CHAPTER FOUR Integration in the Complex Planetaiwan921.lib.ntu.edu.tw/mypdf/Jiang-Ch04.pdf67 CHAPTER FOUR Integration in the Complex Plane 4-1 Contour Integration, Green Theorem and

67

CHAPTER FOUR

Integration in the Complex Plane §4-1 Contour Integration, Green Theorem and Cauchy's Integral Theorem

1. In the real variable analysis, the definite integral is defined as following：

kx

n

k

b

axfdxxf ∆=

=→∆∑∫ )()( 10lim ξ

where 1k k kx xξ− ≤ ≤ , 1−−=∆ kkk xxx , and nkxk ,,2,1,max =∆=∆ . 複變函數之不定積分與實變函數所用之定義相同，因此在 R 區域內若有可解析之函數 )(zf 和

( )F z ，具有下列關係： )()( zfzF =′

則稱 )(zF 為 )(zf 之不定積分（或反導數）寫成

∫= dzzfzF )()( There are some basic properties of definite integral as shown below:

1) dttGdttHdttGtHb

a

b

a

b

a)()()]()([ ∫∫∫ ±=±

2) constant)()( == ∫∫ kdttHkdttHkb

a

b

a

3) dttHdttHa

b

b

a)()( ∫∫ −=

4) cbadttHdttHdttHc

a

c

b

b

a

68

ii) ∫∫ −= )3(cos3cos313cos3sin zdzzdzz

22 3cos

61 cz +−=

因 13cos3sin 22 =+ zz ，故 z3sin 2 和 z3cos2 實際只差一常數而已。

3. Find ∫ zdzz 4sin2 ⇒ ∫ zdzz 4sin2

2

2

1 1 1cos4 (2 ) sin 4 (2) cos44 16 64

1 1 1cos 4 sin 4 cos 44 8 32

z z z z z c

z z z z z c

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − + + +

寫出之原則：各項中前面括號內之量為自 2z 開始，然後逐次微分之導數。而各項後面括號內之量為 z4sin 之逐次積分所得之量，並用正、負相間之符號，但常數項恒寫成正項。

4. Show 211

22 ln21tan1 c

aizaiz

aic

az

aazdz

+⎥⎦⎤

⎢⎣⎡

+−

=+=+

−∫

i) Let 1tan tan zz a u ua

−= ⇒ = and 2secdz a udu=

⇒ ∫ + 22 azdz

∫∫ =+= duauaudua 1

)1(tansec

22

2

11

1 tan11 c

az

acu

a+=+= −

ii) Since ))((

1122 iaziazaz −+=

+

⎥⎦⎤

⎢⎣⎡

+−

−=

iaziazai11

21

⇒ ⎥⎦⎤

⎢⎣⎡

+−

−=

+ ∫∫∫ iazdz

iazdz

aiazdz

21

22

[ ] 2

2

1 ln( ) ln( )21 ln

2

z ia z ia cai

z ia cai z ia

= − − + +

−⎡ ⎤= +⎢ ⎥+⎣ ⎦

5. Line Integrals (Contour Integral) of the Complex Function

There are two kinds of line integrals: i) ( , )

Cf x y dx ⇒∫ 使用在 Scalar field

ii) ( , ) ( , )C

P x y dx Q x y dy+ ⇒∫ 使用在 Vector field For example, let P i Q jφ = +

r x i y j= + ⇒ d r dx i dy j= + ⇒ d r Pdx Qdyφ ⋅ = + In physics, the work is defined as following：

y

x0

C

φ

r

69

C Cw d r Pdx Qdyφ= ⋅ = +∫ ∫

以下我們將分別討論此二類型之積分

6. ( , )C

f x y ds∫ C：curve , s：arc length 參考下列所示之圖，知若 ( ), ( ), andC x t y y t tα β= = ≤ ≤： ⇒ ( , )

Cf x y ds∫

0 1lim ( , )

n

k k kk

f sξ η∆ → =

≡ ∆∑ ------- (1)

Since ss tt

∆∆ = ∆

∆

when 0→∆t , we can obtain

td

dsts=

∆∆

and 2 2

ds dx dyd t d t d t

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

So, equation (1) becomes

[ ]2 2

( , ) ( ), ( )C

dx dyf x y ds f t y t d td t d t

β

α⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫

Example 1 Consider the contour shown in the following figure. Let ( , )f x y xy= . Evaluate (a) ( , )B

Af x y dx∫ ,

(b) ( , )B

Af x y dy∫ , and (c) ( , )

B

Af x y ds∫ .【本小題摘自：A. David Wunsch, Complex Variable

with Applications, 3rd ed., Exercise 4.1(類似題), Problem 1, Pearson Education, Inc., 2005.】

(a) (1,0) (1,0) 2(0,1) (0,1)

12 4

0

( , ) (1 )

12 2 4

B

Ax

x

f x y dx xydx x x dx

x x=

=

= = −

⎡ ⎤= − =⎢ ⎥⎢ ⎥⎣ ⎦

∫ ∫ ∫

(b)

y

tkt

0

)](),([ 11 −− kk tytx

nt=β1−kt0tα = kt ′

)](),([],[ kkkk tytx ′′≡ηξ

x

C)](),([ kk tytx

y

x

A = (0, 1) y = 1 − x2

C

B = (1, 0)

70

(1,0)

(0,1)4( , ) 1

15B

Af x y dy y ydy= − = −∫ ∫

Alternative method for solving the above line integral: On C, with ( , )f x y xy= , 21y x= − , and / 2dy dx x= − , we obtain

( )1 1 2 40 0

( , )

4( , ) ( )( 2 ) 2 215

B

AB

A

f x y dy

dyf x y dx xy x dx x x dxdx

= = − = − + = −

∫

∫ ∫ ∫

(c) Let C: 2( ) , ( ) 1 , 1 0x x t t y y t t t= = = = − ≥ ≥ , then

( )2 2

21 2ds dx dy td t d t d t

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Hence, we have

( )0 0 2 21 1( , ) 1 1 4B

Adsf x y ds xy dt t t t dtdt

⎛ ⎞= = − +⎜ ⎟⎝ ⎠∫ ∫ ∫

Let 21 4u t= + , we have 8du tdt= and 2 14

ut −= . Substituting these into the above integral,

gives

( )

( )( )

0 2 21

51 11 5/ 2 3/ 2

45 81

1 1 4

1 2 5 2 1 50 50 2 101 5 532 5 3 32 5 3 5 3

1 1 150 5 250 5 4432 15

u

t t t dt

udu u u−

− +

×⎡ ⎤ ⎡ ⎤= − = − = − − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= × − +

∫

∫

＊在此應注意：我們所取者為微弦長 kz∆ 才對，而非微弧長 ks∆ ──在複變函數中應是如此。 i) 若 ),(),()( yxviyxuzf += ，其中 u，v 均為實數，則複變函數之線積分可寫成如下式所示：

( , ) ( )( )

[ ] [ ]

C C

C C

f x y ds u iv dx idy

udx vdy i vdx udy

= + +

= − + +

∫ ∫

∫ ∫

ii) 設 )(zf 和 )(zg 為沿著曲線 C 為可積分者，則有

a) [ ( ) ( ) ] ( ) ( )C C C

f z g z dz f z dz g z dz± = ±∫ ∫ ∫

b) ( ) ( ) ,C C

k f z d z k f z d z=∫ ∫ k 為常數

c) ⇒−= ∫∫a

b

b

adzzfdzzf )()( 指線積分之端點為 a 及 b 者，左右項積分路線為同一曲線

d) ⇒+= ∫∫∫b

c

c

a

b

adzzfdzzfdzzf )()()( c 點介於起點 a 與終點 b 之間

Example 2 (a) Compute 1 2 20

i

iz dz

+

+∫ taken along the contour 2 1y x= + (Fig. (a) as shown below).

(b) Perform an integration like that in part (a) using the same integrand and limits, but take as a contour the piecewise smooth curve C shown in Fig. (b).

y

x

y = x2 + 1

C

Fig. (a)

1

2

1

y

x I

Contour C

1

2

1

II

Fig. (b)

71

(a) Put 2 2 2 2( ) ( ) 2f z z x iy x y i xy u iv= = − = − − = + . Thus, with 2 2u x y= − , 2v xy= − , we

have

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

( )( )( ) ( )

1 2 1 2 1 2 1 2 1 22 2 2 2 20 0 0 0 0

1 2 1 2 1 2 1 22 2 2 20 0 0 0

1 220

1

2

2 2 20

21

11

1

2 2

2

2

2 1

2

i i i i i

i i i i ii i i i

i i i i

z dz x y dx xydy i xy dx i x y dy

x y dx xydy i xy dx i x y dy

x dx ydy

i x d

y

yx i y

x

x dy

+ + + + +

+ + + + ++ + + +

+ + + +

= − + + − + −

= − + + − + −

+⎛ ⎞= − +⎜ ⎟⎝ ⎠

−

−⎛ ⎞+ − + − ⎟+ ⎜⎝ ⎠

∫ ∫ ∫ ∫ ∫

∫ ∫ ∫ ∫

∫ ∫

∫2

1

23 32 3 1115 15 2 6

3 105 3

i i

i

= − + − −

= −

∫

(b) Along path I, we have 1y = , so that 2 2 2( ) ( ) 1 2f z z x i x i x u iv= = − = − − = + . Thus, 2 1u x= − , 2v x= − .Since 1y = , 0dy = . The limits of integration along path I are (0, 1)

and (1, 1). Hence, we have

( ) ( )1 120 02( ) 1 23I

f z dz x dx i x dx i= − + − = − −∫ ∫ ∫

Along path II, 1x = , 0dx = , so that 2 2 2( ) (1 ) 1 2f z z iy y i y u iv= = − = − − = + . Thus, 21u y= − , 2v y= − . The limits of integration along path II are (1, 1) and (1, 2). Hence, we

have

( )2 2 21 14( ) 2 1 33II

f z dz ydy i y dy i= + − = −∫ ∫ ∫ The value of the integral along C is obtained by summing the contributions from I and II. Thus,

2 2 4 7 733 3 3 3C

z dz i i i= − − + − = −∫ H.W. 1 Evaluate 2

Cz dz∫ , where C is the parabolic arc 2y x= , 1 2x≤ ≤ . The direction of integration is

from (1, 1) to (2, 4). 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Example 3, Section 4.2, Pearson Education, Inc., 2005.】

2 86 63C

z dz i= − −∫ H.W. 2 Perform the following integrations:

(a) 1

11 dzz

−∫ along 1z = , upper half plane

(b) 1

11 dzz

−∫ along 1z = , lower half plane

(c) 41

iz dz∫ along 1z = , first quadrant

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 8-10, Exercise 4.2, Pearson Education, Inc., 2005.】

(a) 1

11 dz iz

π−

=∫ ; (b) 1

11 dz iz

π−

= −∫ ; (c) 411

3i iz dz −=∫

H.W. 3 (a) Find a parametric representation of the shorter of the two arcs lying along 2 2( 1) ( 1) 1x y− + − =

that connects 1z = and z i= .

72

(b) Find 1

iz d z∫ along the arc of (a), using the parametrization you have found.

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 13, Exercise 4.2, Pearson Education, Inc., 2005.】

(a) 1 itz i e= + + , / 2 tπ π− ≤ ≤ − ; (b) 1

22

iz d z i π⎛ ⎞= −⎜ ⎟

⎝ ⎠∫

7. ML Inequality:

( )C

f z dz ML≤∫ ，其中 Mzf ≤)( ，即 M 為曲線 C 上 )(zf 之上界值，而 L 為曲線 C之長度。

因 Kn

KK

n

Kzz

11 ==∑∑ ≤

故 KKn

KKK

n

Kzfzf ∆≤∆

==∑∑ )()(

11ξξ

KK

n

Kzf ∆=

=∑ )(

1ξ

故有 ( ) ( )C C

f z dz f z dz≤∫ ∫

2 2 2 2

)

2

(

2

( )C

C

Cf z

f z dz

u v dx dy

u v ds

=

= + +

= +

∫

∫

∫

因此，若 )(zf 在曲線 C 上為有界函數，亦即 ( )f z M≤ ，而且曲線 C 的長度為 L 時，則有

( )C C

f z dz M ds ML≤ =∫ ∫ ，故得證。 1) 若 =≡ kzf )( constant，並令 C 為連接 0z 及 nz 之任意曲線。利用線積分之原始定義，首先寫出

1 0 2 1 11

[( ) ( ) ( )]n

n k n nk

s k z k z z z z z z −=

= ∆ = − + − + + −∑

)( 0zzk n −= 而得

00

lim ( )k

n nC zkd z s k z z

∆ →= = −∫

可見線積分只與起點 0z 及終點 nz 有關，而與該線積分之路線無關。若該曲線為封閉曲線則 nzz =0 ，而有

00

lim ( ) 0k

n nC zkd z s k z z

∆ →= = − =∫ or 0C d z =∫

2) 若 zzf ≡)( ，並令 C 為連接 nz 及 0z 之曲線。利用線積分之原始定義，首先寫出

1

n

n k kk

s zξ=

= ∆∑

若取 kk z=ξ 及 1−= kk zξ ，得出

11

n

n k kk

s z z=

= ∆∑ ------------ kk z=ξ 時

)()()( 1122011 −−++−+−= nnn zzzzzzzzz 同樣，

73

2 11

n

n k kk

s z z−=

= ∆∑ ------------- 1−= kk zξ 時

)()()( 11121010 −− −++−+−= nnn zzzzzzzzz

故 2 21 2 0n n ns s z z+ = −

⇒ 2 21 2 0lim ( ) 2n n nCns s zdz z z

→∞+ = = −∫

因此，可得證

( )0

2 20

12

nznz

zdz z z= −∫ 故知積分只與起點與終點 0z ， nz 有關，而與該線積分之路線無關。若該曲線為閉合曲線，則 nzz =0 ，故

∮ 0C zdz = 3) 有圓心在 0z ，半徑為 r 之圓 C，求 Contour Integral：

10( )

nCdz

z z +−∫，其中 n 為整數。

該積分路線 ( ) inC z z z reθθ = − =：

故 θθ dreidz i= 因此，

2

1 100( ) [ ]

i

n i nCdz i re d

z z re

θπθ

θ+ +=−∫ ∫

，

∫ −=π

θ θ2

0de

ri nn

i) 若 0=n ，可得 2

002

Cdz i d i

z zπ

θ π= =−∫ ∫

ii) 若 0≠n ，則得 22

1 0 000

( )i i

n n nCdz i ie d e

z z r in r

ππ θ θθ− −+−

= = =−∫ ∫

綜合上述之結果，可得今後經常使用之公式，如下所示。

a) 10

2 , 00 , 0( )nCi ndz

nz z

π+

=⎧= ⎨ ≠− ⎩

∫ 之整數

＊其中積分路線 C 為圓心在 0z ，半徑為 r 之圓，即 θi

n rezz =−

b) 02 , 1

( )0 , 1

mC

i mz z dz

mπ = −⎧

− = ⎨ ≠ −⎩∫ 之整數

Example 3 Find an upper bound on the absolute value of 0 1 1/

0

i zi i

e dz+

+∫ , where the integral is taken along the contour C, which is the quarter circle 1z = , 0 arg( ) / 2z π≤ ≤ .

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Example 4, Section 4.2, Pearson Education, Inc., 2005.】 Let us first find M, an upper bound on 1/ ze . We require that on C

1/ ze M≤ ----------- (A)

Notice that 2 2 2 2 2 2 2 21/ 1/( ) /( ) /( ) /( ) ( ) /( )z x iy x x y i y x y x x y i y x ye e e e e+ + − + + − += = =

y

x

B = (0, 1) y = 1 − x2

C

A = (1, 0)

74

Hence 2 2 2 2 2 21/ /( ) ( ) /( ) /( )z x x y i y x y x x ye e e e+ − + += =

Since 2 2/( )x x ye + is always positive, we can drop the magnitude sign on the right side of the preceding equation.

On contour C, 2 2 1x y+ = . Thus, 1/ z xe e= on C.

The maximum value achieved by ex on the given quarter circle occurs when x is maximum, that is, at x = 1, y = 0. On C, therefore, xe e≤ . Thus,

1/ ze e= on the given contour.

A glance at Eq. (A) now shows that we can take M equal to e. The length L of the path of integration is simply the circumference of the quarter circle, namely, / 2π . Thus, applying the ML inequality,

0 1 1/0 2i z

i ie dz eπ

+

+≤∫

H.W. 2 Consider Ln( )1

i i zI e dz= ∫ taken along the parabola 21y x= − . Without doing the integration, show that / 21.479I eπ≤ .


8. 複變函數之線積分的第二類型

( , ) ( , )C

P x y dx Q x y dy+∫ 可改成 i) 定積分 ⇒ 非封閉曲線時使用之。 ii) 二重積分 ⇒ 封閉曲線時採用之。

a) 定積分型 ( ), ( ),C x t y y t tα β= = ≤ ≤：

⇒ ( , ) ( , )C

P x y dx Q x y dy+∫

{ }[ ( ), ( )] [ ( ), ( )]dx dyd t d tP x t y t Q x t y t dtβα= +∫ b) 二重積分型 = Green’s Theorem

The conditions of Green’s Theorem： If 1) C is the boundary of the region R. 2) ( , )P x y , ),( yxQ are continuous on C.

3) xQ∂∂

，yp

∂∂ are continuous in R.

⇒ ( , ) ( , )C

P x y dx Q x y dy+∫

R

Q P dxdyx y

⎡ ⎤∂ ∂= −⎢ ⎥∂ ∂⎣ ⎦∫ ∫

＊可改用複數表示法，如下所述：

設有在圍線 C 上及其區域 R 內之連續函數 ( , ) ( , ) ( , )B z z P x y i Q x y= +

故 ( , )C B z z dz∮

( )( )C P iQ dx idy= + +∮

( ) ( )C CPdx Qdy i Qdx Pdy= − + +∮ ∮

y

x0

C R

75

2

R R

R

R

Q P P Qdxdy i dxdyx y x y

P Q P Qi i dxdyx y y x

Bi dxdyz

⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂= − − + −⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦

⎧ ⎫⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂⎪ ⎪= − + +⎨ ⎬⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭∂

=∂

∫ ∫ ∫ ∫

∫ ∫

∫ ∫

因此，可得 Green 氏平面定理之複數表示法：

( , ) 2C RBB z z dz i dxdyz

∂=

∂∫ ∫∮

∫ ∫ ∂∂

=R

dszBi2 ------------ (1)

其中，ds 代表微面積元素 dxdy。＊若函數 ( , )B z z 及 ( , )A z z 在區域 R 及其周界 C 上為連續且具有連續之偏導數，則有另一型式之表

示法：

[ ( , ) ( , ) ] 2 ( )C RB AB z z dz A z z dz i dsz z

∂ ∂+ = −

∂ ∂∫ ∫∮ ----------- (4) ＊若 C 為簡單封閉曲線，所圍成之面積為 A，求證

12 C

A z dzi

= ∮

1 ( )4 C

A z dz zd zi

= −∮

利用式(1)，知

2 2 2C R Rzz dz i ds i ds i Az

∂= = =

∂∫ ∫ ∫ ∫∮

故知 12 C

A z dzi

= ∮ ------------ 得證

利用式(4)，知

( ) 2 ( )

4 4

C R

R

z zz dz zdz i dsz z

i ds i A

∂ ∂− = +

∂ ∂

= =

∫ ∫

∫ ∫

∮

故知 1 ( )4 C

A z dz zd zi

= −∮

＊如下圖所示，區域 R 之面積可由下列各方法求算之：

i) [ ]( ) ( )ba

A g x f x dx= −∫

ii) ∫ ∫= R dxdyA → Double Integral. iii) C CA xdy ydx= = −∮ ∮ → Line Integral. 首先，令 0 ,P Q x= =

⇒ C R

xdy dxdy A= =∫ ∫ ∫ 其次，取 , 0P y Q= =

⇒ C R

ydx dxdy A= − = −∫ ∫ ∫

⇒ CA ydx= −∮

＊此種方法最適合於 C 之 parameter form。

y

x0 a

R

b

)(xgy =

)(xfy =

76

8. If 2cos , 3sin , 0 2C x t y t t π= = ≤ ≤： , find the area bounded by C.

1) 若用定積分，則因 3649 22 =+ yx

⇒ 2

936 2xy −=

故 ∫−−

⋅=2

2

2

29362 dxxA

π6

9362

2

2

=

−= ∫− dxx 2) 若用二重積分法

∫ ∫= R dxdyA 在此我們可以利用 Jacobian 積分來簡化問題，其處理過程中所用之 Jacobian 積分變換公式如下所示： When ),(),( vuyyvuxx == ，

⇒ ∫ ∫xyR

dxdyyxf ),(

∫ ∫=uvR

dudvJvuyvuxf ||)],(),,([

where

vy

uy

vx

ux

vuyxJ

∂∂

∂∂

∂∂

∂∂

=∂∂

=),(),(

因此，

∫∫ =+= 194

22 yx dxdyA --------- (1)

如圖所示：

1( , )( , )

2 00 36

x yJu v

∂=∂

=

=

and

2( , )( , )

cos sinsin cos

u vJx y

rr

r

θ θθ θ

∂=∂

−=

=

故(1)式 ⇒

∫ ∫

∫∫==

=≤+

ππθ

2

0

1

66

622

drd

dudvAvu

3) 若用 Line Integral 法 2cos , 3sin , 0 2C x t y t t π= = ≤ ≤：

⇒ 2

02cos 3cos

CA xdy t t dt

π= = ⋅∫ ∫

ππ

π

602

)22sin(3

cos62

0

2

=+=

= ∫tt

dtt

y

x 0

194

22=+

yx

vyux

32

==

(見下圖)

v

u 0

θθ

sincos

rvru

==

122 ≤+ vu

1

θ

r 0 1

2π

y

x 0

(0, −3)

(0, 3)

(2, 0) (−2, 0)

C

77

9. 由前面所述之定義知

0 1( ) ( )lim

n

x kC kf z dz f zξ

∆ → == ∆∑∫

其中， 1−−=∆ kkk zzz ，且 ||Max kz∆=∆ ，則有下列之性質：

i) If )(zf is continuous on C

⇒ ( )C

f z dz∫ exists

ii) dttigtfb

a)]()([ +∫ dttgidttf

b

a

b

a)()( ∫∫ +=

iii) If )(zf is continuous on C, and βα ≤≤+= ttyitxtz ,)()()( , and 0)()( 22 ≠′+′ tytx , (except for at finitely point), and )(tx′ , )(ty′ are continuous on the interval βα ≤≤ t

⇒ ( ) ( ( )) ( )C

f z dz f z t z t dtβ

α′=∫ ∫

此乃因 dttzdtdtdzdz )(′=⋅= ，且又因 )()()( tyitxtz ′+′=′ ，故知

)()(|)(| 222 tyitxtz ′+′=′

亦即表示 0)()( 22 ≠′+′ tytx ，在此證明被省略。

10. Cauchy-Goursat Theorem Let C be a simple closed contour and let ( )f z be a function that is analytic in the interior of C as

well as on C itself. Then, ( ) 0

Cf z dz =∫

♣ Alternative statement of Cauchy-Goursat Theorem Let ( )f z be analytic in a simply connected domain D. Then, for any simple closed contour C in D, we have

( ) 0C

f z dz =∫ . 1) 0, 0, 1, 2,n

Cz dz n= =∫ , where C is any simple closed contour.

2) 0, 1

2 , 1n

C

nz dz

i nπ≠ −⎧

= ⎨ = −⎩∫ , where C: z r= .

11. Some Examples Example 1

y

x

kz

0

nz

1−kz

0z|| kz∆ C

y

tkt

0

)](),([ 11 −− kk tytx

nt=β1−kt0tα = kt ′

)](),([ kk tytx ′′

x

C

)](),([ kk tytx

78

Evaluate 0C

dzz z−∮

, where 0| | 1z z− = .

解法 I: 0( ) cos sin , 0 2C z t z t i t t π= + + ≤ ≤：

⇒ tittz cossin)( +−=′

故 0

1C

d zz z−∮

dttittit

)cossin(sincos

120

+−⋅+

= ∫π

itdtittiti π

π2

sincos)sin(cos2

0=

++

= ∫ 解法 II:

0( ) , 0 2i tC z t z e t π= + ≤ ≤：

⇒ tieitz =′ )(

故 0C

dzz z−∮

itditdeei

ti

ti

πππ

22

0

2

0=== ∫∫

Example 2

Find C

z d z∫ , where C is the line segment from 0 to 1 i+ . ( ) , 0 1C z t t it t= + ≤ ≤：

⇒ iz +=′ 1 故知

1 1

0 0( )(1 ) 2 1 #

Czdz

t it i dt tdt= − + = =

∫

∫ ∫

Example 3

Find 2| |C

z dz∫ , where 2,C z x i y y x= + =： , z is from 0 to 1+ i .

2( ) , 0 1C z t t it t= + ≤ ≤：故可知 tiz 21+=′ 因此，

2| |C

z d z∫

i

tdtitt

65

158

)21)((1

0

42

+=

++= ∫

Example 4

Find zdzi

∫+

+21

0

2 )1Im( where C is the polygonal line from

0 to 1 to 1+ i .

1 1( ) , 0 1C z t t t= ≤ ≤：

2 2( ) 1 2 , 0 1C z t i t t= + ≤ ≤： Then, we have

,1)(1 =′ tz and itz 2)(2 =′ So,

y

x0

)1( i+

y

x0

)1( i+

y

x0

)1( i+

C2

1 C1

79

zdzi

∫+

+21

0

2 )1Im(

1 2

2 2

1 1 22 20 0

1

0

Im( 1) Im( 1)

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

0 4 2

4 #

C C

i

z d z z d z

t d t ti t id t

t i d t

i

+

= + + +

= + ⋅ ⋅ + + − + ⋅

= + ⋅

=

∫ ∫

∫ ∫

∫

12. Conditions of Contour：

Contour ≡ piecewise regular curve. If the contour C: ( ) ( ) ( ) ,z t x t i y t tα β= + ≤ ≤ , where the conditions of the contour C are described as following：

i) )(tx′ , )(ty′ are continuous, and

ii) 2 2( ) ( ) 0x t y t′ ′+ ≠ , except for at α=t , and t β= . Here, let us find the values of the integral

2C

z dz∫ according to the integral paths shown below:

1) When the contour C is the contour OB from 0=z to iz += 1 ; 2) When the contour C is the contour OAB; 3) When the contour C is simple closed contour OABO.

1) ( ) , 0 1C z t t it t= + ≤ ≤： We have itz +=′ 1)( Then,

2

1 201 2 20

1 20

( ) (1 )

( ) (1 )

( 2 2 )

1 ( 2 2 )3

Cz dz

t it i dt

t i i dt

i t dt

i

= + +

= + +

= − +

= − +

∫

∫

∫

∫

＊本例亦可由直接法求之： 1 12 30 0

1 1 ( 2 2 )3 3

i iz d z z i+ += = − +∫

2) 2 2 2C OA AB

z d z z d z z d z= +∫ ∫ ∫ i32

32+−=

3) 2 2C OA AB BO

z d z z d z+ +

= ∫∮ 2 2

22 2 2 2 1 ( 2 2 )3 3 3 3 3

OA AB BO

BO

z d z z d z

i z d z i i

+= +

= − + − = − + − − +

∫ ∫

∫

Example 5

Find 2 2| 1 | | |C

z d z−∮ , where the contour C: 2|| =z described in the counterclockwise direction. In this problem, || zd means ds , that is, s is the arc length.

Then, we have

y

xO: (0,0)

B: )1( i+

A: (1, 0)

80

2 2ds dx dyd t d t d t

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⇒ 2 2

dx dyds d td t d t

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

And let ( ) 2(cos sin ) , 0 2C z t t i t t π= + ≤ ≤： Hence, we have

2 2 2 2| 1 | | 4cos 8 sin cos 4sin 1 |z t i t t t− = + − − 2

2 2| 4cos2 4sin 2 1 |(4cos 2 1) (4sin 2 )17 8cos2

t i tt t

t

= + −

= − += −

and the arc length ds is

2 2

| | dx dydz ds d td t d t

⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

td

tdtt

2

)cos2()sin2( 22

=

+−=

⇒ 2 2| 1 | | |C

z d z−∮

#68

)25cos817(2

0

π

π

=

−= ∫ td

Example 6

Find the value of the integral cos

Czdz∫

where C is the polygonal line connecting, in the given order, the points 0=z , π=z ,

and iz2ππ += .

Since zzf cos)( = is analytic in the entire complex plane ⇒ cos

Czdz∫

20

sin

sin( )2

sin sinh2 2

iz

i

i

ππ

ππ

π π

+=

= +

= − = −

§4-3 一些有關曲線，區域名稱之定義

【本節相關內容摘自：Dennis G. Zill and Patrick D. Shanahan, A First Course in Complex Analysis with Applications, Ch.1 and Ch. 4, Jones and Bartlett, Inc., 2003.】

1. Definition of Piecewise Smooth Curve (Contour) A piecewise smooth curve is a path made up of a finite number of smooth arcs connected end to end.

2. Definition of Simple Closed Contour (Jordan Contour) A simple closed contour is a contour that creates two domains, a bounded one and an unbounded one;

y

x

C

r = 2

y

π 0

i2ππ +

x

i2π

C

81

each domain has the contour for its boundary. The bounded domain is said to be the interior of the contour.

♣ The integration is said to be performed in the positive sense around the contour if the interior of the contour is on our left as we move along the contour in the direction of integration.

3. Terminology

Suppose a curve C: ( ) ( ) ( )z t x t iy t= + , a t b≤ ≤ , where ( )x t and ( )y t are continuous real function. Let the initial and terminal points of C be ( )( ), ( )A x a y a and ( )( ), ( )B x b y b . ♣ We say a curve C in the complex plane is smooth if ( ) ( ) ( )z t x t iy t′ ′ ′= + is continuous and never

zero in the interval a t b≤ ≤ . ♣ A curve C in the complex plane is said to be a simple if 1 2( ) ( )z t z t≠ for 1 2t t≠ , except possibly

for t a= and t b= . ♣ C is a closed curve if ( ) ( )z z z b= . 1) C is a smooth curve if x′ and y′ are continuous on the closed interval [ , ]a b and

simultaneously zero on the open interval ( , )a b . 2) C is a piecewise smooth curve if it consists of a finite number of smooth curves C1, C2, …, Cn

jointed end to end, that is, the terminal point of one curve Ck coinciding with the initial point of the next curve Ck −1.

3) C is a simple curve if the curve C does not cross itself except possibly at t a= and ( )t b= . 4) C is a closed curve if A = B. 5) C is simple closed curve if the curve C does not cross itself and except possible A = B, hat is, C is

simple and closed.

4. Annulus, Domain, and Regions 1) Annulus The set S1 of points satisfying the inequality 1 0z zρ < − lie exterior to the circle of

radius 1ρ centered at 0z , whereas the set S2 of the points satisfying 0 2z z ρ− < lies interior to the circle of radius 2ρ centered at 0z . Thus, if 1 20 ρ ρ< < , the set of points satisfying the simultaneous inequality

1 0 2z zρ ρ< − < ---------- (A) is the intersection of the sets S1 and S2. The set defined in (A) is called an open circular annulus.

2) Domain If any pair of points 1z and 2z in a set S can be connected by a polygonal line that consists of a finite number of line segments jointed end to end that lies entirely in the set, then the set S is said to be connected. An open connected set is called a domain.

82

Connected set 3) Regions A region is a set of points in the complex plane with all, some or none of its boundary

points. Since an open set does not contain any boundary points, it is automatically a region. A region that contains all its boundary points is said to be closed.

§4-4 Cauchy's Integral Theorem and Its Some Lemma

1. Cauchy's Integral Theorem If 1) )(zf is analytic in simply connected domain D. 2) C is a simple closed curve in D. ⇒ ( ) 0

Cf z d z =∮

在此處，這個定理之證明並不是很嚴密： Let ( ) ( , ) ( , )f z u x y i v x y= + . Since ( )f z is analytic in D, then

⇒ yv

xu

∂∂

=∂∂ and

yu

xv

∂∂

−=∂∂

在點 Dyx ∈),( 時均成立。 Furthermore,

xu∂∂

，yv

∂∂

，xv

∂∂

，yu

∂∂

are continuous in D, ⇒ ( ) ( )( )

C Cf z d z u iv dx idy= + +∮ ∮

[ ] [ ]

( ) ( )

0 0

0

C C

R R

R R

udx vdy i vdx udy

v u u vdxdy i dxdyx y x y

dxdy i dxdy

= − + +

∂ ∂ ∂ ∂= − − − −

∂ ∂ ∂ ∂

= ⋅ + ⋅

=

∫∫ ∫∫

∫∫ ∫∫

∮ ∮

＊注意，此種使用 Green’s Theorem 之證法，其逆證並不真確。

2. 由 Cauchy's Theorem 所得出之幾個之重要理論 1) 若 )(zf 在簡連區域 R 內為單值且可解析之函

數，則簡連區域 R 內兩點 a，b 之曲線 C 而作線積分

( )C

f z dz∫ 與積分路線無關。

From Cauchy’s Theorem, we see that ( ) 0

AEBDAf z d z =∮

or ∫∫ =+ BDAAEB dzzfdzzf 0)()(

y

x0

D

E R

A

a

B b

C1

C2

y

x0

DC

83

Hence, ∫ ∫∫ =−= BDA ADBAEB dzzfdzzfdzzf )()()(

So, we obtain

1 2

( ) ( )C C

f z dz f z dz=∫ ∫ 故知此種積分，只起點 a 及終點 b 有關，而與路線無關。

＊又 C2可認為 C1經由連續變形而成，只要在變形時不通過任意 )(zf 不可解之點（奇異點）時，則上述推論仍屬成立，故此推論稱為路線變形原理。

Example 1

求線積分 2(12 4 )

Cz i z dz−∫

其積分路線 C 為自(1, 1)至(2, 3)之曲線 2( ) ( 1) (2 1),z t t i t= + + + 0 1t≤ ≤

利用路線變形原理將 C1改由 C2和 C3所達成。

2C z x i= +：故知 dxdz = , 0=dy 因此，

2

2(12 4 )C

z i z dz−∫

i

dzixiix

3020

)](4)(12[2

1

2

+=

+−+= ∫

而 3 2C z i y= +：故知 dyidz = , 0=dx 因此，

3

2(12 4 )C

z i z dz−∫

i

dyiiyiiy

8176

)]2(4)2(12[3

1

2

+−=

+−+= ∫

所以，

1

2(12 4 )C

z i z dz−∫

2 3

2(12 4 ) 156 38C C

z i z dz i= + − = − +∫ ∫

2) 若 )(zf 在 simple connected 區域 R 內為單值且可解析之函數，取 R 內兩點 a 及 z 作線積分

( )C

f z dz∫ ----------- (1) 則可寫出

( ) ( )z

aF z f dξ ξ≡ ∫ ------------ (2)

此種寫法說明積分變數局限於所指定之積分路線，而 )(zF 在該區域 R 內為 Analytic，且

( ) ( )F z f z′ = ------------- (3)

(2, 3)

C1

C2 (2, 1) (1, 1)

C3

y

x

y

x0

zz ∆+

az

84

Since )(zf is analytic, then the line integral

∫z

adf ξξ )(

與積分路徑無關，而純由起點與終點 a，z 所決定，a 為定點，z 為變點，則積分所得為 z 之函數，可寫成如(2)式。則用(2)式可寫出

[ ]

( )

1 ( ) ( )

1 ( ) ( ) ( )

1 ( )

1 [ ( ) ( ) ( )]

1 1( ) ( ) ( )]

z z z

a a

z z z z

a a a

z z

z

z z

z

z z z z

z z

F z zz

f d f dz

f d f d f dz

f dz

f f z f z dz

f z d f f z dz z

ξ ξ ξ ξ

ξ ξ ξ ξ ξ ξ

ξ ξ

ξ ξ

ξ ξ ξ

+∆

+∆

+∆

+∆

+∆ +∆

+ ∆∆⎡ ⎤= −⎢ ⎥∆ ⎣ ⎦⎡ ⎤= + −⎢ ⎥∆ ⎣ ⎦

=∆

= − +∆

⎡ ⎤= + −⎢ ⎥∆ ∆⎣ ⎦

∫ ∫

∫ ∫ ∫

∫

∫

∫ ∫

但 z z

zd zξ

+∆= ∆∫

故 z

zFzzF∆

−∆+ )()(

1( ) [ ( ) ( )]z z

zf z f f z d

zξ ξ

+∆= + −

∆ ∫ 若在路線 z 至 zz ∆+ ， max( ) ( )M f f zξ= −

則 1 [ ( ) ( )]

z z

zf f z d M

zξ ξ

+∆− ≤

∆ ∫ 又因 )(zf 具連續性，故當 0→∆z ， 0→M 。即

0

1lim [ ( ) ( )] 0z z

zzf f z d

zξ ξ

+∆

∆ →− =

∆ ∫ 故

0

1 ( ) ( )lim ( )z z

zz

F z z F z f zz z

+∆

∆ →

+ ∆ −=

∆ ∆∫ 與 0→∆z 之方式無關。故得證

ξξ dfzFzz

a)()( ∫

∆+= 為解析函數，且 )()( zfzF = #

Example 2

(a) Find the antiderivative of zze .

(b) Use the result of (a) to find z w

iwe d w∫ .

(c) Verify the above-mentioned theorem for the integral in part (b).

(d) Use the result of (a) to find 1 zi

ze d z∫ .

(a) Using the integration by part, we have ( )z z zze d z ze e C F z= − + =∫

85

(b) Using the result of (b), we have z w z z

iwe d w ze e C= − +∫

We observe that the left side of this equation is zero when z i= . The right side will agree with the left at z i= if we put i iC ie e= − + . Thus,

z w z z i ii

we d w ze e ie e= − + − +∫ (c) The above theorem asserts that

( )z w z z i i zid dwe d w ze e ie e zedz dz

⎛ ⎞ = − + − + =⎜ ⎟⎝ ⎠∫

(d) With ( ) z zF z ze e C= − + , since / zdF dz ze= throughout the z-plane, we have 11 z z z i i

i ize d z ze e C ie e= − + = − +∫

3) If )(zf is analytic on C1 and C2, and the m is the annulus bounded by C1 and C2.

⇒ 1 2

( ) ( )C C

f z dz f z dz=∮ ∮

Let 1 2C C AB C BA= + − + ⇒ C is also a closed curve (see Figure). From Cauchy’s Integral Theorem, we see that

1 2

0 ( )

( ) ( ) ( ) ( )

C

C CAB BA

f z dz

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

=

= + − +∫ ∫

∮

∮ ∮

Hence, we obtained that

1 2

( ) ( )C C

f z dz f z dz=∮ ∮

此一定理又可稱為圍線變形原理，擇要述之如下：解析函數 )(zf 沿任意封閉曲線 C1之圍線積分，與沿其他由 C1連續變形而得之封閉曲線 C2之圍線積分值完全相等，只要變形時並未通過 )(zf 之異點者。

Example 3

What is the value of /C

dz z∫ , where the contour C is the square shown in the figure?

Using the principle of deformation of contours, we have /

/ 2C

C

dz z

dz z iπ′

= =

∫∫

y

x0

m

A

C1

C2 B

y

x

(1, 1)

(− 1, − 1)

C

C’

86

4) If )(zf is analytic on C, C1, C2, in the region R.

⇒ 1 2

( ) ( ) ( )C C C

f z dz f z dz f z dz= +∮ ∮ ∮

Let 1 2C C AB C BA EF C FE′ = + − + + − + ⇒ C' is a closed curve (see the above Figure) From Cauchy’s Integral Theorem, we see that

1

2

0 ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

C

C C BAAB

EF C FE

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

( ) ( ) ( )C C C

f z dz f z dz f z dz= +∮ ∮ ∮

3. Some important Examples

Example 1

Evaluate 1C

dzz −∮

where ( ) 3cos sin , 0 2C z t t i t t π= + ≤ ≤： .

Let ( ) 1 , 0 2i tC z t e t π′ = + ≤ ≤：

⇒ 1C

dzz −∮

1

2

0

1

2

C

it

it

dzzie d te

i

π

π

=−

=

=

∫

∮

Example 2 Show that

2 2

cos cosh2 0

( 16)( 25)

zzdz

z zΓ

+=

+ −∫ where the simple closed contour

2|| =Γ z： .

Since the function

y

x0

(0, −4)

(0, 4)

(3, 0) (−3, 0)

C' C

y

x0

5

2|| =Γ z：

−5

4 i

−4 i

y

x0

AEF

C

B

D

C1

C2

87

)25)(16(2

coshcos)( 22 −+

+=

zz

zzzf

is analytic for every point in the region of 2|| ≤z ,

⇒ 2 2cos cosh

2 0( 16)( 25)

zzdz

z zΓ

+=

+ −∫ Example 3

Show that 2sinh 0cos

z dzzΓ

=∫ where the simple closed contour 1|| =Γ z： .

Since 0cos =z ,

⇒ ππ nz +=2

, ,2,1,0 ±±=n

⇒ zzzf 2cos

sinh)( = is analytic on Γ and the inside of Γ.

By the Cauchy’s Integral Theorem, we see that

2sinh 0cos

z dzzΓ

=∫

Example 4

Evaluate ?1C

dzz

=+∮ where 1|1| =−zc： .

Since 1

1)( 2 +=

zzf

is analytic in the region of 1|1| ≤−z ,

by the Cauchy’s Integral Theorem we see that

2 01Cdz

z=

+∮

Example 5

Evaluated2 4

Cz dz

z+

∮ , where | | 1C z =： .

Since ( ) , 0 2itC z t e t π= ≤ ≤： , we have

2 4C

z dzz+

∮

i

ttiti

tdtiti

tdei

tdeie

e

ti

titi

ti

π

πππ

π

π

π

8

02

402

22cos

02

22sin

)42sin2(cos

)4(

4

2

0

2

0

2

2

0

2

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+−=

⎥⎦⎤

⎢⎣⎡ ++=

+=

⋅⋅+

=

∫

∫

∫

y

x 0

| 1 | 1C z − =： i

−i

88

Example 6

Evaluate 2 2( 1)( 4)Cdz

z z+ +∮ , where

3| |2

C z =： .

Since

2 2

2 2

( 1)( 4)1 1 13 4 1

C

C

dzz z

dzz z

+ +

−⎡ ⎤= +⎢ ⎥+ +⎣ ⎦

∮

∮

2

0

21 13 31 4C C

dz dzz z

=

= −+ +

∮ ∮

thus, we have

1 2

1 2

1 2

2

2 2

1 13 11 13 31 11 1 1 1 1 1 13 2 2161 [2 2 ] 06

C

C C

C C

C C

zdz dz

z z

dz dzi z i z i i z i z idz dz

i z i z i

i ii

π π

+

= ++ +

⎡ ⎤⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎢ ⎥− + − +⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤= −⎢ ⎥− +⎣ ⎦

= − =

∮

∮ ∮

∮ ∮

∮ ∮

Hence,

2 2( 1)( 4)Cdz

z z+ +∮

2 21 13 31 40

C Cdz dz

z z= −

+ +=

∮ ∮

H.W. 1 (a) Find z

Ce d z∫ , where : 1C z = .

(b) Show that ( )2 cos0

cos sin 0e dπ θ θ θ θ+ =⎡ ⎤⎣ ⎦∫ and ( )

2 cos0

sin sin 0e dπ θ θ θ θ+ =⎡ ⎤⎣ ⎦∫

(c) Show that ( )2 sin0

cos cos 0ne n dπ θ θ θ θ− =∫ and ( )

2 sin0

sin cos 0ne n dπ θ θ θ θ− =∫

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 12-15, Section 4.3, Pearson Education, Inc., 2005.】

(a) 0zC

e d z =∫ H.W. 2 The contour is the square centered at the origin with the corners at (2 2 )i± ± . Find

(a) 4( )Cdz

z i−∫; (b) ( 1)

m

mCz d z

z+

∫ , where 0m ≥ is an integer; (c) ( 1)m

mCz d z

z −∫, where 0m ≥

is an integer 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 19 and 21-22, Section 4.3, Pearson Education, Inc., 2005.】

(a) 4 2( )Cdz

z iπ= −

−∫; (b) ( 1) 2

m

mCz d z im

zπ+ =∫ ; (c) 2( 1)

m

mCz d z im

zπ=

−∫

4. Special Topics 【本小節相關內容摘自：James Ward Brown and Ruel V. Churchill, Complex Variable and

Applications, 7th ed., Sec. 79, McGraw-Hill, Inc., 2004.】

y

x0

−2 i

2 i

)0,23(

C1

)0,23(−

C2

C

89

♣ A function ( )f z is said to be meromorphic in a domain D if it is analytic throughout D except for poles.

1) Basic concept: Consider a function ( )f z that is analytic and nonzero everywhere on a simple closed contour C. Assume that ( )f z is analytic in the domain inside C, except possibly at a finite number of pole singularity.

2) Suppose we write ( )arg ( )( ) ( ) i f zf z f z e= ---------- (1)

3) Define arg ( )C f z∆ ≡ the increase in argument of ( )f z (final minus initial value) as the contour C is negotiated once in the positive sense. Principle of the argument: Suppose that (a) a function ( )f z is meromorphic in the domain interior to a positively oriented

simple closed contour C; (b) ( )f z is analytic and nonzero on C; (c) counting multiplicities, Z is the number of zeros and P is the number of poles of

( )f z inside C. Then

1 1 ( )arg ( )2 2 ( )C C

f zf z Z P dzi f zπ π

′∆ ≡ − = ∫

Let ( )z z t= , a t b≤ ≤ be a parametric representation for C, so that ( ) [ ( )] ( )( ) [ ( )]

b

C af z f z t z tdz dtf z f z t′ ′ ′

=∫ ∫ Since, under the transformation ( )w f z= , the image Γ of C never passes through the origin in the w-plane, the image of any point ( )z z t= on C can be expressed in exponential form as

( ) exp[ ( )]w t i tρ φ= . Thus, [ ( )] ( ) exp[ ( )]f z t t i tρ φ= , a t b≤ ≤

⇒ { } ( ) ( )[ ( )] ( ) [ ( )] ( ) exp[ ( )] ( ) ( ) ( )i t i td df z t z t f z t t i t t e i t e tdz dz

φ φρ φ ρ ρ φ′ ′ ′= = = +

⇒ ( ) ( ) ( ) ln ( ) ( )( ) ( )

b b b ba aC a a

f z tdz dt i t dt t i tf z t

ρ φ ρ φρ

′ ′′= + = +∫ ∫ ∫

But ( ) ( )a bρ ρ= and ( ) ( ) arg ( )Cb a f zφ φ− = ∆

Hence ( ) arg ( )( ) CC

f z dz i f zf z′

= ∆∫ ----------- (2) If ( )f z ha s a zero of order 0m at 0z , then

0( ) ( ) ( )mf z z z g z= −

where ( )g z is analytic and nonzero at 0z . Hence 0 010 0 0 0( ) ( ) ( ) ( ) ( )

m mf z m z z g z z z g z−′ ′= − + −

⇒ 00

( ) ( )( ) ( )

mf z g zf z z z g z′ ′

= +−

----------- (3)

Since ( ) / ( )g z g z′ is analytic at 0z , it ahs a Taylor series representation about that point; Eq. (3) tell us that ( ) / ( )f z f z′ has a simple pole at 0z with residue m0. If ( )f z ha s a pole of order

pm at 0z , then

0( ) ( ) ( )pmf z z z zφ−= − ---------- (4)

where ( )zφ is analytic and nonzero at 0z . Similarly, in this case ( ) / ( )f z f z′ has a simple pole at

0z with residue pm− . Applying the residue theorem, then, we find that

( ) ( )( )C

f z dz i Z Pf z′

= −∫

90

⇒ 1 1 ( )arg ( )2 2 ( )C C

f zf z Z P dzi f zπ π

′∆ ≡ − = ∫

§4-5 Cauchy's Integral Formulas

1. Cauchy's Integral Formula If

1) )(zf is analytic in simply connected domain D. 2) C is a simple closed curve in D. 3) 0z is an interior point of C

⇒ 00

( ) 2 ( )C

f z d z i f zz z

π=−∮

Since

0

0

0

0

0

)()()()(zz

zfzz

zfzfzzzf

−+

−−

=−

0

( )C

f z d zz z

⇒−∮

0 0

0 0

00

0

( ) ( ) ( )

( ) ( ) 2 ( )

C C

C

f z f z f zd z d zz z z z

f z f z d z if zz z

π

−= +

− −

−= +

−

∮ ∮

∮

Here, we need to prove that 0

0

( ) ( ) 0C

f z f z dzz z−

=−∮

Since )(zf is analytic at 0zz = , then ⇒ )(zf is continuous at 0zz = ⇒ For every 0> , there exists 0>δ such that

0| |z z δ− < We have

0| ( ) ( ) |f z f z− < Then, we take a circle whose radius is ρ, and let

δρ

91

0

0 0

0 0

( )

( ) ( ) ( )

C

C C

f z dzz zf z f z f zdz dz

z z z z

−−

= +− −

∮

∮ ∮

)(2)(20 00 zfizfi ππ =+= That is,

00

1 ( )( ) #2 Cf zf z dz

i z zπ=

−∮

2. Some Examples

Example 1

Show that22sinh 3cosh 3 6

Cz z dz i

zπ+ =∮ ,

where C is a simple closed contour having 0=z in its interior. Since zzzf 3cosh3sinh2)( 2 += is analytic in the entire complex plane, from Cauchy's

integral formula we see that 2

0

2sinh 3cosh 3

2 ( ) 2 3 6C

z z dzz

i f z i iπ π π

+

= ⋅ = ⋅ =

∮

Example 2

Show that2

29 4 18

( 1)Cz i z dz iz z

π− + =+

∮ ,

where | | 2C z =： .

Since

))((49

)1(49 2

2

2

izizzziz

zzziz

−++−

=++−

izizz −+

++=

234

⇒ 2

29 4

( 1)Cz i z dzz z− +

+∮

4 3 2

(4 3 2) 218

C C Cdz dz dz

z z i z ii

iπ

π

= + ++ −

= + + ⋅=

∮ ∮ ∮

*** Find ℒ −1⎭⎬⎫

⎩⎨⎧

−−+

)2()1(13

2 sss

Since

)2()1(13

2 −−+

sss

2)1(1 2 −+

−+

−=

sC

sB

sA ---------- (1)

⇒ 1

3 1 42 s

sBs =+⎡ ⎤= = −⎢ ⎥−⎣ ⎦

2 2

3 1 7( 1) s

sCs =

⎡ ⎤+= =⎢ ⎥

−⎢ ⎥⎣ ⎦

y

x0

2|| =z

i

−i

92

but 1

3 12 s

d sAds s =

+⎡ ⎤= ⎢ ⎥−⎣ ⎦

2 1

( 2) 3 (3 1) 1( 2)

7s

s ss =

− ⋅ − + ⋅=

−= −

Here, we can also use another method to find A. When we find A, we can multiply by (s − 1) in both sides of equation (1) and then let s →∞ . Hence,

∞→−−

+−

+=∞→−−

+ss

sCs

BAsss

s2

)1(1)2()1(

132

⇒ 7A C= − = −

故 ℒ −1⎭⎬⎫

⎩⎨⎧

−−+

)2()1(13

2 sss

=ℒ⎭⎬⎫

⎩⎨⎧

−+

−−

+−−−

27

)1(4

17

21

sss

ttt etee 2747 ++−−=

Example 3 Let iC z e θ=： be the unit circle, where πθπ ≤≤− .

a) Show that

2 ,Kz

Ce dz i K R

zπ= ∈∮

b) Using the result of part (a), show that

πθθθπ

=⋅∫ dKe K )sincos(cos0

a) Since Kzezf =)( is analytic in the entire complex plane and 0=z is an interior point of C. By Cauchy's integral formula, we see that

02 [ ]

2

KzKz

C z

e d z i ez

i

π

π=

= ⋅

=

∮

b) Since

2Kz

Cei d z

zπ =∮

⇒ iKz K e

iiC

e edz ie dz e

θ

π θθπ

θ⋅

−= ⋅∫∮

(cos sin )

cos

cos cos

[cos( sin ) sin( sin )]

cos( sin ) sin( sin )

K i

K

K K

i e d

i e K i K d

i e K d e K d

π θ θππ θππ πθ θπ π

θ

θ θ θ

θ θ θ θ

+−

⋅−

− −

=

= +

= ⋅ − ⋅

∫

∫

∫ ∫

Since )sinsin( θK is an odd function, and θcosKe is an even function, hence

)sinsin(cos θθ KeK ⋅ is an odd function still. Thus, we have

0)sinsin(cos =⋅∫− θθθ

π

πdKeK

Then,

93

⇒ πθθθπ

π2)sincos(cos =⋅∫− dKe

K

i.e. πθθθπ

2)sincos(2 cos0

=⋅∫ dKeK ⇒ πθθθ

π=⋅∫ dKeK )sincos(cos0

H.W.1 Evaluate 21 Ln( )

2 9Cz d z

i zπ +∫ around 4 3z i− = .


21 Ln( ) 1 Ln 3

2 12 69Cz d z i

i zπ

π= −

+∫

§4-6 Derivatives of Analytic Functions

1. If

i) )(zf is analytic in a simply connected Domain D. ii) C is a simple closed curve in D. iii) 0z is an interior point of C.

⇒ ( ) 0 10

( )( )2 ( )

nnC

n f zf z dzi z zπ +

=−

！∮

證明此定理時須使用到下述之定理，即

If ( ) ( , )C

F z f z dξ ξ= ∫ ⇒ ( ) ( , )CF z f z dz ξ ξ∂⎡ ⎤′ = ⎢ ⎥∂⎣ ⎦∫

Since )(zf is analytic in a simply connected domain and 0z is an interior point of C, where C is a simple closed curve in D. By Cauchy's integral formula,

⇒ 00

1 ( )( )2 C

f zf z dzi z zπ

=−∮

⇒ 00 0

1 ( )( )2 C

f zf z dzi z z zπ

⎡ ⎤∂′ = ⎢ ⎥∂ −⎣ ⎦∮

20

1 ( )2 ( )C

f z dzi z zπ

=−

∮

⇒ 0 30

2 ( )( )2 ( )C

f zf z dzi z zπ

′′ =−

！∮

0 40

( )0 1

0

3 ( )( )2 ( )

( )( )2 ( )

C

nnC

f zf z dzi z z

n f zf z dzi z z

π

π +

′′′ =−

=−

！

！

∮

∮

** 以上之證明缺乏數學上之嚴密性，只能算是一個說明而已，請參閱課本說明。


Show that

94

( )3

4

2

3coshzC

e z dzz iπ+

−∮

where C is any simple closed contour containing iz20π

= in its interior and the integral along C is

taken in the positive direction.

( )3

4

2

3coshzC

e z dzz iπ+

−∮

33

32

3

2

2 3cosh3

27 3sinh3

zz i

zz i

i d e zdz

i e z

π

π

π

π

=

=

⎡ ⎤= +⎣ ⎦

⎡ ⎤= +⎣ ⎦

！

3 3 327 cos sin 3sinh3 2 2 2

27 3sin3 2

( 24 )3

8

i ii

i i i

i i

π π π π

π π

π

π

⎡ ⎤⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤= − +⎢ ⎥⎣ ⎦

= −

=

H.W. 1 (a) Show that 12!

az n

nCe a id z

nzπ

+ =∫ , where C: 1z = .

(b) Show that ( )2 cos0

cos sin 2 / !a ne a n d a nπ θ θ θ θ π− =∫ and ( )

2 cos0

sin sin 0ae a n dπ θ θ θ θ− =∫

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 15, Exercise 4.5, Pearson Education, Inc., 2005.】 H.W. 2 (a) If a is a real number and 1a = show that

220

1 cos 21 2 cos

a da a

π θ θ πθ

−=

− +∫

(b) Evaluate the above integral for the case 1a > . 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 17, Exercise 4.5, Pearson Education, Inc., 2005.】

3. Cauchy's Inequality For )( 0)( zf n

Let C be a circle with center at 0z and radius r , and let function f be analytic in an open set D containing C and its interior. Then, we have

,2,1,0,)( 0)( =≤ n

rMnzf n

n ！

where M is the least upper bound of ( )f z .

The equation of the circle C is given by

0| | , 0 2iC z z r e θ θ π− = ≤ ≤：

Since ( ) 0 10

( )( )2 ( )

nnC

n f zf z d zi z zπ +

=−

！∮

Clearly, such a constant M（> 0）exist.

y

x0

C

z0

r

95

⇒ ( ) 0 10

( )( )2 ( )

nnC

n f zf z d zi z zπ +

=−

！∮

10

1

( )2 ( )

22

nC

n

n

n f z d zz z

n M rr

Mnr

π

ππ

+

+

=−

≤ ⋅

=

！

！

！

∮

i.e. ( ) 0( )n n Mf z

r≤

！

4. Liouville's Theorem

If function f is analytic and bounded for all values of z in the complex plane (C), then f is a constant.

Since f is bounded Cz ∈∀ , there exists a constant 0>M such that

CzMzf ∈∀≤|)(| Let 0z be any point in C, and let Γ be a circle with center 0z and radius r. Clearly, Γ and its interior are contained in C. Since f is analytic Cz ∈∀ , then by Cauchy's Inequality for )( 0

)( zf n with 1=n

⇒ r

Mzf ≤′ )( 0

Let ∞→r , then 0( ) 0f z′ = ⇒ 0)( 0 =′ zf Since 0z is any point in C (entire complex plane). ⇒ 0)( 0 =′ zf for all z. ⇒ Czf =)( 0 is a constant.

5. Morera's Theorem If )(zf is continuous in a simply connected domain D, and if

0)( =∫ Γ dzzf for every closed contour Γ in D. ⇒ )(zf is analytic in D.


Find z

dzKiKi∫− for any contour joining the points )0(to >− KKiKi and lying in the domain of { | 0 , 0}D C z z x i x= − = + ≤ .

y

x0

C

z0

0| |r z z= −

96

Method I:

i

iKiK

KiKi

zz

dzKi

KiKi

Ki

π

ππ

=

+−+=

−=

−=∫−

2ln

2ln

)Ln(Ln

Ln

Method II:

Since z

zf 1)( = is analytic in D,

⇒ z

dzKiKi∫− is independent of the path, then we choose that

( ) ,2 2

i tC z t K e tπ π′ = − ≤ ≤：

Hence, we have

idti

KdteieK

dzz

titi

Ki

Ki

ππ

π

π

π

==

⋅=

∫

∫∫

−

−−

2

2

2

2

11

Example 3

Evaluate 2cos

( 4)Cz dz

z z +∮ ， | | 1C z =： .

Let 2cos( )

4zf z

z=

+.

⇒ )(zf is analytic on C and interior of C, then by Cauchy's Integral Formula, we obtain that

2

2 0

cos( 4)

2 (0)

cos2( 4)

2

C

z

z d zz zi f

ziz z

i

π

π

π

=

+= ⋅

⎡ ⎤= ⎢ ⎥

+⎢ ⎥⎣ ⎦

=

∮

Example 4

Find 11 ,

2 ( )

n az

na e dz a C

i n zπ +Γ∈∫

！.

The given simple closed contour Γ is taken as the boundary of a square whose sides lies along the line 7±=x and 7±=y , and is described in the positive direction.

y

x0

Ki C'

−Ki

C

y

x0

1|| =z

2 i

−2 i

C

97

( )

1

2

2

12 ( )

1 202

( )

n az

n

naz

n

n

a e dzi n z

i a dn ezi n n dz

an

π

ππ

+Γ

= ⋅ ⋅ ⋅=

=

！

！！

！

∮

7. Gauss’ Mean Value Theorem

Let ( )f z be analytic in a simply connected domain. Consider any circle lying in this domain. The value assumed by ( )f z at the center of the circle equal s the average of the values assumed by ( )f z on its circumference. If 0z is the center of the circle and r its radius, this is equivalent to

2

0

1( ) ( ) .2

io of z f z re d

π θ θπ

= +∫

C: 0iz z re θ= + , 0 2θ π≤ ≤ ⇒ idz re dθ θ=

From Cauchy Integral Theorem, we have

00

1 ( )( )2 C

f zf z d zi z zπ

=−∫

⇒ 2

0

1 ( )( )2

iio

o if z ref z ire d

i re

θπ θθ θπ+

= ∫

⇒ 2

0 00

1( ) ( )2

if z f z re dπ θ θ

π= +∫

The expression on the right of the above Equation is the arithmetic mean (average value) of ( )f z on the circumference of the circle.

If let ( ) ( , ) ( , )f z u x y iv x y= + , it follows 2 2

0 0

1( ) ( ) ( ) ( )2 2

i io o o o

iu z iv z u z re d v z re dπ πθ θθ θ

π π+ = + + +∫ ∫

Taking 0 0 0z x iy= + and equating corresponding parts of each side of above equation, we obtain 2

0 0 00

1( , ) ( ) .2

iu x y u z re dπ θ θ

π= +∫

2

0 0 00

1( , ) ( ) .2

iv x y v z re dπ θ θ

π= +∫

Example 5 Using Gauss’ mean value theorem, evaluate

2

0cos(cos sin )i d

πθ θ θ+∫

By identifying the real and imaginary parts of the integrand, what identities are obtained?

y

x0

(−7,7)

(−7,−7) (7,−7)

(7,7)

y

x

θ

zz 0

r θ∠

98

Since : cos sinC z iθ θ= + is a unit circle, we can take 0 1z = and 1r = . Then, we have 2

001 cos( ) cos

2i

ze d zπ θ θ

π ==∫

⇒ 2

0cos(cos sin ) 2i d

πθ θ θ π+ =∫

Since cos cos cosh sin sinhz x y i x y= − , the last equation can be rewritten as

[ ]20

cos(cos )cosh(sin ) sin(cos )sinh(sin ) 2i dπ

θ θ θ θ θ π− =∫ Equating corresponding parts (real and imaginary) on both sides of the equation, we have

2

0cos(cos )cosh(sin ) 2d

πθ θ θ π=∫

and 2

0sin(cos )sinh(sin ) 0d

πθ θ θ =∫

Also, we have

cos(cos )cosh(sin ) 2dπ

πθ θ θ π

−=∫

and sin(cos )sinh(sin ) 0dπ

πθ θ θ

−=∫

The last result can be checked if we verify the odd symmetry of the integrand.

H.W. 3 Using gauss’ mean value theorem, prove the following:

(a) cos cos(sin ) 2e dπ θπ

θ θ π−

=∫

(b) 2cos 2

1 2 cosa n d

aa a n

π

πθ πθ

θ−+

=+ +∫

, where 1a > , n integer.

(c) ( )2 20

Ln 1 2 cos 4 Lna a n d aπ

θ θ π⎡ ⎤+ + =⎣ ⎦∫ , where 1a > , n integer. 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 2 and 4-5 , Exercise 4.6, Pearson Education, Inc., 2005.】

(a) Using 2

01 1

2iee dθπθ

π=∫ ; (b) Using

1( ) nf z z a=

+; (c) Using ( )

22 1 2 cos ina a n a e θθ+ + = +

H.W. 4 In this problem, we derive one of the four Wallis formulas. They allow one to evaluate

[ ]/ 20

( )m

f dπ

θ θ∫ where 0m ≥ is an integer and ( ) sinf θ θ= or cosθ . The cases of odd and even m must be considered separately. We will consider m even. (a) Using the binomial theorem, show that

( )2 2 12211

0

(2 )!(2 )! !

n knn

zk

n zz zn k k

− −−

=+ =

−∑ , 0, 1, 2,n =

(b) Using the above result, a term-by-term integration, and the extended Cauchy integral formula (i.e., Derivatives of Analytic Functions), show that

( )211 2(2 )!2 ( !)n

znz z dz i

nπ− + =∫

where the integration is around 1z = .

(c) With iz e θ= , 0 2θ π≤ ≤ , on the unit circle, show from (b) that 22

20(2 )!(2cos ) 2( !)

n ndn

πθ θ π=∫

(d) Noting the symmetry of cosθ , and that 2n is even ( 0, 1, 2,n = ), explain why 2/ 2

20(2 )!(cos )

2 ( !) 2

n

nnd

n

π πθ θ =∫

This is one of Wallis’s formulas.

99

(e) Find 2/ 2

0(sin )

nd

πθ θ∫ , where 0, 1, 2,n = .

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 17 , Exercise 4.6, Pearson Education, Inc., 2005.】

§4-7 Introduction to Dirichlet Problems: The Poisson Integral Formula for the Circle and Half Plane

♣ Dirichlet Problem Suppose that D is a domain in the plane and that g is a function defined on the boundary C of D. The problem of finding a function ( , )x yφ , which satisfies Laplace’s equation in D and which equals g on the boundary C of D is called a Dirichlet problem.

1. The Dirichlet Problem for a Circle

1) Consider a circle of radius R whose center lies at the origin of the complex w-plane.

2) Let ( )f w be a function that is analytic on and throughout the interior of this circle. 3) Cauchy Integral formula:

1 ( )( ) .2

f wf z dwi w zπ

=−∫ ------------ (1)

Suppose ( ) ( , ) ( , )f z U x y iV x y= + . We would like to use the preceding integral to obtain the explicit expressions for U and V. Define the point 1z as

21 /z R z= . We have

2

1 .R Rz Rz z

= =

Since z R< (z is an arbitrary interior point of the circle),

we have 1z R> (exterior point). However, 1arg argz z= . The function 1( ) /( )f w w z− is analytic in the w-plane on and inside the given circle. Hence, from the Cauchy integral formula, we have

21

1 ( ) 1 ( )0

2 2 ( / )f w f w

dw dwi w z i w R zπ π

= =− −∫ ∫ ------------ (2)

Subtracting Eq. (2) from Eq. (1), we obtain

2

1 1 1( ) ( )

2 ( / )f z f w dw

i w z w R zπ= −

− −⎡ ⎤⎢ ⎥⎣ ⎦

∫

( )2

2

1 ( / )( ) .

2 ( ) ( / )z R z

f w dwi w z w R zπ

−=

− −

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

∫ ------------ (3)

Changing to polar coordinate, let iw Re φ= and iz re θ= . Thus, iz re θ−= . Along the path of integration, idw Re dφ φ= and 0 2φ π≤ ≤ . Rewriting the right side of Eq. (3), we have

( )( )( )

( )( )( )( )

22

20

22

20

/1( , ) ( , )

2 ( ) /

/1( , )

2 ( ) /

i ii

i i i i

i i i

i i i i

re R ref r f R Re id

i Re re Re R re

re R e r Ref R d

Re re Re R e r

θ θπ φ

φ θ φ θ

θ θ φπ

φ θ φ θ

θ φ φπ

φ φπ

−

−

−=

− −

−=

− −

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

∫

∫

If we multiply the two terms in the denominator of the preceding integral together and then multiply numerator and denominator by ( )( / ) ir R e θ φ− +− , we can show, with the aid of Euler’s identity, that

w-plane

R

z

z 1

100

2 22

2 20

1 ( , )( )( , ) .

2 2 cos( )f R R r d

f rR r Rr

π φ φθ

π φ θ−

=+ − −∫ ------------ (4)

Since ( , ) ( , ) ( , )f R U R iV Rφ φ φ= + and ( , ) ( , ) ( , )f r U r iV rθ θ θ= + , Eq. (4) becomes

[ ] 2 222 20

( , ) ( , )1( , ) ( , ) .

2 2 cos( )

U R iV R R r dU r iV r

R r Rrπ φ φ φ

θ θπ φ θ

+ −+ =

+ − −

⎡ ⎤⎣ ⎦∫ ------------- (5) By equating the real parts on either side of this equation, we obtain the following formula:

2 22

2 20

( , )1( , ) .

2 2 cos( )

U R R r dU r

R r Rrπ φ φ

θπ φ θ

−=

+ − −

⎡ ⎤⎣ ⎦∫ -------------- (6) 2 2

2

2 20

( , )1( , ) .

2 2 cos( )

V R R r dV r

R r Rrπ φ φ

θπ φ θ

−=

+ − −

⎡ ⎤⎣ ⎦∫ The formula yields the value of the harmonic function

( , )U r θ (or ( , )V r θ ) everywhere inside a circle of the radius R, provided we know the values ( , )U R φ (or ( , )V R φ ) assumed by U (or V) on the circumference of the circle.

Example 1

An electrically conducting tube of unit radius is separated into two halves by means of infinitesimal slits. The top half of the tube ( 1, 0R φ π= < < ) is maintained at an electrical potential of 1 volt while the bottom half ( 1, 2R π φ π= < < ) is maintained at −1 volt. Find the potential at an arbitrary point ( , )r θ inside the tube (see the shown figure). Assume there is a dielectric material inside the tube.

Since the electrostatic potential is a harmonic function, the Poisson integral formula is applicable. From above discussion (Eq. (6)), with 1R = , we have

2 22

2 20 0

1 (1 ) 1 (1 )( , )2 21 2 cos( ) 1 2 cos( )

r d r dU rr r r r

π πφ φθπ πφ θ φ θ

− −= −

+ − − + − −∫ ∫ (7) In each integral, we make the change of variable x φ θ= − ; from a standard table of integrals, we find

the following formula, which is valid for 2 2 0a b> ≥ : 2 2

1

2 2

2 tan( / 2)tan .cos

dx a b xa b x a ba b

−⎡ ⎤−

= ⎢ ⎥+ +− ⎢ ⎥⎣ ⎦

∫

Using this formula in Eq. (7) with 21a r= + , 2b r= − , we obtain

1 1

1

1 1 1( , ) 2 tan tan tan tan1 2 2 1 2

1tan tan1 2

r rU rr r

rr

π θ θθ ππ

θ

− −

−

⎡ + +⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎢ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣⎤+⎛ ⎞⎛ ⎞− −⎜ ⎟ ⎥⎜ ⎟− ⎝ ⎠⎝ ⎠⎦

(8)

Conclude that 1 ( , ) 1U r θ− ≤ ≤ when 1r ≤ . ♣ Recall that tan( ) tannπ θ θ+ = , where θ is any angle and n any integer. We can recast eq. (8) as

follows:

1 12 1 1( , ) tan tan tan tan .1 2 2 1 2 2

r rU rr r

π θ θ πθπ

− −⎡ ⎤+ +⎡ ⎤⎛ ⎞ ⎡ ⎤= − + ±⎜ ⎟⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎝ ⎠ ⎣ ⎦⎣ ⎦⎣ ⎦ (9)

where the minus sign is to be used in the front of / 2π when 0 θ π< < and the plus sign when 2π π π< < .

Poisson Integral Formula (For interior of a circle)

y

x1R =

1 volt, top half

r φ

θ

−1 volt, top half

gap

gap

101

♣ MATLAB Commands: % for figure of Example 1 clear fig clear r=[.1 .3 .5 .7 .9]; thet=linspace(0,2*pi,200); for j=1:length(r) q=(1+r(j))/(1-r(j)); p2=pi/2; u1=2/pi*(atan(q*tan(p2-thet/2)))

u2=2/pi*(atan(q*tan(thet/2))) u3=sign(thet-pi); u=u1+u2+u3 plot(thet,u);hold on;

end grid on

2. The Dirichlet Problem for a half plane (Infinite Line Boundary) 1) Find a function ( , )u vφ that is harmonic in the upper half

complex w-plane (the domain 0v > ). 2) B.C. of ( , )x yφ : ( ,0)uφ at 0v = . 3) Let ( ) ( , ) ( , )f w u v i u vφ ψ= + be a function that is analytic

for 0v ≥ . Also, let z be a point inside this semicircle. 4) Cauchy Integral formula:

1 ( )( ) .2 C

f wf z dwi w zπ

=−∫ ------------ (10)

Since z lies inside the semicircle, the point z must lie outside the semicircle. Hence the function ( ) /( )f w w z− is analytic on and interior to the contour C. We have

1 ( )0 .2 C

f w dwi w zπ

=−∫ ------------- (11)

Let us subtract Eq. (11) from eq. (10): 1 1 1 1 ( ) ( )( ) ( ) .

2 2 ( )( )C Cz z f wf z f w dw dw

i w z w z i w z w zπ π−⎛ ⎞= − =⎜ ⎟− − − −⎝ ⎠∫ ∫ (12)

Note that the contour C ≡ the base ( 0v = , R u R− ≤ ≤ , denoted by → ) + the arc with radius of R (denoted by ∩ ). Thus, Eq. (12) can be expressed as

1 ( ) ( ) 1 ( ) ( )( ) .2 ( )( ) 2 ( )( )

z z f w z z f wf z dw dwi w z w z i w z w zπ π→ ∩

− −= +

− − − −∫ ∫ -------------- (13) Consider the Cartesian representations z x iy= + , z x iy= − , and w u iv= + . We see that

2z z iy− = and w u= along the base. Hence the first integrand on the right in the preceding equation can be rewritten as

[ ][ ] 2 2( ) ( ) 2 ( ) 2 ( ) .

( )( ) ( ) ( ) ( )z z f w iyf u iyf u

w z w z u x iy u x iy u x y−

= =− − − + − − − +

so that Eq. (13) becomes

2 2( ) ( )( ) .

( ) ( )( )R

R

y f u du y f w dwf zu x y w z w zπ π→− ∩

= +− + − −∫ ∫ ----------- (14)

As R →∞ , we have ( ) 0

( )( )f w dw

w z w z∩→

− −∫ Thus, Eq.(14) simplifies to

v

uR

z

Arc

R−

The closed contour C

Base

102

2 2( )( ) .

( )y f u duf z

u x yπ+∞

−∞=

− +∫ ------------ (15) With ( ) ( , ) ( , )f z x y i x yφ ψ= + and ( ) ( , ) ( , )f w u v i u vφ ψ= + , we obtain, from Eq. (15),

2 2( ,0) ( ,0)( , ) ( , ) .( )

y u i ux y i x y duu x y

φ ψφ ψπ

+∞

−∞

++ =

− +∫ By equating the real parts on either side of this equation, we obtain the following formula:

2 2( ,0)( , ) .

( )y ux y du

u x yφφ

π+∞

−∞=

− +∫ ----------- (16)

2 2( ,0)( , ) .

( )y ux y du

u x yψψ

π+∞

−∞=

− +∫ Eq. (16) will yield the value of a harmonic function ( , )x yφ anywhere in the upper half plane provided φ is already known over the entire real axis.

Example 2

As indicated in the figure, the upper half-space Im 0w > is filled with a heat-conducting material. The boundary

0, 0v u= > is maintained at a temperature of 0 while the boundary 0, 0v u= < is kept at temperature at T0. Find the steady-state distribution of temperature ( , )x yφ throughout the conducting material.

Since the temperature is a harmonic function, the Poisson integral formula is directly applicable. We have 0( ,0) , 0u T uφ = < and ( ,0) 0, 0u uφ = > . Thus,

00

2 2 2 20

0( , )( ) ( )

y T du y dux yu x y u x y

φπ π

∞

−∞= +

− + − +∫ ∫ In the first term, make the change of variables p x u= − . Thus,

1 10 0 02 2( , ) tan tan .2x x

T y dp T p T xx yp x y y

πφπ π π

∞∞ − −⎡ ⎤= = = −⎢ ⎥+ ⎣ ⎦∫ -------------- (17)

From the trigonometric identity 1 1tan / 2 tan (1/ )θ π θ− −= − , we that the expression in the brackets on

the right side of Eq. (17) is 1tan ( / )y x θ− = , where θ is the polar angle associated with the point ( , )x y . If we take θ as the principal polar angle in the space 0y ≥ . Thus,

0( , ) , 0Tx yφ θ θ ππ

= ≤ ≤

H.W. 1 (a) An electrically conducting metal sheet is perpendicular to the y-axis and passes through y = 0, as

shown in Fig. A. The right half of the sheet, x > 0, is maintained at an electrical potential of V0 volts while the left half, x < 0, is maintained at a voltage − V0. Show that in the half space, y ≥ 0, the electrostatic potential is given by

( )10 00 02 2( , ) tan Im LnV Vyx y V V z

xφ

π π− ⎛ ⎞= − = −⎜ ⎟⎝ ⎠

where 10 tan ( / )y x π−≤ ≤ .

Poisson Integral Formula (For the Upper half plane)

v

u

(x ,y)

0T 0

y

x0T 0

034T 0

4T

02

T

103

(b) Sketch the equipotential lines (or surfaces) on which 0 0( , ) / 2, ( , ) 0, ( , ) / 2x y V x y x y Vφ φ φ= = = −

(c) Find the components of the electric field Ex and Ey at x =1, y = 1, and draw a vector representing the field at this point.

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 6 , Exercise 4.7, Pearson Education, Inc., 2005.】

y

x

(x ,y)

0V− V0

Infinitesimal gap Fig. A

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