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 z 0  S 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.
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
Fig 3
Closed
Neither
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.
S
z1
Fig 4
z2
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 S,
is the set of S together with all of its boundary. Thus
S  S  B(S.)
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.
Open Ball
It is open.
Prove it.
Is it connected?
Yes
Fig 5
Problems

The graph of |z – (1.1 + 2i)| < 0.05 is shown
in Fig 6 It is an open set.
Is it connected?
Yes
Fig 6
Problems

The graph of Re(z)  1 is shown in Fig 7.It is
not an open set.
It is a closed
set.
Prove it.
Is it connected?
Fig 7
Yes
Problems

Fig 17.10 illustrates
Both aresome
open. additional open
sets.
Prove it.
Are they connected?
Yes
Fig 8
Problems
Both are open.
Prove it.
Are they connected?
Fig 9
Yes
Problems
It is open.
Prove it.
Is it connected?
Yes
Fig 10
Problems

The graph of |Re(z)|  1 is shown in Fig 11.It
is not an open set.
X= -1
Is it connected?
Fig 11
No
Polar Form

Polar Form
Referring 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
Express1  3i in polar form.
Solution
See Fig 13 that the point lies in the fourth
quarter.
r  z  1  3i  1  3  2
 3
5
tan  
,  arg( z ) 
1
3
5
5 

z  2 cos  i sin 
3
3

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.
w  f ( z )  u ( x, y )  iv( x, y )
Graph of Complex Function
y
w = f(z)
v
x
z-plane
domain of
definition
Fig 15
u
range
w-plane
Example 1
Find the image of the line Re(z) = 1 under f(z) =
z2.
f ( z )  z 2  ( x  iy) 2
Solution
2
2
u ( x, y )  x  y , v( x, y )  2 xy
y  v / 2, then u  1  v / 4
2
Now Re(z) = x = 1, u = 1 – y2, v = 2y.
Fig 16
Complex Exponential Function
Def : e  e
n
z
x iy
 e (cos y  i sin y)
x
Evaluate e1.7+4.2i.
e1.7 4.2i  e1.7 (cos 4.2  i sin 4.2)
 2.6873  4.7710 i
Solution:
We can easily prove
z1
z z
z z e
z z
e e  e , z  e , e z z  e 0  1
e
e iy  1
1
2
1
2
1
2
2
You prove them.
Periodicity
e
e e
z  i 2
z
i 2
 e (cos 2  i sin 2 )  e
z
z
e
e e
 e (cos 2n   i sin 2n  )  e ,
where n is any integer
z  i 2n 
z
i 2n 
z
Show that
i)  no z s.t. e z  0 ii) e z  1  z  2n  i
iii) e
z1
e
z2
 z 2  z 1  2n  i iv )the function
is not onto. v) no one-to-one and vi) find images
of both axes under this function.
z
Polar From of a Complex Number
Revisited
z  r (cos   i sin  )  re
i
Arithmetic 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 :
z 1  x 1  iy 1  r1e
i1
z 2  x 2  iy 2  r2e

 x
i2
 
  i y
z 1  z 2  x 1  iy 1  x 2  iy 2

1
 x2
z 1z 2  r1e
i1
 r1r2e
magnitudes multiply!
1
 y2
r e 
i2
2
i (1 ( 2 ))
phases add!


Easier with normal
form than polar form
Easier with polar form
than normal form
For a complex number z2 ≠ 0,
z1
z2

r1e
r2e
i1
i2

magnitudes divide!
z1
r1

z2
r2
r1
r2
e
i (1 ( 2 ))

r1
r2
e
i (1 2 )
phases subtract!
z  1  ( 2 )  1   2
Some Exercises (Example2)
Ex. 1: Express cos 3 and sin 3
of cos  and
sin 
in terms of powers
cos 3  i sin3  (cos  i sin ) 3
 (cos3   3 cos sin2  )  i ( 3 sin cos2   sin3  )
 cos 3  cos3   3 cos sin2   4 cos3   3 cos
 sin3  3 sin cos2   sin3   3 sin  4 sin3 
Ex. 2:
zn 
1
in
 in

e

e
 (cosn  i sinn )  (cosn  i sinn )
n
z
 2 cos n
1
z  n  2i si nn
z
n

find the nth roots of unity
z n  1  e i 2 k
ze
i 2 k / n
 z1, 2, 3.....,n  1, e
i 2 / n
,........,e
i 2 ( n  1 ) / n
k is an integer
Ex. : Find the solutions to the equation 3
z 1
z  e i 2k / 3
Imz
z2  e i 2 / 3
 z1  e i 0  1
z1  1
 z2  e i 2 / 3  1 / 2  i 3 / 2
 z3  e i 4 / 3  1 / 2  i 3 / 2
Ex. :
2
The three roots of z 3  1 are 1,  , 
  3  1 and 1     2  0
Rez
z3  e i 4 / 3
Proof:
  e i 2 / 3 ,  2  e i 4 / 3  e  i 2 / 3
 1     2  1  e i 2 / 3  e  i 2 / 3  1  2 cos(2 / 3)  0
Ex: Solve the polynomial equation
f ( z )  z 6  z 5  4z 4  6z 3  2z 2  8z  8  0
try z  1  f ( z  1)  0  z  1 is a root
 ( z 5  4 z 3  2 z 2  8)(z  1)  0
 ( z 3  2)(z 2  4)(z  1)  0
(1)
z  2  2e
3
i 2 k
k  0,1,2,........
z2  4  0  z  i 4
 z 4  2i
(2)
z  21 / 3 e i 2 k / 3
 z 5  2i
 z1  2
1/ 3
 z2  21 / 3 e i 2 / 3  21 / 3 ( 1 / 2  i 3 / 2)
 z3  21 / 3 e i 4 / 3  21 / 3 e  i 2 / 3  21 / 3 (1 / 2  i 3 / 2)
(3)
z  1  0  z6  1
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.
y
domain
f(z)
x
v
range
u
Example 4
Describe 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
  3 
2i
2e
 8e


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.
i 2
3
w = z3
y
v
8
2
-2
x
2
-8
8
-8
Fig 17
u
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)= z
i)
D={(x,x)|x is a real number}
ii)
D={|z|<4| z is a complex number}
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
x  iy  eu iv  eu (cos v  i sin v)  eu cos v  ieu sin v
We have
and also
x  eu cos v, y  eu sin v
e2u  x 2  y 2  r 2 | z |2 , u  log e | z |
y
tan v  , v    2n ,   arg z, n  0,  1,  2,...
x
Logarithm of a Complex Number
DEFINITION
For z  0, and  = arg z,
ln z  log e | z | i (  2n ) , n  0,  1,  2, 
Example
6
Find the values of (a) ln (−2) (b) ln i, (c) ln (−1 – i ).
Solution
(a)   arg( 2)   , log e | 2 | 0.6932
ln(2)  0.6932  i (  2n )
(b)   arg(i ) 


2
, log e 1  0
ln(i )  i (  2n )
2
5
(c)   arg( 1  i )  , log e | 1  i | log e 2  0.3466
4
5
ln(1  i )  0.3466  i (  2n )
4
Example 7
e z  3  i.
Find all values of z such that
z  ln( 3  i ), | 3  i | 2, arg( 3  i ) 
Solution

z  ln( 3  i )  log e 2  i (  2n )
6

 0.6931  i (  2n )
6

6
Principal Value
Ln z  log e | z | iArg z

Since Arg z  ( ,  ] is unique, there is only
one value of Ln z for which z  0.
Example 8

The principal values of example 6 are as
follows.
(a) Arg(2)  
Ln(2)  0.6932  i


(b) Arg(i ) 
2
, Ln (i )  i
2
5
(c) Arg(1  i ) 
is not the principal value.
4
3
Let n  1, then Ln(1  i )  0.3466  i
4
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:
ln(z 1z 2 )  ln z 1  ln z 2
ln(
z1
)  ln z 1  ln z 2
z2
Example 9
Suppose z1 = 1 and z2 = -1. If we take ln z1 =
2i, ln z2 = i, we get
ln( z1z2 )  ln(1)  ln z1  ln z2  3i
z1
ln( )  ln(1)  ln z1  ln z2  i
z2