COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions

ME 210 Applied Mathematics for Mechanical Enginee rs

Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 1/39

COMPLEX ANALYSIS

Complex Numbers

Complex numbers are useful in the fields such as:

� Solutions of some types of linear differential equations

� Electrical circuit analyses

� Inverse transformations

� Solutions of field problems

� etc.



The general form of a complex number z, is z = x + i y or z = a + i b

where

x = Re (z) is a real number and called as the real part of z .

y = Im (z) is a real number and called as the imaginary part of z .

Another representation of a complex number is its polar form which is obtained by

applying usual rectangular to polar coordinate transformation as

x = r cosθ & y = r sinθ

z = r [cosθ + i sinθ] = r ei θ = r θ

i = -1

r = |z| = = mod(z) is called as the modulus (magnitude) of z22 yx +

xy

θ = arctan = arg(z) is called as the argument (angle) of z

Using EulerUsing Euler’’s identity s identity eeiiθ = = coscosθ + i + i sinsinθ



y

x

y

z

x 0

r

θ

z–plane

The figure below denotes the geometric representation of a complex number in a

plane called complex z-plane , or Argand diagram , defined by a Cartesian

coordinate system whose abscissa is used to represent the real part of z, and

ordinate is used to represent the imaginary part of z.

[ ]

2 2

1

z = x + i y

r = x + y x = r cos( )

y = tan y = r sin( )

x

z = r cos( ) + i r sin( )

z = r cos( ) + i sin( )

θ

θ θ

θ θθ θ

−

z = r eiθ

z = r θ



Note the following :

Geometrically, the argument θ is the directed angle measured in radians from

positive x–axis in counterclockwise direction. For z = 0, this angle is undefined.

For a given z ≠ 0, this angle is determined only up to integer multiples of 2π.

The value of θ that lies in the interval –π < θ ≤ π is called as the principle value

of the argument of z and is denoted by Arg(z) , with capital A. Thus, – π < Arg(z) ≤ + π

It becomes extremely important to consider the quadrant of the z–plane in which

the point z lies when determining the value of the argument by using the above

equation.

The modulus r, as the distance of the point z to the origin, is a non-negative quantity;

i.e., r ≥ 0



Example:

Determine the modulus and argument of the following complex numbers.

z = - 1 + i ⇒

z = 3 - 4 i ⇒

z = - 5 - 12 i ⇒

211r =+= )(135rad2.3564

3π1

1arctanθ o≅=

−=

5169r =+= )53.1(rad0.92734

arctanθ o−−=

−=

1314425r =+= )112.6(rad1.9655

12arctanθ o−−=

−−=



There are two fundamental rules for the manipulation of complex numbers:

1. A complex number z = x + i y is zero iff both its real and imaginary parts are zero;

i.e., z = 0 iff x = 0 & y = 0

it follows that two complex numbers z1 = x1 + i y1 and z2 = x2 + i y2 are equal

iff both their real and imaginary parts are equal;

i.e., z1 = z2 iff x1 = x2 & y1 = y2

Example :

What are the x and y values that satisfy the equation

(x2y – 2) + i (x + 2xy – 5) = 0

x2y – 2 = 0 and x + 2xy – 5 = 0 → x = 1 and y = 2x = 4 and y = 1/8



2. Complex numbers obey the ordinary rules of algebra

(a + i b) ± (c + i d) = (a ± c) + i (b ± d)

(a + i b) (c + i d) = (a c – b d) + i (a d + b c)

(a + i b)2 = a2 – b2 + i (2 a b)

with the addition that

i2 = – 1 , i3 = – i , i4 = +1 , i5 = i

Addition and Subtraction

z = z1 ± z2 = (x1 ± x2) + i (y1 ± y2)

� Easier to perform in Cartesian form

� Similar to addition and subtraction of vectors in a plane

Triangle Inequality:

Length of any one side of a triangle is less than or equal to the sum of the lengths

of other two sides. That is r ≤ r1 + r2

, i6 = – 1 , i7 = – i , ..... modularity



Minus, Conjugate, Addition, and Subtraction in Polar Coordinates

x

y

z = a + b i

_

z = a - i b- z = - a - b i

x

y

z1

z2z1 + z2

- z2

z1 - z2

conjugate



Multiplication

z = z1 z2 = (x1 x2 – y1 y2) + i (x1 y2 + y1 x2) = r1r2 θ1 + θ2

r = r1 r2

θ = θ1 + θ2

� Easier to perform in polar form

� Not similar to multiplication of vectors in plane

( )( )

( )

1 1 1 1

2 2 2 2

1 2 1 2 1 2 1 2

z = r cos( ) + i sin( )

z = r cos( ) + i sin( )

z z = r r cos( + ) + i sin( + )

θ θ

θ θ

θ θ θ θ



Division

1

2

1 2 1 2 1 2 1 22 2 2 22 2 2 2

z x x + y y y x - x yz = = + i

z x + y x + y

r = r1 / r2

θ = θ1 – θ2

� Easier to perform in polar form

� Not similar to any operation of vectors in plane

[ ][ ]

[ ]

1 1 1 1

2 2 2 2

1 11 2 1 2 2

2 2

z = r cos( ) + i sin( )

z = r cos( ) + i sin( )

z r = cos( - ) + i sin( - ) , r 0

z r

θ θ

θ θ

θ θ θ θ ≠

212

1 θθ rr −=



Complex conjugate

The complex conjugate of a complex number z = x + i y is defined as_

z = x - i y

Another practical way to obtain division of two complex numbers is to use the

conjugate of the denominator:

Example:

2 2 2 2

2 2

a + i b a + i b c - i d a c + b d b c - a d = = + i

c + i d c + i d c - i d c + d c + d

c + d 0

≠



Note the following:

Letz = x + iy → = x – iy

then(x + iy)2 + 2(x – iy) = x2 – y2 + i2xy + 2x – i2y = –1 + i6

which gives the following two real equationsx2 – y2 + 2x = –1 → y = ± (x+1)

2xy – 2y = 6 → y(x–1) = 3whose solutions are found as

x = 2 & y = 3 and x = –2 & y = –1Hence, the required solution for z is found as

z1 = 2 + i3 and z2 = – 2 – i

2 z + z

Re(z)x2x = z + z ==→i2

z zIm(z)yi2y = z z

−==→−

222 |z| = y+ x=z z 2121 zz)z(z ±=± 2121 zz)z(z =2

1

2

1

zz

zz =

Example :

Solve the following (find z): i6 1 z2 z2 +−=+

z



Integer powers of complex numbers (de Moivre’s formu la)

The idea of product of two complex numbers can be extended to the product of n complex numbers

If all the complex numbers multiplied are the same, the above expression gives an important result

zn = rn einθ = rn [cosθ + i sinθ]n = rn [cos(nθ) + i sin(nθ)]

where n is either an integer or a rational number (that is, n=p/q where p and q areintegers) with the condition that z ≠ 0 for n = –1.

For r = 1, the expression reduces the form

(cosθ + i sinθ)n = cos(nθ) + i sin(nθ)

called as the de Moivre’s formula named after Abraham de Moivre (1667-1754)

∑

=…=…= =∏∏

=

+++

=

n

1kk

n21

θin

1k

k)θ...θi(θ

n21n21

n

1k

k erer rr z zzz



Example: Find z5 for z = 1 + i 3

Solution 1 : Taking the power of a binomial

( )55 2 3 4 5z = 1 + i 3 = 1 + 5 ( i 3) + 10 ( i 3) + 10 ( i 3) + 5 ( i 3) + ( i 3)

= 1 + i 5 3 - 30 - i 30 3 + 45 + i 9 3

= 16 - i 16 3

Solution 2 : Using de Moivre’s formula

( )5 55

5 i 5 3

5i 5 5 3

1 3z = 1 + i 3 = 2 + i = 2 cos + i sin = 2 e

2 2 3 3

5 5 = 2 cos + i sin = 2 e

3 3

1 3 = 32 - i = 16 - i 16 3

2 2

π

π

π π

π π



Let

z = wn where w = R eiφ = ?

thenthen

z = Rn (cos nφ + i sin nφ) = Rn einφ

Hence

r (cos θ + i sin θ) = Rn (cos nφ + i sin nφ) or r eiθ = Rn einφ

De Moivre’s formula may also be employed to evaluate the nth root of a complex

number z = r (cos θ + i sin θ)

n 1/n nr = R or R = r = r

Therefore

1/n 1/n θ + 2 k θ + 2 k z = r cos + i sin , k = 0, 1, ..., n-1

n nπ π

and cos θ = cos nφ & sin θ = sin nφ

1n , 1, 0, k,n2kπθ

or2kπθn −…=+=φ+=φ

w =w =



Example: Find 3 8

Let z = 8 + i 0

1/3 1/3 0 + 2 k 0 + 2 k z = 8 cos + i sin , k = 0, 1, 2

3 3π π

1/31z = 8 cos(0) + i sin(0) = 2

1/32

0 + 2 0 + 2 1 3z = 8 cos + i sin = 2 - + i = - 1 + i 3

3 3 2 2π π

1/33

0 + 4 0 + 4 1 3z = 8 cos + i sin = 2 - - i = - 1 - i 3

3 3 2 2π π



Note that all solutions have a common modulus of 2, but their arguments differ

from each other, and they are located on a circle of radius of 2 about the origin of

the z–plane, equally spaced around it with an incremental angle of 2π/3 as

illustrated

(81/3)2

Im

Re

(81/3)1

i2

0 2

2π/3

z–plane

(81/3)3

4π/3



Example: Find i

Let i /2z = 0 + i = 1 e π

1/2 1/2 /2 + 2 k /2 + 2 k z = 1 cos + i sin , k = 0, 1

2 2π π π π

i /41

2 2z = 1 cos + i sin = e = + i

4 4 2 2ππ π

i 5 /42

5 5 2 2z = 1 cos + i sin = e = - - i

4 4 2 2ππ π



Complex Functions

Let z and w are two complex variables defined as z = x + i y and w = u + i v

where x, y, u, and v are real variables.

If, for each value of z in some portion of the complex z–plane, one or more value(s)

of w are defined, then w is said to be a complex function of z.

w = f(z) = u + i v = f(x + i y) = u(x,y) + i v(x,y)

This complex functional relationship between z and w may be regarded as a

complex mapping or complex transformation of points P within a region in

the z–plane (called as the Domain ) to corresponding image point(s) Q within a

region in the w–plane (called as the Range ).



w = f(z) mapping

y

x

domain

P

z–plane v

u

range

Q

w–plane

Complex Functions as Complex Mapping



Example:

Find the ranges, R, in the complex w–plane of the following complex functions

corresponding to their domains, D, in z–plane. Plot D and R regions.

Use the standard notation, z = x + i y = r eiθ & w = u + i v

(a) w = f(z) = i z , D: Re(z) ≥ 0

w = i z = i (x + i y) = – y + i x

u(x,y) = – y & v(x,y) = x

Re(z) = x ≥ 0 => v(x,y) ≥ 0



(b) w = f(z) = 3 z – π , D: – π ≤ Re(z) ≤ π

w = 3 z – π = (3x – π) + i 3 y

u(x,y) = 3 x – π & v(x,y) = 3 y

–π ≤ x ≤ π => – 4 π ≤ u(x,y) ≤ 2 π

(c) w = f(z) = z2 , D: |z| ≤ 1 and 0 ≤ Arg(z) ≤ π /4

w = z2 = (r ei θ)2 = r2 ei 2θ

|w| = r2 = |z|2 & Arg(w) = 2 Arg(z)

|z| ≤ 1 => |w| ≤ 1

0 ≤ Arg(z) ≤ π /4 => 0 ≤ Arg(w) ≤ π /2



In dealing with complex functions, it is possible to distinguish the following two cases.

� Complex valued functions of a real variable

� Complex valued functions of a complex variable

Complex valued functions of a real variable

In this case, to each value of a real variable, t (a ≤ t ≤ b), one or more complex

value(s) of z are assigned, which can be shown as z(t) = x(t) + i y(t) = f(t)

Some examples for this type of complex function are:

z = ei t (0 ≤ t ≤ 2 π) => z = cos(t) + i sin(t) => x(t) = cos(t) & y(t) = sin(t)

z = t + i t2 (for all t) => x(t) = t & y(t) = t2

z = (1 – i ) (1 + i 2 ) (t ≥ 0) => x(t) = (1 + 2 t) & y(t) = tt t t t



Example:

For the slider-crank mechanism shown in the figure, it is desired to represent the

position of the point Q as a (complex valued) function of the horizontal position t

(a real variable) of the slider P as z(t) = x(t) + i y(t) (h – r ≤ t ≤ h + r)

y

x

P

Q

θ

r h

β

t

This form can conveniently be used to represent the parametric equations of planar

curves in complex z–plane.



Note that as P moves on a straight line, Q moves on a circle.

For every position of P denoted by t, it is possible to find two positions for Q denoted by z.

Therefore, the relationship between the positions of P and Q can be considered as

a mapping of points lying on a straight line to points lying on a curve (circle).

By using either θ or β angle, the coordinates x and y of Q can be related to the

position t of P.

y

x

P

Q

θ

r h

β

t



y

x

P

Q

θ

r h

β

t

x(t) = r cos(θ) = t – h cos(β)

y(t) = r sin(θ) = h sin(β)

Since x2 + y2 = r2 => [t – h cos(β)]2 + [h sin(β)]2 = r2

2 2 2t + h - rcos(β) =

2 h t

22 2 2t + h - rsin( ) = 1 -

2 h tβ

±



The parametric representation of the position of Q in the complex z–plane becomes

( )

−+−±

+−=

−+−±

−+−=

t2

rhtth4i

t2

rht

ht2

rht1hi

t2

rhtt)t(z

222222222

2222222

y

x

P

Q

θ

r h

β

t



For r = 1 and h = 4, x(t) and y(t) are shown for 3 ≤ t ≤ 5 in the Figure.

-1.00

-0.75

-0.50

-0.25

0.00

0.25

0.50

0.75

1.00

3.0 3.5 4.0 4.5 5.0Real variable t

y(t)

x(t)

Position of Q as a function of Slider Position



Complex valued functions of a complex variable

In this case, to each value of a complex variable z in a domain D of the z-plane,

one or more complex values w are assigned, which can be shown as w = f(z)

For a given complex variable z, a complex value w may be obtained regardless

of the function, f, itself being a real function or a complex function.

Some examples for this type of complex function are:

w = i z → i (x + i y) = – y + i x → u = – y & v = x

w = z2 → (x + i y)2 = (x2 – y2) + i 2 x y → u = x2 – y2 & v = 2 x y

w = z1/4 → (r eiθ)1/4 = R eiΦ → R = r1/4 & Φ = (θ + 2 k π) / 4

w = ez → ex + iy = ex (cos y + i sin y) → u = ex cos y & v = ex sin y



Note that once w = f(z) is known, it is a straightforward algebra to obtain u & v (or R

& f) in terms of x & y (or r & q).

However, if u(x,y) & v(x,y) are given in turn, to express w as a function of z is not a

trivial problem at all.

� In fact, it may even not have a solution.

� If there is a solution, then a suitable manipulation must be carried out in order

to come up with the correct w = f(z) expression.

Example :

Given w = u(x,y) + i v(x,y) = (x2 + x – y2 + 1) + i y (2x + 1) express w as a function

of z = x + i y

Rearranging gives

w = x2 + x – y2 + 1 + i 2 x y + i y

= x2 + i 2 x y – y2 + x + i y + 1

= (x + i y)2 + (x + i y) + 1

= z2 + z + 1



Some elementary functions of z

Exponential function: Using the basic definition of exponential function for real

variables, one gets

( )1

n 2 3 4z z z z z

f z = e = = 1 + z + + + +…n! 2! 3! 4!

∞

=∑k

u(x,y) = ex cos(y) & v(x,y) = ex sin(y) Mod(ez) = |ez| = ex & Arg(ez) = y

Note that ( )1

n 2 3 4- z (-z) z z z

f z = e = = 1 - z + - + - …n! 2! 3! 4!

∞

=∑k

( )1

n 2 3 4i z (i z) z i z z

f z = e = = 1 + i z - - + - …n! 2! 3! 4!

∞

=∑k

( )1

n 2 3 4- i z (- i z) z i z z

f z = e = = 1 - i z - + + - …n! 2! 3! 4!

∞

=∑k



Hyperbolic functions:

cosh(z) = (ez + e–z)/2 = cosh(x) cos(y) + i sinh(x) sin(y)

sinh(z) = (ez – e–z)/2 = sinh(x) cos(y) + i cosh(x) sin(y)

If z = 0 + iy => cosh(iy) = cos(y) sinh(iy) = i sin(y)

Familiar laws for the hyperbolic functions:

cosh2(z) – sinh2(z) = 1

cosh(z1 + z2) = cosh(z1) cosh(z2) + sinh(z1) sinh(z2)

sinh(z1 + z2) = sinh(z1) cosh(z2) + cosh(z1) sinh(z2)

cosh(2 z) = cosh2(z) + sinh2(z) = 1 + 2 sinh2(z) = 2 cosh2(z) – 1

sinh(2 z) = 2 sinh(z) cosh(z)



Trigonometric functions:

cos(z) = (eiz + e–iz)/2 = cos(x) cosh(y) – i sin(x) sinh(y)

sin(z) = (eiz – e–iz)/2i = sin(x) cosh(y) + i cos(x) sinh(y)

If z = 0 + iy => cos(iy) = cosh(y) sin(iy) = i sinh(y)

Familiar laws for the trigonometric functions:

cos2(z) + sin2(z) = 1

cos(z1 + z2) = cos(z1) cos(z2) – sin(z1) sin(z2)

sin(z1 + z2) = sin(z1) cos(z2) + cos(z1) sin(z2)

cos(2 z) = cos2(z) – sin2(z) = 1 – 2 sin2(z) = 2 cos2(z) – 1

sin(2 z) = 2 sin(z) cos(z)



Logarithmic function:

The logarithm of z = r eiθ that is defined implicitly as the function w = ln(z) which

satisfies the equation z = ew or r eiθ = eu + i v = eu eiv

Hence, eu = r or u = ln(r), and v = θ

Thus, w = u + i v = ln(r) + i θ = ln |z| + i arg(z)

If the principal argument of z is denoted by Arg(z), then this equation can

be rewritten as ln(z) = ln |z| + i [Arg(z) + 2 k π] k = 0, ±1, ±2, …

This indicates that complex logarithmic function is infinitely multi–valued.

For k = 0, the part (branch) of the logarithmic function is called as the principal

value .



Familiar laws for the logarithms of real quantities all hold for the logarithms of

complex quantities in the following sense:

ln(z1z2) = ln(z1) + ln (z2)

ln(z1/z2) = ln(z1) – ln (z2)

ln(zk) = k ln(z) k = 0, ±1, ±2, …

Example:

Compute w(z) = ln(1 – i)

ln(z) = ln |z| + i [Arg(z) + 2 k π] k = 0, ±1, ±2, …

( ) L2,1,0,k,2kπ4π

i2lni)ln(1 ±±=

+−+=−

( ) ( ) ( ) L,4

9πi2ln,

47π

i2ln,4π

i2lni)ln(1 −+−=−

–iπ/4

ln( )2

i7π/4

i15π/4

–i9π/4

i23π/4

–i17π/4

w-plane

Re(w)

Im(w)



General Powers of z

w(z) = zc , c is any number, real or complex

( )0c ln(r) + i + 2 k c ln(z)z = e = e , k = 0, 1, 2, ...c θ π ± ±

If c = n:( ) ( )0 0 0

ln rn ln(r) + i i n i n n ln(z)

i 2 n k

z = e = e = e e = r e

e = 1

nn nθ θ θ

π

Because for all k’s

If c = m/n: ( )

( ) ( )

0i m/n + 2 k / /

/0 0

z = r e

m m = r cos + 2 k + i sin + 2 k

n n

k = 0, 1, 2, ...

m n m n

m n

θ π

θ π θ π

, n, n--11



If c is an irrational number, not expressible in the form m/n

( )

( )( ) ( )( )0i c + 2 k

0 0

z = r e

= r cos c + 2 k + i sin c + 2 k

k = 0, 1, 2, ...

c c

c

θ π

θ π θ π

± ±

If c is complex, i.e., c = a + i b

( )

( ) ( )

( ) ( )( )( )( )

0

00

0

(a + i b) ln(r) + i + 2 k

i b ln(r) 4 a + 2 k a ln(r) - b + 2 k

0a ln(r) - b + 2 k

0

z = e

= e e

cos b ln(r) + a + 2 k = e

+ i sin b ln(r) + a + 2 k

k =

c θ π

θ πθ π

θ πθ π

θ π

0, 1, 2, ... ± ±

+



Example:

Find all possible values of (1)i

Summary: zc is

single valued if c is an integer

n-valued if c = 1/n

n-valued if c = m/n

infinite valued if c is real and irrational

infinite valued if Im(c) ≠ 0

→→ r = 1 , r = 1 , θθoo = 0 , c = i = 0 , c = i

Example:

Find all possible values of (i)i →→ r = 1 , r = 1 , θθoo = = ππ/2/2 , c = i , c = i

...2,1,0,k,eee(1) 2kπ)]2ki(0i[ln(1)ln(1)ii ±±==== −π++

...2,1,0,k,eee(i) 21)4k)]2k2

i(i[ln(1)ln(i)ii ±±==== π+−π+π+ /(



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COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions