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
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 �
[email protected] [email protected]
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
§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 [email protected]
§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
�
( )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
(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)
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
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
� �
� �