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Complex Variables
Mohammed Nasser Acknowledgement:
Steve Cunningham
Open Disks or Neighborhoods
Definition. The set of all points z which satisfy the inequality |z – z0|<, where is a positive real number is called an open disk or neighborhood of z0 .
Remark. The unit disk, i.e., the neighborhood |z|< 1, is of particular significance.
1
Fig 1
Interior Point
Definition. A point is called an interior point of S if and only if there exists at least one neighborhood of z0 which is completely contained in S.
Sz 0
z0
S
Fig 2
Open Set. Closed Set.
Definition. If every point of a set S is an interior point of S, we say that S is an open set.
Definition. S is closed iff Sc is open. Theorem: S` S, i.e., S contains all of its limit points S is closed set.
Sets may be neither open nor closed.
Open ClosedNeither
Fig 3
Connected
An open set S is said to be connected if every pair of points z1 and z2 in S can be joined by a polygonal line that lies entirely in S. Roughly speaking, this means that S consists of a “single piece”, although it may contain holes.
SS
z1 z2
Fig 4
Domain, Region, Closure, Bounded, Compact
An open, connected set is called a domain. A region is a domain together with some, none, or all of its boundary points. The closure of a set S denoted , is the set of S together with all of its boundary. Thus .
A set of points S is bounded if there exists a positive real number r such that |z|<r for every z S.
A region which is both closed and bounded is said to be compact.
S
)(SBSS
Open Ball
Fig 5
It is open.
Prove it.
Is it connected?
Yes
Problems
The graph of |z – (1.1 + 2i)| < 0.05 is shown in Fig 6 It is an open set.
Fig 6
Is it connected?
Yes
The graph of Re(z) 1 is shown in Fig 7.It is not an open set.
Problems
Fig 7
It is a closed set.
Prove it.
Is it connected?
Yes
Fig 17.10 illustrates some additional open sets.
Problems
Fig 8
Both are open.
Prove it.
Are they connected?
Yes
Problems
Fig 9
Both are open.
Prove it.
Are they connected?
Yes
Problems
Fig 10
It is open.
Prove it.
Is it connected?
Yes
The graph of |Re(z)| 1 is shown in Fig 11.It is not an open set.
Problems
Fig 11
Is it connected?
No
X= -1
Polar Form
Polar FormReferring to Fig , we have
z = r(cos + i sin ) where r = |z| is the modulus of z and is the argument of z, = arg(z). If is in the interval − < , it is called the principal argument, denoted by Arg(z).
Fig 12
Example
SolutionSee Fig 13 that the point lies in the fourth quarter.
form.polar in 31 Express i
35
sin3
5cos2
35
)arg(,1
3tan
23131
iz
z
izr
Fig 13
Review: Real Functions of Real Variables
Definition. Let . A function f is a rule which assigns to each element a one and only one element b , . We write f: , or in the specific case b = f(a), and call b “the image of a under f.” We call “the domain of definition of f ” or simply “the domain of f ”. We call “the range of f.” We call the set of all the images of , denoted f (), the image of the function f . We alternately call f a mapping from to .
Real Function
In effect, a function of a real variable maps from one real line to another.
f
Fig 14
Complex Function
Definition. Complex function of a complex variable. Let C. A function f defined on is a rule which assigns to each z a complex number w. The number w is called a value of f at z and is denoted by f(z), i.e., w = f(z). The set is called the domain of definition of f. Although the domain of definition is often a domain, it need not be.
Remark Properties of a real-valued function of a real variable are
often exhibited by the graph of the function. But when w = f(z), where z and w are complex, no such convenient graphical representation is available because each of the numbers z and w is located in a plane rather than a line.
We can display some information about the function by indicating pairs of corresponding points z = (x,y) and w = (u,v). To do this, it is usually easiest to draw the z and w planes separately.
),(),()( yxivyxuzfw
Graph of Complex Function
x u
y v
z-plane
w-plane
domain ofdefinition
range
w = f(z)
Fig 15
Example 1
Find the image of the line Re(z) = 1 under f(z) = z2.
Solution
Now Re(z) = x = 1, u = 1 – y2, v = 2y.
xyyxvyxyxu
iyxzzf
2) ,( ,) ,(
)()(22
22
4/1 then ,2/ 2vuvy
Fig 16
: (cos sin )n z x iy xDef e e e y i y
Evaluate e1.7+4.2i.
Solution:
i
iee i
7710.46873.2
)2.4sin2.4(cos7.12.47.1
1
1 2 1 2 1 2
2
0
We can easily prove
, , 1
1
zz z z z z z z z
z
iy
ee e e e e e
ee
Complex Exponential Function
You prove them.
Periodicity
2 2 (cos2 sin2 ) ,
where n is any integer
z i n z i n z ze ee e n i n e
2 2 (cos2 sin2 )z i z i z ze ee e i e
1 2
2 1
Show that
i) no z s.t. 0 ii) 1 2
iii) 2 )the function
is not onto. v) no one-to-one and vi) find images
of both axes under this function.
z z
z z
e e z n i
e e z z n i iv
Polar From of a Complex Number Revisited
(cos sin ) iz r i re
Arithmetic Operations in Polar FormArithmetic Operations in Polar Form
The representation of z by its real and imaginary parts is useful for addition and subtraction.
For multiplication and division, representation by the polar form has apparent geometric meaning.
Suppose we have 2 complex numbers, z1 and z2 given by :
1
2
1 1 1 1
2 2 2 2
i
i
z x iy re
z x iy r e
1 2 1 1 2 2
1 2 1 2
z z x iy x iy
x x i y y
1 2
1 2
1 2 1 2
( ( ))
1 2
i i
i
z z re r e
r r e
Easier with normal form than polar form
Easier with polar form than normal form
magnitudes multiply! phases add!
For a complex number z2 ≠ 0,
1
1 2 1 2
2
( ( )) ( )1 1 1 1
2 2 22
ii i
i
z re r re e
z r rr e
magnitudes divide!phases subtract!
2
1
2
1
r
r
z
z 2121 )( z
Ex. 1:
Express and in terms of powers of and
3cos 3sincos sin
332
323
3223
3
sin4sin3sincossin33sin
cos3cos4sincos3cos3cos
)sincossin3()sincos3(cos
)sin(cos3sin3cos
i
ii
Ex. 2:
n
ninnineez
z ininn
n
cos2
)sin(cos)sin(cos1
niz
z nn sin2
1
Some Exercises (Example2)
Ex. : Find the solutions to the equation 13 z Imz
Rez
11 z
3/22
iez
3/43
iez 2/32/1
2/32/1
1
3/43
3/22
01
3/2
iez
iez
ez
ez
i
i
i
ki
Ex. : The three roots of are13 z2,,1
011 23 and
Proof:
0)3/2cos(2111
,
3/23/22
3/23/423/2
ii
iii
ee
eee
find the nth roots of unity
nninin
nki
kin
eezez
ez/)1(2/2
.....,3,2,1/2
2
,........,,1
1
k is an integer
Ex: Solve the polynomial equation
088264)( 23456 zzzzzzzf
rootaiszzfztry 10)1(1
0)1)(4)(2(
0)1)(824(23
235
zzz
zzzz
(1) kiez 23 22 ,........2,1,0k
)2/32/1(222
)2/32/1(22
2
2
3/13/23/13/43/13
3/13/23/12
3/11
3/23/1
ieez
iez
z
ez
ii
i
ki
(2)
iz
iz
izz
2
2
404
5
4
2
(3) 101 6 zz
Example 3
Describe the range of the function f(z) = x2 + 2i, defined on (the domain is) the unit disk |z| 1.
Solution: We have u(x,y) = x2 and v(x,y) = 2. Thus as z varies over the closed unit disk, u varies between 0 and 1, and v is constant (=2).
Therefore w = f(z) = u(x,y) + iv(x,y) = x2 +2i is a line segment from w = 2i to w = 1 + 2i.
x
y
u
vf(z)
domain
range
Example 4Describe the function f(z) = z3 for z in the semidisk given by |z| 2, Im z 0.
Solution: We know that the points in the sector of the semidisk from Arg z = 0 to Arg z = 2/3, when cubed cover the entire disk |w| 8 because
The cubes of the remaining points of z also fall into this disk, overlapping it in the upper half-plane as depicted on the next screen.
iiee 2
33
282
2-2 x
y
u
v
8-8
-8
8
2
w = z3
Fig 17
If Z is in x+iy form
z3=z2..z=(x2-y2+i2xy)(x+iy)=(x3-xy2+i2x2y+ix2y-iy3 -2xy2)
=(x3-3xy2) +i(3x2y-y3)
If u(x,y)= (x3-3xy2) and v(x,y)=(3x2y-y3), we can write
z3=f(z)=u(x,y) +iv(x,y)
Example 5
f(z)=z2, g(z)=|z| and h(z)=
i) D={(x,x)|x is a real number}
ii) D={|z|<4| z is a complex number}
z
Draw the mappings
Logarithm Function
Given a complex number z = x + iy, z 0, we define
w = ln z if z = ew Let w = u + iv, then
We have
and also
vieveviveeiyx uuuivu sincos)sin(cos
2 2 2 2 2| | , log | |
tan , 2 , arg , 0, 1, 2,...
uee x y r z u z
yv v n z n
x
veyvex uu sin ,cos
For z 0, and = arg z,
DEFINITION
,2,1,0,)2(||logln nnizz e
Logarithm of a Complex Number
Example 6
Find the values of (a) ln (−2) (b) ln i, (c) ln (−1 – i ).
)22
()ln(
01log ,2
)arg( )(
)2(6932.0)2ln(
6932.0|2|log ,)2arg( )(
nii
ib
ni
a
e
e
)24
5(3466.0)1ln(
3466.02log|1|log ,4
5)1arg( )(
nii
iic ee
Solution
Example 7
Find all values of z such that
Solution
.3 iez
)26
(6931.0
)26
(2log)3ln(
6)3arg(,2|3|),3ln(
ni
niiz
iiiz
e
Principal Value
Since Arg z is unique, there is only one value of Ln z for which z 0.
zizz e Arg||logLn
( , ]
Example 8
The principal values of example 6 are as follows.
2
)(Ln ,2
)(Arg )(
6932.0)2(Ln
)2(Arg )(
iiib
i
a
iin
ic
43
3466.0)1(Ln then ,1Let
value.principal not the is 4
5)1(Arg )(
Important Point
Each function in the collection of ln z is called a branch. The function Ln z is called the principal branch or the principal logarithm function.
Some familiar properties of logarithmic function hold in complex case:
1 2 1 2
11 2
2
ln( ) ln ln
ln( ) ln ln
zz z z
zz z
z
Example 9
Suppose z1 = 1 and z2 = -1. If we take ln z1 = 2i, ln z2 = i, we get
izzzz
izzzz
212
1
2121
lnln)1ln()ln(
3lnln)1ln()ln(