Visualization of Visualization of Complex IntegraComplex Integrationtion

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By By Shigeki Shigeki OgoseOgose ((生越生越 茂樹茂樹))

ATCM ATCM 2019 2019 in in 楽山楽山

§1. Basic Calculus in the complex plane

2

)

( , ), same as

Definition of the

a point in

| | distance between O and

arg( ) angle of the radius

Complex plane

( ,

O

R

x y R

z x

z z

z

iy y

z

x

⇕

⋯

⋯

1 2 1 2

1 2 1 2

2

1

1

1

2

Multiplication&Division

| | | ||

arg(

'To multiply ' is

around O by arg( )

|

) arg( ) arg( )

to

'rotate z &

magnify the distance from O by | | times.'

Example.

z z z z

z

z

z

z z z

z

z

'To multiply ' is ' to rotate by

( )

' 90

a bi i b ai

i

1 1 1 2 2 2

1 2 1 1 2 2

1 1 1

1 2 1 2

2

( ) ( )

Addition&Subtraction

W

Same as the sum of 2 vectors.

hen , ,iy x iy

z z x y x y i

x x x x

y y y

z x z

y

⇕

1 2z z

Basic.ggb

§2. Complex function as a vector field

, )

( )corresponds to a vector field by

A complex function ( ) (

assigning a vector to each p

) ( ) ( ( ),

oin

( )

t)

.(

Cf z u z v z i R

u zz

v z

z u z v z

2

2

2

2

1,

1 1

2 2

2 2 4,

2(1 )

Vector field of ( ) .

( 1) 1, ( )

( ) 4, ( ) 4

4

Example.

f z z

f f i i

f i

f f i i

f i

i i

i

⋮ ⋮

Vector field.ggb

§3. Definition of complex integration

1 2 3

+1 ( 1,2,3 ).

, ,

is the maximum of

is a piec

| |

ewise smooth oriented curve in the complex plane. , , are

points on .

Whe

converges to 0 as increases, thenn the line integ ral of

n

k k k n

C z z

n

z

z z

z

C

⋯

⋯

+1

1 1

( ) over is

( ) ( )( ) lim ( ) z

n n

k k k k kC n

k k

f z C

f z dz f z z z f z

Line Integral Contour Integral

1 . Then is

called a closed curve, and the integral is called a contour in

The initial point and

tegral.

Contour integra

the termi

l is often

nal point z can

expressed as

coincid

( ) .

en

C

C

f z dz

z

� template@auto.ggb

Basic properties of complex integration

1 2 1 2

(2) Adding (or partitioning) of path

( ) ( ) ( )C C C Cf z dz f z dz f z dz

1 2 is closed. C C

1 2 (semicircle)C C C

They correspond to these properties in integral calculus.

( ) ( ) ,

( ) ( ) ( )

f x dx f x dx

f x dx f x dx f x dx

(1) Orientation reversal. ( ) means the reversed curve of .

( ) = ( )

C C

C C

f z dz f z dz

§4. Fundamental theory of complex integration

3

1

2 22

3

3 2 2 2

) ( = 1, 2, 3, ),

1 1 1 1 1 1 , = + , = + , z , , ,

2 2 2

These integrals are independent from the path

Example

Because of (z

1

.

3

.

3

s

2

z z

n

z z

n

z

nz n

dz dz dz zdz z dzz

z z z z

z

⋯

⋯ ⋯

is a domain in the complex plane. ( ) and ( ) are complex functions in .

If for any curve ( ) which joins & z,

'( ) ( ) everywhere in

, C

D f z F z D

F z f Dz D

( ) ( ) ).

The value of integral only depends on & . i.e. It is independent from .

Therefore if (i.e. if is closed),

( ) (

z

CF z Ff z dz F z

z C

z C

( ) 0cf z dz �

template@auto.ggb rectangle@auto.ggb

1 1Note: There is no primitive for . Because can change with a path from .

z

dz C zz z

1overz.;basic.ggb

§5. Integration of 1/(z-a) around a circleWhen is on the cirlcle of radius , centered at a, anticlockwised, then is parameterized by

sin )

(

sin cos ) ( )

co

( s i

dz r

z C r z

z

i d i z a d

a r

(!)dz

idz a

⋯ rotate90

| | |

( )

It doesn't depend on .

|

r

dz z a d

z

d

a dz

r

fig2

0

2

0

line integral along (fig2) is,

1

T

Especially when is a single circle,

herefore,

21

C

C

C

dz id iz a

C

dz id iz a

�

fig1

fig3

+2

0'

1 isn't uniquely defined by .

Line integral along ' (fig3) is

1 +2

It can be ( 2 ) ( 0

But (

, 1, 2 ) f sameor the .

)z

a r

C

dz zz a

C

dz id i iz a

i n n

F z

z

⋯

1overz;2circles.ggb

1overz@circle.ggb

§6. Cauchy ’s Integral theorem

A singular point is a point where (

If ( ) is analytic everywhere inside a curve (i.e. if there is no singular point inside )

( ) 0c

f

f z C C

f z dz

�) is not analytic. z

C

( )

Therfore for any closed curve

which doesn't

1 is

encl

an

Example

ose ,

alytic except .

01

.

f Oz

d

z

C

z

O

z

�

　

1

2

2

| | 1

's theorem can be applied to .

But it can't be applied to . For example,

01

is not implied by 's theorem.

It is implied by the existance of a primitive.

zCa

Caushy C

ushyz

C

dz

�( ) is analytic in the shaded area.f z

1overz;cut plane@auto.ggb

§7. Deformation of the path(Changing end points.)

1

2

1 22

1( ) is analytic inside a closed curve except . Devide into and make 2 small circles

around , all anticlockwise, 4 pathes as below. Then applying 's t

,

,

heorem,

a b

k

f z C z C C C

C z z T Caushy

z

C

1 2 3 4

1

1 2

21 2 43

( ) 0

Because line integral is additve,

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0

a b

a b

C T C T C T C T

C CC T T TC T

f z dz

f z dz f z dz f z dz f z dz f z dz f z dz f z dz f z dz

�

� �

1 2

1 2 3 4

Because ( ) is continuous,

( ) ( ) 0, ( ) ( ) 0 , ( ) ( ) ( )

Therefore,

( ) ( ) ( )

a b C

C C C

T T C CT T

f z

f z dz f z dz f z dz f z dz f z dz f z dz f z dz

f z dz f z dz f z dz

�

� � �

§8. Cauchy ’s integral formulaaAssume ( ) is analytic inside a closed curve C, points a,z are inside C. Take a small circle

o . Then by a path deformationf radius around , antic ,

lo

ckwise

Cf z

r a

( ) ( )

(1

( ) ( ) ( ) '( ) (on )

Therefore

)

Because ( ) is analytic, when is small

)

(

aC C

a

f z f zdz dz

z a z a

f z r

a

f zd

z a

f z f a z f a C

⋯� �

( )( ) ( ) '(

'

)) 1

(a a a aC C C C

az dz f a dz f a dz

z a z a

f a z f a

� � � �

What will happen when

0r

2 ( ) 0· '( )

Because for any , (1) holds true,

if a f a

r

0

" 's integral formula" is normally written as the upper r

( ) ( ) lim 2 ( )

That is,

( ) 1 ( ) 2 ( ) ( )

2

aC Cr

C C

Cauchy

f z f zdz dz if a

z a z a

f z f zdz if a f a dz

z a i z a

� �

� �

1 2

( ) ( )t's easy to extend this to , ,

( )( ) ( z )( z ) ( z )

ight form.

I n

f z f z

z a z b z z z ⋯

⋯

�2

2 2

:semicircle of radius 2

+segment , anticlockwise

= ( 1)( 4) 3C

C rAB

BA

zI dz

z z

�Example1

2

1 2

2

2 2

Integrand ( ) has 2

,

singular points & 2 inside .( )( )( 2 )( 2 )

Make small circles around , 2 , anticlockwise. Then by a pat

h deformation,

( 1)( 4)

i i Cz i z i z i

zf z

C

z i

C i i

z

z z

2 2

1 2

2 2

2 2

2 2 2 2

2 2

2 2

21

/ { } / { }( 4)

( 1)( 4) ( 1)( 4)

Here ( ) can be written in two ways.

( ) ( 2 ) = =

( 1)( 4) 2

( ) (That is: ( )

( 1)

C C C

z

z zdz dz dz

z z z z

f z

z z z i z z i

z z z i z i

zf z

f

z i

f

z

� � �

1

2 2

1 2 2

1 2

2 2

12 2

2

), where ( ) , ( )

2 ( )( 4) ( 1)( 2 )

Since ( ) , ( ) are analytic, by

2 (

's for

) 2 · , ( 1

mula,

)( 4) ·3

2 3C

z z zf z f z

z i z i z z z i

f z f z

zdz i f

Cauch

i iz z

z

y

i

i

�

2

2 2

2 22

(2 )

( 3) 4

22 (2 ) 2 ·

( 1)( 4) ·

2

3

3 3

3

C

idz i f

z ii i

z

I

�

template@auto.ggb

( )B r ( )A r

2 2

2 2 2 2

( 1)( 4) 3 1)( 4) 3(C

x

x

zI dz dx

z z x

�

�

2

2 2

( )

li

This integral holds true for any ( 2) , thus

m( 1)( 4) 3

But for (cos sin ) on the semicircle , when 1

Cr

Continued

zdz

z z

z r r

r

i AB

≫

�

� �

2

2 2 2

2

2 2 0

2 2

2 2 2

2

A 2

1 | ( ) | , | | ,

·

( ) | | · 0( 1)( 4)

It follows that

lim =lim( 1)( 4) ( 1)( 4)

1rAB AB

Cr r B

rf z dz rd

r r r

zdz f z dz rd

z z r

z zdz dz

z z

x

z z

����2

2 2

1)( 4)( 3d

xx

x

What will happen when

r( )B r ( )A r

Detour. Euler ’s formula

2 3 4 5 7

3

6

0

2 4 5 6 7

For any z ,

1! 2! 3! 4! 5! 6! 7!

Therefore for R,

( ) ( ) ( ) ( ) ( ) ( )

is defined as a power series:

12! 3! 4! 5

! 6! 7

z

nz

n

i

C e

z z z ze z

n

i i i i i ie

z z z

i

⋯

62 4 3 5 7

+!

1 + +2! 4! 6! 3! 5! 7!

cos sin

i

i

⋯

⋯ ⋯

, ,

cos sin ,

· (cos sin )

For

i

x iy x iy x

e

e

x y R

i

e e e y i y

It's very easy. This will work fine

.

sum

[Check it w

( sequence(

ith Geo

/ !, ,0,50)

G br

)

e a]

nxe x n nEuler.ggb

2 2

is the same curve

as example 1, but 1

cos

1 1C

iz C

re

e xI dz dx

z x e

�Example 2

1

1

2 2

( ) around , then,

by a path defo

has a singular point inside . Make a circle ( )( )

rm

1 1

We separate ( ) to

ation

,

C C

iz

iz iz

f ze

i C Cz i z i

e edz dz

z z

f z

i

� �

1

11

1

1

2

analytic and divergent part.

( ) ( ) , ( )

( )( )

Because ( ) is analytic in , by 's formula,

( )

/ ( )

1

iz

C

iz iz

iz

ee f z ez if z f z

z i z i z i z i z i

f z C

e f zdz d

z

Cau y

z i

ch

�

1

1

1

2 2

2

2 ( ) 22

That is

1 1

It follows that when nears (same example 1 as )

cos + sin =

1

C

C C

iz iz

ez i f i i

i e

e eI dz dz

z z e

r

x i xI dx

x e

�

� �

2

cos

1

xdx

x e

template@auto.ggb

(Fresnel's integral)

Example 3

Example 4

§9.miscellaneous

00

sin

2

iz

h C g d

e xdz dx

xz

�

h

c

g

d

c

2 2 2

0 0

2 cos( 0 ) sin( = )

4g C h

z dz x dx x dxe

�

g

h

(integral on c is too tiny to see)

fresnel@manual.ggb

e^(iz)over z.ggb

c

g

h

d

Example 5

0

1 ( 1)log1

integrate complex function on the real axis

,( )

s s xx s x

s e x dxe

4s 4s i 4 2s i 4 3s i 4 4s i

( function)

Example 6

1When the plane is cut by the negative real axis,

is uniquely defined and analytic in the cut plane.

1 /0

z

h C g d

Log z zdzLog z dz

�

C

Log z@auto.ggb

gamma@auto.ggb

2 2

0 0( integral)

2 cos( sin( = )

4) FresnelI x dx x dx

Example 3

2

(g,h are segments from O to A,B, C is an arc from A to B) ( ) 0

It is proven that (

Since ( ) is analytic everywhere, by Caushy's theor

pro

em,

fs a

o

z

g C hf z dz

f z e

�

2

0(Gauss integral)

re not easy )

lim ( ) 0 & lim ( ) 2

It follows that

lim ( ) lim ( ) (1)2

On h, is

x

r r

r r

C g

h g

f z dz f z dz e dx

f z dz f z dz

z

⋯

2 2 2 2

2 2

0

2 2

0

represented as (0 ) , then cos( ) sin( ), 2 2

lim ( ) (cos sin ) 22

1 1

1

(cos sin )

z it

r h

z t t r e t i t dz dt

f z

i ie

idz t i t dt

t i t d

「 」

2 2

0 0

2 2· (1 )

1 2 4

By comparing the real and imaginary part on both side, we get

2 cos sin =

4

t ii

x dx x dx

1

1 1 1

-1

1

1

-1

1

1

2 1 3 2 1

1

0

) ( ) )( ) )( )

lim )( )

= ) ( )

= ) ( ) ) ( )

Since '( ) ( ), w

) ( )

= )

hen ,

( '( (

( )

(

(

( ( (

(

k k

k

n n

n

k k k k k k k

n

k k kn

k

n

k k

k

z

F z z z z z

z

F z f z z

F z F z f z

f z

F z

F z F z F z

F z

z

F z

F z F z F z

F z

⋯

Rough proof of fundamental theorem

Rough proof of Cauchy’s theorem

1 2

Devide the interia of a closed anticlockwised curve into

many small regions which only share the smooth borders

with neighbors, and orientalize these borders anticlockwise

and name the bord r

, ,

e

nD D

C

D ⋯

1

curve along .

Then on the border between & , the line integral along the border

is canceled. Which happens in all the borders, and thus

( ) (

)

k k

k k

Cf

D as C

D D

z dz f z

�1

is very small,

( )

In each ,however, because ( ) is analytic , when

) ( ) (where )

Therefore,

( ) '( is a fixed poin

t in

k

n

Ck

k k

k k k k kf z

dz

D

f a

f z D

z a Daf a

�

( ) ) ( )

( ) '(

( ) ) (

'(

0

)

k kk k k

C C

k k kC C

f z dz z a dz

dz z a

f a f a

f a df a z

� �

� �