Chapter 7 Functions of a Complex Variable II
October 21 Residue theorem
7.1 Calculus of residues
Suppose an analytic function f (z) has an isolated singularity at z0. Consider a contour
integral enclosing z0 .

C
f ( z )dz  
C


 a ( z  z ) dz   a  ( z  z )
n
n  
n
0
n  
n
C
0
n
dz
 ( z  z ) n 1 z '
0

a
 0,
n  1
n

n
n

1
an  ( z  z0 ) dz  
z'
C

z'
a1 ln( z  z0 ) z '  2ia 1 , n  1

C
z0
f ( z )dz 2ia 1  2i Re s f ( z0 )
The coefficient a-1=Res f (z0) in the Laurent expansion is called the residue of f (z) at z = z0.
If the contour encloses multiple isolated
singularities, we have the residue theorem:

C
f ( z )dz 2i  Re s f ( z n )
z0
z1
n
Contour integral =2i ×Sum of the residues
at the enclosed singular points
1
Residue formula:
To find a residue, we need to do the Laurent expansion and pick up the coefficient a-1.
However, in many cases we have a useful residue formula(Problem 6.6.1):
For a pole of order m,

1
d m 1
Re sf ( z0 ) 
lim m 1 ( z  z0 ) m f ( z )
(m  1)! z  z0 dz

Particular ly, for a simple pole ,
Re sf ( z0 )  lim ( z  z0 ) f ( z ).
z  z0
Proof :


1
d m 1
1
d m 1 
m
lim m 1 ( z  z0 ) f ( z ) 
lim m 1  an ( z  z0 ) n  m
(m  1)! z  z0 dz
(m  1)! z  z0 dz n   m

d m 1 
nm
lim m 1  an ( z  z0 )
 lim  an (n  m)( n  m  1)  (n  2)( z  z0 ) n 1  a1 (m  1)!
z  z 0 dz
z  z0
n m
n  1

1
d m 1
 a1  Re sf ( z0 ) 
lim m 1 ( z  z0 ) m f ( z )
(m  1)! z  z0 dz

2
Proof Method #2 :
f ( z) 

 a (z  z )
n m
n
( z  z0 ) f ( z ) 
m
n
0

 a (z  z )
n m
n
nm
0
Because ( z  z0 ) m f ( z ) is analytic, by Taylor expansion



1
dk
( z  z0 ) f ( z )   bk ( z  z0 ) , bk  lim k ( z  z0 ) m f ( z ) .
k! z  z0 dz
k 0
m
k


1
d m 1
Also bk  ak  m . Pick up k  m  1 gives us a1 
lim m 1 ( z  z0 ) m f ( z ) .
(m  1)! z  z0 dz
We actually proved that ther e is a way to find all the a coefficien ts :
ak  m


1
dk
 lim k ( z  z0 ) m f ( z ) , k  0.
k! z  z0 dz
3
Fact :
If lim ( z  z0 ) f ( z ) exists and is not equal to 0, then
z  z0
1) f ( z ) has a simple pole at z  z0 .
2) Re sf ( z0 )  lim ( z  z0 ) f ( z ).
z  z0
Proof :
Let f ( z ) 

 a ( z  z ) , then
n
n n
n
0
lim ( z  z0 ) f ( z )  lim
z  z0
z  z0

 a (z  z )
n  
n
0
n 1
exists and  0
an  0 for n  2.
  f ( z ) has a simple pole at z  z0 .


lim ( z  z0 ) f ( z )  a1  0 Re sf ( z0 )  lim ( z  z0 ) f ( z ).
 z  z0
z  z0
 
Examples :
1  1
1
1

lim ( z  i )  2   , therefore 2
has a simple polse at z  i with a residue of .
z i
z  1 2i
z 1
2i

1 
1

lim  z 


1
,
therefore
has a simple polse at z  0 with a residue of  1.

z
z
z 0
1

e
1

e


4
Residue at infinity:
Stereographic projection:
Residue at infinity:
Suppose f (z) has only isolated singularities,
then its residue at infinity is defined as
1
1
Re s f () 
f
(
z
)
dz


f ( z )dz
~
2i C
2i C
  Re s f ( zn )
~
C
n
Another way to prove it is to use Cauchy’s integral theorem. The contour integral for a
small loop in an analytic region is


f
(
z
)
dz

0


2

i
Re
s
f
(
z
)

Re
s
f
(

)
n


Cz0
n

One other equivalent way to calculate the residue at infinity is
let z 1 / Z 1
1
Re s f () 

~ f ( z ) dz

C

2i
2i C0
 1  1 
f    2 dZ  Res
 Z  Z 
 1  1 
f   2  at Z  0.
 Z  Z 
1
Example : f ( z )  , Re s f ()  1.
z
By this definition a function may be analytic at infinity but still has a residue there.
5
Read: Chapter 7: 1
Homework: 7.1.1
Due: October 28
6
October 24 Cauchy principle value
Cauchy principle value:
Suppose f (z) has an isolated singularity z0 lying on a closed contour C. The contour
integral  f ( z )dz is then not well defined. To solve this problem, we can remove a small
C
segment of the contour for a distance of  on each side of the singularity and create a
new contour C(). We define the Cauchy principle value of C f ( z )dz as
P  f ( z )dz  lim  f ( z )dz (If the limit exists.)
 0 C (  )
C
Let us first see on what condition this limit exists. We draw a
semicircle path S with radius  around z0. Suppose contour
C=C()+S does not enclose any singularity, then

C ( )

S

z0
S
C
f ( z )dz   f ( z )dz  0.
f ( z )dz 
 a  (z  z )
n  

 0 
 an 
n  
C()
S

0
n
0
S
i n
i
n

dz (let z  z0  ei )
(e ) ie d 

 0 
 ani n1 
n  
0
ei ( n 1) d
 i ( n 1)  0 
0, n  odd  1.
i ( n 1)( 0  )
i ( n 1) 0
i ( n 1) 0
i ( n 1)
e
e

e
e
e

1


 0  i ( n 1)


  2i i ( n 1) 0

i (n  1)
i (n  1)
, n  even
0 e d   i(n  1) 0
 n  1 e

  , n  1.
7



C ( )
f ( z )dz    f ( z )dz  ia1 
S
The condition for lim

 0 C ( )
 an
n  even
2 n 1 i ( n 1) 0
 e
n 1
f ( z )dz to exist is then an  0 for negative even n.
Under this condition,
P  f ( z )dz  lim
1
f ( z )dz  ia1  i Res f ( z0 )   2iRes f ( z0 ).
2

 0 C ( )
C
More about Cauchy principle values:
1) If C originally encloses isolated singularities, then
n
P  f ( z )dz  i Re sf ( z0 )  2i  Re sf ( zi ).
C
i 1
2) We assumed that the contour is smooth at z =z0. If not the value should be i Res f ( z0 ).
3) It is easy to remember: when the singularity is on the contour, it contributes
half as if it were inside the contour.
4) Similar definitions exist for open contour integrals. Example:
b
x0 
b
P  f ( x)dx  lim  
f ( x)dx   f ( x)dx 
a
x0 
 0   a

Also for infinity integration limits. Example:

P  f ( x)dx  lim


a
a   a
f ( x)dx
8

1
dx
  ( x 2  1)( x  1)
R
1
I  lim P 
dx
 R ( x 2  1)( x  1)
R 
Example: P 
 R


1
1
1

 lim  P 
dx

dz

dz

  ( z 2  1)( z  1) 
 R ( x 2  1)( x  1)
 ( z 2  1)( z  1)
R 




1
 P 2
dz
C ( z  1)( z  1)
(an  0 for negative even n.)
 iRes f (1)  2iRes f (i )
1
Res f (1)  lim ( z  1) f ( z )  ,
z 1
2
i 1
Res f (i )  lim ( z  i ) f ( z ) 
.
z i
4
1
i 1

I  i   2i 
 .
2
4
2
C
i
-i
1
R
9
Read: Chapter 7: 1
Homework: 7.1.4
Due: November 4
10
October 26 Evaluation of definite integrals -1
7.1 Calculus of residues
Cauchy’s integral theorem and Cauchy’s integral formula revisited:
(in the view of the residue theorem):

Analytic function : f ( z )   am ( z  z0 ) m  f ( z0 )  f ' ( z0 )( z  z0 )  
m 0
1)

2)
f ( z0 )
f ( z)
f ( z)
dz  2if ( z0 )
 f ' ( z0 )    

C
z  z0
z  z0 z  z0
C
f ( z )dz  2iRes f ( z0 )  0

f ( n ) ( z0 )
f ( z)
f ( z)
m  n 1
dz  2ia n  2i

 a ( z  z0 )
3)
n 1
C 
n!
z  z0 n1 m0 m
z  z0 
3' )
f ' ( z0 )
f ( z0 )
f ( z)
 . It is a pole of order n  1.


z  z0 n1 z  z0 n1 z  z0 n
According to the residue formula, its residue at z  z0 is
f ( z )  f ( n ) ( z0 )
d ( n 1) 1 
1
n 1
.

lim
 z  z 0 
n 1 
z  z 0 ( n  1  1)! dz ( n 1) 1
n!
 z  z0  

11
L’Hospital’s rule:
If lim f ( z )  lim g ( z )  0 or  , and lim
z  z0
( or  )
z  z0
( or  )
z  z0
( or  )
f ' ( z)
f ( z)
f ' ( z)
exists, then lim
 lim
.
z

z
z

z
0
0
g ' ( z)
g ( z ) ( or  ) g ' ( z )
( or  )
f ( z)
f ' ( z)
f ' ' ( z)
f (n) ( z)
Repeatedly , lim
 lim
 lim
   lim ( n )
until the limit exsists.
z  z0 g ( z )
z  z0 g ' ( z )
z  z0 g ' ' ( z )
z  z0 g
( z)
Example :
e z  z 1
e z 1
ez 1
lim
 lim
 lim
 .
2
z 0
z

0
z

0
z
2z
2 2
I. Integrals of trigonometric functions :

2
0
f (sin  , cos  )d
C
This can be done by using a contour integral around the unit circle.
r=1
dz
z 1/ z
z 1/ z
, sin  
, cos  
.
iz
2i
2
  z 1/ z z 1/ z  1 
 z  1 / z z  1 / z  dz
f
,
,
  2i  Res  f 
 .
2  iz
2
i
2
 2i

 iz 
unit circle

z  e i , dz  ie i d , d 

2
0
f (sin  , cos  )d  
C
12
Example 1 :
2
1
1
dz
dz
d  
 2i  2
0 2  cos 
C
C z  4z 1
z  1 / z iz
2
2
z 2  4 z  1  0  z   2  3 (inside the circle) ,
I 
C
z-
z+
r=1
z   2  3 (out of the circle).


1
1
1
Res f ( z  )  lim  z  z  


z  z
z  z z  z  z  z 2 3

1
2
I  2i  2i 

.
2 3
3
Example 2 :
4
2
 z  1 / z  dz 1 z  2 z  1
2
I   cos d   (of course)   
 
dz

0
C
z3
 2  iz 4i C
1 1 2
1

   3   z dz   2i  2   .
4i C  z
z
4i

2
2
C
r=1
13
Example 3 :
d
,  is real and |  | 1. (From our textbo ok.)
0 1   cos 
1
dz 2
dz
2
dz
I 
 
 
C
 z  1 / z  iz i C z 2  2 z  1 i C ( z  z  )( z  z  )
1  


 2 
1 1
We have 2 simple poles, z 2  2 z /   1  0  z    
1  2 .
I 
2

C
z-
z+
r=1

| z  z  | 1, | z  || z  | z  is in the circle, z  is out of the circle.


1
1

1
Res f ( z  )  lim  z  z  


.

2
z  z
(
z

z
)(
z

z
)
z

z
2
1 

 



2

1
2
I   2i 

.
2
i
2 1  2
1 
14
Example 4 :
sin d
, a is real and | a | 1.
0 1  a sin 

z  1 / z  / 2i dz 1 z 2  1
1
1
1 z2
I 
 
dz  
dz
C 1  a  z  1 / z  / 2i iz
i C z  az 2  2iz  a
ia C z ( z 2  2iz / a  1)
i
We have 3 simple poles, z0  0, z 2  2iz / a  1  0  z   1  1  a 2 .
a
| z  z  | 1, | z  || z  | z  is in the circle, z  is out of the circle.
I 
2


z+
z- C
z0


1 z2
1
Res f (0)  lim  z

.

z 0
 z ( z  z  )( z  z  )  z  z 
r=1


1 z2
1  z 2
Res f ( z  )  lim  z  z  
.

z  z
z ( z  z  )( z  z  )  z  ( z   z  )

1  1 1  z 2 
1  z z
1  (1)
 
Res f (0)  Res f ( z  )   

z  z z  z  z ( z  z ) i
  2i

1 1 a2  
1 a2 
a
 a




1 
a2
1 a
2
 1 a
2
,
1
a2
2a
I   2i 

ia
1 1 a2 1 a2 1 1 a2 1 a2




15
Read: Chapter 7: 1
Homework: 7.1.7,7.1.8,7.1.10
Due: November 4
16
October 31 Evaluation of definite integrals -2
7.1 Calculus of residues
II. Integrals along the whole real axis:


f ( x)dx

Assumption 1: f (z) is analytic in the upper or lower half of the complex plane, except
isolated finite number of poles.



R
f ( x)dx  lim   f ( z )dz   f ( z )dz    f ( z )dz   lim  f ( z )dz  lim  f ( z )dz
 R C

R  
R  
 
  R
Condition for closure on a semicircular path:
lim

R  
 lim
f ( z )dz  lim


R  0


R  0
f ( R ei )iR ei d  lim


R  0
f ( R ei )iR ei d
f ( R ei ) Rd    lim Rf max   0  lim f ( z ) ~
R 
R 
1
z1
∩
,   0.
R
Assumption 2: when |z|, | f (z)| goes to zero faster than 1/|z|.
Then,



f ( x)dx  lim

R  C
f ( z )dz  2i  Residues of f ( z ) on the upper half plane.
17
Example 1 :

dx
 x 2  1
I 
1
on the upper half plane
z2 1


1
 2iRes f (i )  2i lim  z  i 
 .

z i
( z  i )( z  i ) 

I  2i  Residues of

dx


arctan
x
 .

 x 2  1
Or 
Example 2 :
I 


x
dx
2
a

2 2
, a  0.
I  2i  Residues of
z
1
2
a

2 2
on the upper half plane


1

2
 2iRes f (ia )  2i lim  z  ia 
'

.
2
2
3
z  ai
( z  ai ) ( z  ai )  2a

18
III. Fourier integrals:


f ( x)eikx dx, k  0.

Assumption 1: f (z) is analytic in the upper half of the complex plane, except isolated finite
number of poles.




f ( x)eikx dx  lim
R  C
f ( z )eikz dz  lim

R  
f ( z )eikz dz
∩
Jordan’s lemma:
If lim f ( z )  0, then lim

R  
z 
f ( z )eikz dz  0 with k  0.
R
Proof :
For any given   0 there exists M so that when z  M we have f ( z )   . Let R  M , then


f ( z )e dz    e dz  
ikz

ikz

 2R 
 /2
0
e
 kR sin
d  2R 

0
 /2
0
e
e
ikR cos  kR sin

iR e d  R  e  kR sin d
i
0
 kR 2 / 
Since  can be any given number, lim
1  e  kR 

d  2R 

(1  e  kR ) 
2kR / 
k
k

R  
f ( z )eikz dz  0.
Assumption 2 : lim f ( z )  0.
z 
Then,



f ( x)eikx dx  lim

R  C
f ( z )eikz dz  2i  Residues of f ( z )eikz on the upper half plane.
19
Example 1 :
e ikx dx
I  2
, k  0, a  0.
 x  a 2
eikz
I  2i  Residues of 2
on the upper half plane
z  a2

   ka
eikz
 2iRes f (ia )  2i lim  z  ia 
 e .
z ia
( z  ia )( z  ia )  a


Question: How about



f ( x)e ikx dx, (k  0) ?
Answer: We can go the lower half of the complex plane using a clockwise contour.



f ( x)e ikx dx  lim

R  C
f ( z )e ikz dz  2i  Residues of f ( z )e ikz on the lower half plane.
Example 2 :
e ikx dx
I  2
, k  0, a  0.
 x  a 2
e ikz
I  2i  Residues of 2
on the lower half plane
z  a2

   ka
e ikz
 2iRes f (ia )  2i lim z  ia 
 e .
z   ia
(
z

ia
)(
z

ia
)

 a

20

Question: How about  f ( x) cos kxdx,




f ( x) sin kxdx, (k  0) ?
Answer:






f ( x) cos kxdx  Re 






f ( x)eikx dx,  f ( x) sin kxdx  Im 


f ( x)eikx dx.



1 
1 
ikx
ikx
f ( x) cos kxdx   f ( x) e  e dx,  f ( x) sin kxdx   f ( x) eikx  e ikx dx.

2 
2i 
Example 3 :

cos kxdx
, k  0, a  0.
 x 2  a 2
ikx
 e dx
  ka
I  Re  2

e .
 x  a 2
a
I 
Example 4 :

sin x
dx.
0
x
ix
 e dx
1
I  Im P 
,

2
x
ix
 e dx
 eix 
P
 i Res f (0)  i lim  x   i .

x 0
x
 x 
I  P
I

2
.
21
Read: Chapter 7: 1
Homework: 7.1.11,7.1.12,7.1.13,7.1.14,7.1.16
Due: November 11
22
November 2 Evaluation of definite integrals -3
7.1 Calculus of residues
IV. Rectangular contours: Exponential and hyperbolic forms
e x  e x  2i is a periodic function. We may use rectangular contours
and hope that the integral reappears in some way on the upper
contour line.
i
R
Example 1 :
e ax dx
I 
, 0  a  1.
 1  e x
a ( x  2i )
R e
 R e ax dx 2 e a ( R iy )idy
e az dz
dx 0 e a (  R iy )idy 

lim
 lim  



R  iy
( x  2i )
 R  iy 
0
R
2

R   1  e z
R   R 1  e x
1 e
1 e
1 e



 R e ax dx 2ia  R e ax dx 
e az
2ia
  (1  e ) I  2i  Residues of
 lim  
e 
in the rectangle
z
R 1 ex 
R   R 1  e x
1

e



e az 
 2iRes f (i )  2i lim  z  i 
 2i  (e ia )

z
z  i
1 e 

 2i  e ia

I

1  e 2ia
sin a
23
Example 2 :
eiaxdx
I 
, a  0.
  cosh x
ia ( R  iy )
ia ( x  2i )
2 e
R e
0
 R e iaxdx
eiaz dz
idy
dx
eia(  R iy )idy 
lim
 lim  



0 cosh( R  iy )
R cosh( x  2i ) 2 cosh(  R  iy ) 
R   cosh z
R   R cosh x



iax
 R e dx 
 R eiaxdx
 2a
  (1  e  2a ) I
 lim  
e 
R cosh x
R   R cosh x


e iaz
 2i  Residues of
in the rectangle
cosh z
i
i3 

 2i Res f ( )  Res f (
)
2
2





i  eiaz 
i3  eiaz  
 2i  lim  z  

  limi 3  z 

i
2
cosh
z
2
cosh
z
z

z







 
 2 
2
 2i  (ie  a / 2  ie 3a / 2 )  2 (e  a / 2  e 3a / 2 )
i3/2
i/2
R
2 (e  a / 2  e 3a / 2 )

I

1  e  2a
cosh( a / 2)
This integral can also be done by using the line y=i and the fact cosh( z  i )   cosh z.
24
V. Sector contours
For functions involving x n , because x n  ( xei 2 / n ) n , we may use sector contours and hope
that the integral reappears in some way on the upper radius of the sector.
Example 1 :

dx
, a  0, n is an integer and n  2.
0 xn  an
2 / n
0
 R dx
dz
iR e i d
e i 2 / n dr 
lim
 lim   n


i n
n
0
R ( re i 2 / n ) n  a n 
R   z n  a n
R  0 x  a n
(
R
e
)

a


I 
0
dr 
 R dx
i 2 / n
i 2 / n
 lim   n

e

(
1

e
)I

n
n
n

R r a
R  0 x  a


1
 2i  Residues of n
in the sector
n
z a
 e i / n 
1 

i / n
i / n
 2iRes f ae
 2i limi / n  z  ae
 2i    n 1 
n
n
z  ae
z

a


 na 



2/n
R

 e i / n 
2i    n 1 

 na  
I
1  e i 2 / n
na n 1 sin(  / n)
25
VI. Contours avoiding branch points
When the integrands have branch points and branch cuts, contours need to be designed to
avoid the branch points and the branch cuts.
Example 1 :

x dx
I  2
, a  0.
0 x  a2
z  0 is a branch point. We use the positive x axis as the branch cut.


R
z dz
z dz
z dz
x dx
x dx 
 lim 



contour 2

shown
z  a 2 R0  C z 2  a 2 C z 2  a 2 R x 2  a 2  x 2  a 2 
C
ia
L+
L-ia
z dz
z
1

0
,
because
~
when R  .
2
2
3/ 2
R   C z 2  a 2
z a
z
0
z dz
 e i  / 2  i  e i d 
lim  2
 lim 
 0.
  0 C z  a 2
  0 2
 2 e i 2  a 2
i

 r e dr
R
x dx
r dr
lim  2

lim

lim
 I.
R  R x  a 2
R   R r 2  a 2
R    r 2  a 2
 0
 0
 0
lim
z dz
z

2
I

2

i
R
esidues
of
in the contour
contour 2

2
2
2
shown
z a
z a
26
z dz
z

2
I

2

i
R
esidues
of
in the contour
contour 2

shown
z  a2
z2  a2
 2iRes f ia   Res f  ia 
 

z 
z  
 2i lim  z  ia  2
 lim  z  ia  2
2
2 
z ia
z   ia
z

a
z

a
 


 
C
ia
L+
L-ia
1 i 
2
 1 i
 2i  



a
2
2
a
2
2
a


I

2a
27
Read: Chapter 7: 1
Homework: 7.1.17 (b),7.1.18,7.1.19,7.1.21,7.1.22,7.1.25,7.1.26
Due: November 11
28
Reading: Dispersion relations
7.2 Dispersion relations
Suppose f (z) = u(z)+iv(z) is analytic, and lim f ( z )  0 in the
| z | 
upper half plane, then

 R f ( x)dx
f ( z )dz
f ( z )dz 
lim P
 lim  P 

 if ( x0 )
R x  x
 zx 
R  C z  x
R 

0
0
0 


i
i
f ( z )dz
f ( R e )iR e d

lim 
 lim 

0

R   z  x
R  
R ei  x0
0

 f ( x ) dx
 u ( x )  iv ( x ) dx
P
 if ( x0 )  P 
 iu ( x0 )  iv ( x0 ) 
 x  x

x  x0
0
C
x0
R
1  v( x)dx

u
(
x
)

 0  P  x  x

0

v( x )   1 P  u ( x)dx
 0
  x  x0
These are the dispersion relations. u(z) and v(z) are sometimes called a Hilbert
transform pair.
29
Symmetry relations:
Suppose f ( x)  f * ( x) (Fourier t ransform of a real function),
then u ( x)  u ( x), v( x)  v( x).
u ( x0 ) 

2

P
1


0
2

 0 v( x)dx  v( x)dx  1
P 


 x  x
0 xx 

0
0 

  v( x)dx  v( x)dx 
P 

0 xx
0 xx 
0
0 

xv( x)dx
x 2  x02
v( x0 )  


v( x)dx 1

 x  x

0
P
P
1


0

u ( x)dx
1

 x  x

0
P
 0 u ( x)dx  u ( x)dx 
1
P 



 x  x
0 xx 

0
0 

 u ( x ) dx 
  u ( x)dx
P 

0 xx
0 xx 
0
0 

x0u ( x)dx
x 2  x02
30
In optics, the refractive index can be chosen to be a complex number where its real part is
the normal refractive index, determined by the electric permittivity (w), and its
imaginary part is responsible for absorption, determined by the electric conductivity s(w):
n 2 (w )   (w )  i
4s (w )
w
 f (w )
2  w Im f (w )dw
 
Re
f
(
w
)

P
0
 
0
 0

w 2  w02
 


x u ( x)dx  
2
2  w0 Re f (w )dw
v( x0 )   P  0 2
Im
f
(
w
)


P
0
2
0



x  x0  
 0
w 2  w02
u ( x0 ) 
2
P

xv( x)dx
x 2  x02
These are the Kramers-Kronig relations. These relations state that knowledge of the
absorption coefficient of a material over all frequencies will allow one to obtain the
(normal) refractive index of the material for any frequency, and vise versa.
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Hi, Everyone:
My previous student Mahendra Thapa brought to me this problem, which may be in
Arfken 3rd edition:
3
 x
0 e x  1 dx
It is used in the black body radiation:
If you can crack this integral (using contour integral on the complex plane) by yourself,
please come to show me.
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