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(