v

y. .

0

This

FHGIKJ

FHGIKJ

FHGIKJ

FHGIKJ

u

x

v

x

u

y

v

y

2 2 2 2

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

and

u

x

u

y

v

x

v

y. . 0 .......... (2)


( 66 )

from (2)

u

xv

y

v x

u y

/

– / (suppose)

u

x

v

y and

v

x

u

y– .......... (3)

Putting this in (1),

2

FHGIKJ

FHGIKJ

FHGIKJ

FHGIKJ

v

y

u

y

u

y

v

y

2

2

2 2 2

or, (2 – 1)

FHGIKJ

FHGIKJ

LNMM

OQPP

u

y

v

y

2 2

0–

2 – 1 = 0

= ± 1

Using this in (3),

u

x

v

y

u

y

v

x, – where = 1 .......... (4)

u

x

v

y

u

y

u

x– , where = –1 .......... (5)

The equation (4) are Cauchy-Riemann equations and hence = f(z) is an

analytic function. The equation (5) are reduced to (4) by writing -hv in place of v,

i.e by taking as image figure obtained by the reflection in the real axis of -

plane.

The mappings = zn (n being a +ve integer) dw

dz = n zn – 1 = 0 at z = 0 so that

the above mapping is conformal at every point except at the point z = 0

where dw

dz = 0.

It is also to be noted as z = 1/n therefore for a given value of we will have n

value of we will have n values of z showing therefore that the correspondence

between z and w is also not one-to-one but n to 1.


( 67 )

Writing z = x + iy = zei and w = u + iv = ei, transformation can be written as

ei = in ein. Equating modulus and argument on both sides,

we get = rn and = n ... (1)

It is now easy to verify the following facts about the mapping.

(A) From the relation = rn, we see that the concentric circles r = in the z-

plane for different values of correspond to the concentric circles = n in the -

plane.

(B) The straight lines = through origin for varying values of change over

the straight line = n through the origin. Here the angle between the two

straight lines through the origin is not preserved but is multiplied by n. Thus

the conformal property does not hold at z = 0. It follows that a circular sector

(wedge) with origin as vertex is transformed into a circular sector with origin as

vertex but central angle becomes n times.

(C) The interior of the circular sector defined 0 < arg z < /n is mapped

conformally on to 0 < arg < . Onto the entire upper half plane I () > 0.

Also the interior of the circular sector defined by

0 < arg z < 2

n

is mapped conformally onto the entire -plane cut along the entire positive

real axis. The cut is introduced to avoid the origin where the conformal property

does not hold. Such a a cut is often called a slit and the plane is returned to as

slit along positive real axis.

2.10. The mapping = zn, = z2 and inverse mapping = z½.

2.10.1 The transformation = z2.

Here this transformation will be discussed in detail.

= z2 d

dz

= 2z = 0 at z = 0

So that the conformal property does not hold at the origin. Here it is convenient

to consider the cartisian form z = x + iy, = u + iv.

Substituting these values in the given transformation we have

u iv = (x + iy)2 = x2 – y2 + 2i xy.


( 68 )

Hence equating the real and imaginary parts, we get u = x2 – y2 and v = 2xy.

(i) We first consider the images of lines x = x1, x = x

2, the line x = x

1 > 0 is

transformed into the curve u = x12 – y2, v = 2x

1y.

Where y is regarded as a parameter. Eliminating y, we get

u = xv

x12

2

124

–

or, v2 = –4x12 u x 1

2d i

Which is a parabola in the -plane with its vertex at (x12, 0), focus at the

origin and symmetrical about the real axis. Observe that the line x = x1 is also

transformed into the same parabola.

It is also evident that the two parts of the line x = ± x1 lying above and below

the real axis seperately correspond to the two parts of the parabola lying above

and below the real axis in the w-plane.

(ii) Both the infinite strips defined by x1 < x < x

2 and x

2 < x < –x

1 are conformally

mapped on the domain in the -plane included between the parabolas.

u2 = –4x12 (u – x

12), v2 = –4x

22 (u – x

22).


z-plane w-plane

y

x

x=-x2

o

u

v

x=-x1 x=x1 x=x2

oo

y


( 69 )

If we make x2 tend to infinity, we see that the half plane to the right of x = x

1

i.ee the domain defined by R(z) > x

1 is mapped conformally on the exterior of the

parabola v2 = –4x12 (u – x

12) not containing the origin. The half plane defined by x

< x1 i.e., R (z) < x

1 is also mapped on the same exterior as shown on the previous

page.

It is thus shown that the transformation w = z2 maps the half plane R (z) > x1

on to the exterior of the parabola v2 = –4x12 (u – x

12) where x

1 > 0.


(iii) We now study the interior of this parabola. If we consider x = x1 and make

x1 tend to zero. We easily see the domain 0 < x < x

1 is conformally mapped on to

the whole interior of the parabola u2 = –4x12 (u – x

12) with a cut along the negative

real units from – to 0 as shown below.

Combining the discussions given in (II) and (III) we can say that the half plane

defined by R(z) > 0 is mapped onto the whole plane cut along the negative real

axis from 0 to –. The half plane R(z) < 0 is also mapped onto the same whole cut

plane.

xo

u

v

oo

y

z - Plane w - Plane


( 70 )

(iv) We leave it to the reader to verify that the image of the line y = y1 is the

parabola v2 = 4y12 (u+ y

12). Whose vertex is at (–y

12, 0) and whose focus at the

origin. Details similar to those in (i), (ii) and (iii) can be easily disussed. It will be

seen as a final result that the half plane f(z) > 0 is mapped on the whole -plane

cut along the positive real axis from 0 to . The half plane I(z) < 0 is also mapped

on the same cut plane.

(v) The two families of parabolas.

v2 = –4x12 (u – x

12) and v2 = 4y

12 (u + y

12).

Corresponding to two orthogonal families of straight line x = ±x1 and y = ± y

1

for varying values of (x1, y

1) themselves from an orthogonal net, as is well known

from analytic geometry. The families of straight lines and parabolas are shown

in the above figures.

2.10.2 Transformation = z

= z 2 = z (u + iv)2 = x + iy

u2 – v2 + i uv = x + iy

u2 – v2 = x, ... (1)

2uv = y ... (2)

Case I : When x = constant.

(a) From (1) it is clear that line x = a, where a > 0 in the z-plane corresponds

to the hyperbola u2 – v2 = a in -plane.

(b) The slit in z-plane between the lines x = a, x = b where a, b > 0 and b > a

is mapped upon the area included between the hyperbolas u2 – v2 = a,

u2 – v2 =b.

z - Plane w - Plane


( 71 )

We have u2 – v2 = x, Hence x = a, x = b gives u2 – v2 = a and u2 – v2 = b. The

lines x = in z-plane as takes values continuously from a to b. This shaded

area of z-plane is mapped upon the shaded area between the hyperbola in -

plane which is developed by the hyperbola u2 – v2 = as varis from a to b.

(c) To prove that the domain in the z-plane to the right of the line x = a is

mapped upon the interior of the hyperbola in -plane.

We have seen that the lines x = in z-plane corresponds to the hyperbola u2

– v2 = in plane. Suppose takes the values continuously from a to . Then

the area in z-plane developed by the line n = will be the entire of z-plane lying

to the right of the line x = a. This area will be mapped upon the area interior to

the hyperbola u2 – v2 = a in -plane which is developed by u2 – v2 = .

Case II : When y-constant.

(a) The line y = a corresponds to the rectangular hyperbola 2uv = a for 2uv =

y, y = a 2uv = a.

(b) The slit in z-plane between the lines y = a and y = b where a, b > 0 and b

> a is mapped upon the area included between the rectangular hyporbolas 2uv

= a and 2uv = b

w - plane

z - plane


( 72 )

complete the proof. The proof is similar to (b) part of case I

2.11. Summary

In this unit, we have discussed.

(i) Bilinear Transformation.

(ii) Resultant or Product of two bilinear transformation.

(iii) Cross Ratio.

(iv) Preservation of family of straight lines and circles.

(v) Fixed points and normal form of a bilinear Transformation.

(vi) Some special bilinear Transformation.

(vii) Conformal Mapping.

(viii) The mapping w = zn, w = z2 and the inverse mapping w = z½.

2.12. Solved examples on conformal mappings.

Example 1 : Find the fixed point and the normal form of the following billinear

transformation and classify their nature.

(a) w = 3 1iz

z i

(b) w =

3 4

1

z

z

–

–

Solution : (a) Putting w = z, then fixed points are given by

z = 3 1iz

z i

z2 + iz – 3iz – 1 = 0

z2 – 2iz – 1 = 0 z2 – 2iz + i2 = 0

(z – i)2 = 0

or, z = i, i.

w - plane


( 73 )

Hence there is only one distinct fixed points namely i.

Now – i = 3 1 3 1 2iz

z ii

iz iz i

z i

–

– –

= 2 1 1 2 2iz

z i

iz

z i

1

2 2 2

1

2

2

– ( – )

–

–i

z i

iz

z i

i z i i

z i i

z i

LNM

OQP

= 1

21

2 1

2

1

i

i

z i i z iLNM

OQP

– –

or,1 1

2 – –i z ii which is the normal form. The transformation is parabola.

Solution: (b) = 3 4

1

z

z

–

–

Putting = z, z = 3 4

1

z

z

–

–

z(z – 1) = 3z – 4 z2 – z – 3z + 4 = 0

(z – 2)2 = 0.

Thus z = 2 is the only fixed point so the transformation is parabolic. For

normal form – 2 = 3 4

12

2

1

z

z

z

z

–

––

–

–

1

2

1

2

2 1

2

1

21

–

–

–

–

– –

z

z

z

z z

Above is the required normal form.

Example 2. Find the bilinear transformation which maps the points z1 = 2, z

2

= i and z3 = –2 into the points

1 = 1,

2 = i and

3 = –1.

Solution : The required transformation is given by

( – )( – )

( – )( – )

( – )( – )

( – )( – )

1 2 3

1 2 3

1 2 3

1 2 3

z z z z

z z z z.... (1)

Substituting the given values as above (1) becomes


( 74 )

( – )( )

( – )(–1 – )

( – )( )

( – )(–2 – )

1 1

1

2

0

i

i

z i z

z z

–( – )( )( _

( )( – )( )–

( – )( )( )

( )( – )( )

1 1 1

1 1 1

2 2 2

2 2 2

i i

i i

z i i

z i i

( – )( )

( )[ – ]

( – ) ( )

( ) ( – )

1 1

1 1

2 2

2 2

2

2

2

2 2

i

i

z i

z i

( – ).

( ).

( – )( )

( ).

1 2

1 2

2 3 4

2 5

i z i

z

– ( – )( )

( )

1

1

1 3 4

2 5

z i

z i

( – )

( )

( – ) ( – )

( )

1

1

2 4 3

5 2

z i i

z i

– ( – )( – )

( )

1

1

2 4 3

5 2

z i

z

Using componendo & dividendo, we get

– ( )

( ) – ( – )

( – )( – ) ( )

( ) – ( – )( – )

1 1

1 1

2 4 3 5 2

5 2 2 4 3

z i z

z z i

2

2

4 8 3 6 5 10

5 10 4 8 3 6

w z i z i z

z z iz i

– –

– –

= 9 3 2 6

3 18 6

3 3 2 1 3

1 3 6 3

z iz i

z iz i

i z i

z i i

–

–

( – ) ( )

( ) ( – )

= 3 2

1 3

31 3

36

3 2

6

zi

ii

iz

z i

iz

FHGIKJ

( )

–

–

which is required transformations.

Example 3. Find the bilinear transformation which carries 0, i, –i and 1, –1,0.


( 75 )

Solution : Consider z1, z

2, z

3 as 0, i, –i and

,

2,

3 as 1, –1, 0. By cross ratio

we have

( – )( – )

( – )( – )

( – )( – )

( – )( – )

z z z z

z z z z1 2 3

1 2 3

1 2 3

1 2 3

( – )( )

( – )(– – )

( – )(–1 – )

( )( – )

z i i

i i z

0

0

1 0

1 1 0

2 1iz

i i z( )

–( – )

–2

4 z = ( – 1) (z + i)

4 z = ( – 1)z + ( + 1) i

4 z – z + z = + i – i

(4z – z – i) = –z – i

(3z – i) = – z – i

= –( )

–(–3 ) –3

z i

z i

z i

z i

... (1)

Which is of the form = f(z) = az b

cz d

, at ad – bc 0

The relation (1) is called bilinear transformation.

Example 4. Show that the transformation = 2 3

4

z

z

– transforms the circle x2 + y2

– 4x = 0 into the straight line 4u + 3 = 0 and explain the curve obtained is not a circle.

Solution : We have = 2 3

4

z

z

– or (z – 4) = 2z + 3

z – 4 = 2z + 3 z ( – z) = 3 + 4

z3 4

2

w

–

The equation of the circle in z-plane is

x2 + y2 – 4x = 0

z z –2(z + z ) = 0


( 76 )

By virtue of the given transformation this is transformed to

( )( )

( – )( – )–

– –

3 4 3 4

2 22

3 4

2

3 4

20

LNM

OQP

or, 9 + 16 + – 2[(3 + 4) ( – 2) + (3 + 4) ( – 2) = 0

or, 9 + 16 + 12 ( + ) – 2 [(6 + 4– 3 – 84 + 3 – 6 – 8)] = 0

or, 22 ( + ) + 33 = 0

or, 2 ( +

or, 4u + 3 = 0 which is the eqn. of a straight line in -plane.

Hence curve thus obtained is not a circle.

Example 5. Find the image of circle |z – 2| = 2 under the Mobius

transformation = z

z 1.

Solution : = z

z 1 (z + 1) = z z ( – 1) =

z =

1 –... (1)

Now |z – 2| = 2 gives (z – 2) (z – 2) = z2

By (1), z – 2 =

12

2 2

1

3

1––

– –

–

–

–

z

z – 2 =

12

3 2

1––

–

–

Putting these values in (2)

3 2

1

3 2

14

–

–

–

–

FHG

IKJFHG

IKJ

or, 9 + 4 – 6 ( + ) = 4 [1 + – ( + )]

or, 5 – 2 ( + ) = 0

or, 5 (u2 + v2) – 2.2u = 0 where = u + iv

or, u2 + v2 – 4

5u = 0


( 77 )

Hence it is a circle with centre 2

50,

FHGIKJ and radius = (g2 + f2 – c)½ =

2

5.

So Image is a circle with centre 2

50,

FHGIKJ and radius =

2

5 in -plane.

Example 6. Discuss the application of transformation = iz

z

1

1 to the areas in

the z-plane which are respectively inside and outside the unit circle with its

centre at the origin.

Solution : The given transformation is

= iz

z i

i x iy

x iy i

ix y

x i y

1 1 1

1

( ) –

( )

u + iv = ( – )

( )

– ( )

– ( )

1

1

1

1

y ix

x i y

x i y

x i y

LNM

OQP

LNM

OQP

= – ( – ) ( – ) ( )

( )

i y ix x y x y

x y

1 1 1

1

2 2

2 2

Equating imaginary parts both sides only,

v = x y

x y

x y

x y

z

x y

2 2

2 2

2 2

2 2

2

2 2

1

1

1

1 1

– ( – )

( )

–

( )

| | –1

( )

v = | | –1

( )

z

x y

2

2 21 ... (1)

z-plane

exterior

interior


( 78 )

The denominater of this fraction is essentially positive. From (1) it is clear that

(i) v = 0 (i.e., real axis of -plane) corresponds to |z|2 – 1 = 0 or |z| = 1

(boundary of the circle |z| = 1).

(ii) v > 0 (upper half of -plane) corresponds to |z|2 – 1 > 0 or |z| > 1 (exterior

of the circle |z| = 1).

(iii) v < 0 (lower half of -plane) corresponds to |z| < 1 (interior of the circle

|z| = 1).

This concludes the problem.

Example 7. Show that the transformation = 5 4

4 2

–

–

z

z transform the circle

|z| = 1 into a circle of radius unity in -plane and find the centre of the circle.

Solution : We have = 5 4

4 2

–

–

z

z from which 4z – 2 = 5 – 4z

or, z = 2 5

4 4

|z| = zz = 1 corresponds to

2 5

4 1

2 5

4 11

LNM

OQP

LNM

OQP

( ) ( ( )

or, 4 + ) = 16 (+ + + 1)... (1)

But = u + iv, = u – iv,

u2 + v2 and + = 2u.

Hence (1) becomes 4 (u2 + v2) + 25 + 10. 2u

= 16 [u2 + v2 + 2u + 1]

4u2 + 4v2 + 25 + 20u – 16u2 – 16v2 – 32u – 16 = 0

–12u2 – 12v2 – 12u + 9 = 0

12 [u2 + v2 + u – 9/12] = 0

u2 + v2 + u – 3/4 = 0 ... (2)

Comparing this with x2 + y2 + 2gx + 2fy + c = 0 we get centre as

(–g, –f) = – ,1

20

FHGIKJ and radius = (g2 + f2 – c)½, i.e., radius =

1

40

3

41

FHG

IKJ

½

.


( 79 )

Hence (2) represents circle with its centre as – ,1

20

FHGIKJ and of radius one in -plane.

Problem on special Transformation.

Example 8. Discuss the application of the transformation = z2 to the area in

the first quadrant of z-plane bounded by the axes and the circle |z| = a, |z| = b,

(a > b > 0).

Is the transformation conformal ?

Solution :

The given transformation is = z2.

Taking = Rei, z = Rei we get

Rei= r2 ei2

R = r2, = 2

In view of this, r = a R = a2 and 0

2 0

Now the quadrants |z| = a, 0

2

and |z| = b, 0

2in the z-plane,

are transformed into the semicircles || = a2, 0 and || = b2, 0

respectively of -plane.

From this it follows that the area/included in the first quadrant of z-plane is

transformed into the area included between the semi-circles of -plane as shown

in the figure.

The transformation = z2 is conformal.

For f (z) = d

dzZ

2 0 for any z in the given area.

w - plane


( 80 )

Example 9. To show that the transformation

= tan2

42

FHGIKJ

transforms the interior of units circle || = 1 into the interior of the parabola

r = sec2

2.

Solution : The transformation is = tan2t where t =

4z

= 1 2

1 2

– cos

cos

t

t ... (1)

2t = 2

4 2 2

1

2 2

z r e a ibi

, say

Then a =

2 2r cos

b =

2 2r sin

cos 2t = cos (a + ib) = cosa cosb – sina sin ib = cos a cos hb – i sin a sin hb.

Now (1) becomes

= 1

1

– cos cos sin sinh

cos cos – sin sin

a hb i a b

a hb i a hb

|| 1 ( – cos cos ) sin sin

( cos cosh ) sin sin

1

11

2 2 2

2 2 2

a hb a b

a b a h b

or, (1 – cosa cos hb)2 + sin2a sin h2b (1 + cos a cos hb)2 + sin2a sin2b

or, cos a cos hb 0, but cos hb = e eb b

–

20

Hence, cos a 0 or, a

2

or,

2 2 2r cos


( 81 )

or, r cos2

21

or r sec2

2

or r (1 + cos) 2

or,2

1r cos ............ (2)

Which represents interior and boundary of the parabola

r = sec2

2

or,

2

r = 1 + cos.

This concludes the problem.

Example 10. If = 2z + z2, prove that the circle |z| = 1 correspond to a

cardioid in -plane.

Solution : The given transformation is expressible as + 1 = (1 + z)2.

By writing + 1 = Rei, i.e., by taking the pole on -plane at = –1, we get

Rei = (1 + z)2 ..... (1)

Any point on the circle |z| = 1 is expressible as z = ei, By the transformation

(1), this is transformed to the curve Rei = (1 + ei)2

= e e eii

i

/–

/2 2 2

2

FHG

IKJ

LNMM

OQPP

= 4 cos2

2. ei = 2 (1 + cos) ei

or, Rei = 2 (1 + cos) ei

Equating the modulus and argument

R = 2 (1 + cos), =

R = 2 (1 + cos)

Thus the unit circle z = ei in z-plane corresponds to the cardioid R = 2 (1 +

cos) in -plane.

Example 11. Show that the transformation (z + i)2 = 1 maps interior of the

circle |z| = 1 in the z-plane on the exterior of the parabola in -plane.

Solution : Suppose z = rei, = Rei.

Now the given transformation is expressible as

z + i = 1


( 82 )

or, rei + i = 1

R e–i /2

Equating the real and imaginary parts

r cos = 1

2Rcos

r sin + 1 = –1

Rsin

2

1

21

1

2

2 2

2

R Rrcos sin

FHG

IKJ FHG

IKJ

1

12

22

R Rr sin

... (1)

Therefore the unit circle |z| = r = 1 is z-plane corresponds to the region of -

plane defined by 11 2

21

R Rsin

or,1

R = –2 sin

2

1

42

2

R sin

or,1

2 1R

( – cos )

This represent a parabola in -plane. Next we examine about the

correspondence between the interior of the circle and exterior of the parabola.

If we suppose |z| = r < 1, then the interior of the circle |z| = 1 corresponds to

the region -plane defined by

11 2

2

R Rsin

= r2 < 1, by (1)

or, 1

22R

sin

or,1

R < 2 (1 – cos)


( 83 )

or, R > 1

2 1( – cos )

This represents domain outside the parabola.

R = 1

2 1( – cos ),

i.e., 1

R = 2 (1 – cos)

This completes the proof.

2.13 Model Questions

1. Find the fixed points and the normal form of the following bilinear

transformations.

(a) = z

z – 2,

(b) = 3 1iz

z i

,

(c) = z

z

–1

1

(d) = ( ) –2 2

i z

z i,

(e) = z

z2 –

Is any of these transformations hyperbolic, elliptic or parabolic ?

2. Find the bilinear transformation which maps the points z = –2, 0, 2 into the

points = 0, i, – i respectively.

3. Find the bilinear transformation which maps the points z = 0, –1, into the

points = –1, –2, –i, i respectively.

4. Find linear maps for z = 0, –i, –1 and corresponding values of are = i, 1, 0.

5. I f z = 2 + 2, then show that || = 1 corresponds to a cardioid of z-plane.

6. Find bilinear transformation which maps 0, i, –i of z-plane to 1, –1, 0 of

-plane.

7. Prove that the relation = iz z

z i

4 transforms the real axis in the z-plane into


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a circle in the -plane. Find the centre and radius of the circle and the point in

the z-plane which is mapped on the centre of the circle.

8. Show that the transformation 5 4

4 2

–

–

z

z transforms the circle |z| = 1 into a

circle of radius unity in the w-plane and find the centre of this circle.

9. Discuss the application of the transformation w = iz

z i

1 the areas in the z-

plane which are respectively inside and outside the unit circles with centre at

the origin.

Answers

1. (a) k = 1

2ei

, map is loxodromic

(b) 1 1

2 – –

–

i z i

i which is the normal form. The transformation is

parabolic,

(c) k = –i = e–i

2, hence transformation is elliptic,

(d) u = 5 ei so that transformation is loxodromic.

(e) k = 1

2 > 0 and u 1. Hence transformation is hyperbolic.

2. = i 2

2 3

FHG

IKJ

z

z–, 3. =

iz

z

– 2

2 4. = –

( )

–

i z i

z

1 5. r = 2 (1 + cos) which

represents cardioid. 6. =z i

z i

–3, 4u 0 represents a bilinear transformation.

7. The real axis is z-plane is transformed on to a circle of plane whose

centre is transformed on to a circle of plane whose centre is at –7

8

i and

radius = 9

8 and the centre of the circle corresponds to the point z =

i

4 in z-plane.


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8. Centre – ,1

20

FHGIKJ and of radius one in -plane.

9. If v > 0 i.e., upper of plane corresponds to |z| > 1 i.e., exterior of the

circle |z| = 1.

If v < 0, i.e., lower half of plane corresponds to |z| < 1 i.e interior of the

circle |z| = 1.

2.14. References :

1. Functions of a Complex Variable — Goyal & Gupta.

2. Complex Analysis — by J.B. Conwary.

3. Complex Function Theory — D. Sarason.

4. Function of a Complex Variable — J.N. Sharma.

5. Functions of a Complex Variable — M.L. Khanna.


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UNIT - 2 BILINEAR TRANSFORMATION