CHAPTER THREE Basic Transcendental Functions of Complex …taiwan921.lib.ntu.edu.tw/mypdf/Jiang-Ch03.pdf · 2014-05-22 · 37 CHAPTER THREE Basic Transcendental Functions of Complex

37

CHAPTER THREE Basic Transcendental Functions of Complex Variables

§3-1 The Exponential Function

1. If iyxz += , the complex exponential ez is given by

(cos sin )

z x iy x iy

xe e e e

e y i y

+= =

= +

上述之定義中，使用到 Euler's Formula： θθθ sincos ie i += ， R∈θ

在此須先證明 e i y 是否亦符合 Euler's Formula，使得 yiyeiy sincos += ------ (1)

所以，底下我們將先舉出在實變函數中 Exponential 函數的性質，而後再證明之。＊在實變微積分中，若定義

axexf =)( 則其有下列之性質

i) 2121 xxxx eee =+ ii) 1)0( =f

iii) )()( xafaexf ax ==′ Using the above mentioned properties, we want to prove that the Equation (1) holds for any real number y.

yiyeiy sincos +=

Let iyeyf =)( )()( yhiyg +=

where both g(y ) and h (y) are real functions and satisfy the properties mentioned above, i.e., )()( yfiyf =′

1)0( =f Thus, when

)()()( yhiygyf ′+′=′ ( )

[ ( ) ( )]( ) ( )

i f yi g y i h yi g y h y

== += −

We know that when the equality holds any y, there must have Real part = Real part and Im = Im

So, we obtain the following equations： ( ) ( ) (2)( ) ( ) (3)

g y h yh y g y′ = − − − − − − −⎧

⎨ ′ = − − − − − −⎩

Differentiating equation (2) with respect to y, gives )()()( ygyhyg −=′−=′′

⇒ 0)()( =+′′ ygyg ⇒ )4(sincos)( 21 −−−−−−+= ycycyg From equation (2), we know that

1 2

( ) ( )sin cos (5)

h y g yc y c y

′= −= − − − − − − −

And then we use the property 1)0( =f : ⇒ 1)0()0()0( =+= higf This implies that

38

(0) 1(0) 0

gh

=⎧⎨ =⎩

⇒ 12

10

cc=⎧

⎨ =⎩

⇒ ( ) cos( ) sin

g y yh y y

=⎧⎨ =⎩

⇒ )()()( yhiygyfeiy +== yiy sincos +=

♣ Some basic properties of the exponential function are discussed as below.

2. If, 111 yixz += and 222 yixz += are two complex numbers, then 221 zzzz zeee +=

)()( 2121121 yixyixzz eeee ++=

[ ] [ ][ ][ ]

1 2

1 2

1 2

1 2 1 2

1 2 1 2

1 2

1 1 2 2

1 2 1 2 1 2 2 1

1 2 1 2( ) ( )

( ) ( )

cos sin cos sin

(cos cos sin sin ) (sin cos sin cos )

cos( ) sin( )

x x

x x

x x

x x i y y

x x i y y

z z

e y i y e y i y

e y y y y i y y y y

e y y i y y

e eee

+

+

+ +

+ + +

+

= + ⋅ +

= − + +

= + + +

=

=

=

3. If ]sin[cos)( yiyezf x += is analytic in the entire complex plane, then zezf =′ )(

＊為了證明上述結果，我們必須先了解下列的性質。 If ),(),()( yxviyxuzf += ，且其

yv

xu

∂∂

=∂∂

， yu

xv

∂∂

−=∂∂

而且 xu∂∂

，yv

∂

∂，

xv

∂∂

，yu

∂∂ 均為連續 ⇔ )(zf is analytic

Since ]sin[cos)( yiyezf x +=

⇒ yeyxu x cos),( =

yeyxv x sin),( = It is easily seen that

yexu x cos=∂∂

yeyv x cos=

∂∂

⇒ yv

xu

∂∂

=∂∂ -------- (1)

And yexv x sin=

∂∂

， yeyu x sin−=

∂∂

⇒ yu

xv

∂∂

−=∂∂ -------- (2)

Furthermore, xu∂∂

，yu

∂∂

，xv

∂∂ and

yv

∂

∂ are continuous everywhere in the xy-plane, and

satisfy the Cauchy-Riemann conditions for all finite values of z everywhere in this plane. ⇒ )(zf is analytic for all z and is therefore an entire function.

39

The derivative ( )f z′ is easily found as

xvi

xuzf

∂∂

+∂∂

=′ )(

cos sin[cos sin ]

x x

x

z

e y i e ye y i ye

= +

= +

=

♣ Chain Rule: If ( )g z is also an analytic function, we have ( ) ( )( ) ( ) ( )g z g zd e e g zdz

′= .

4. zx ee 和之比較如下表所示：

xe ze

1) 0>xe ze 沒有大小

2) 0≠xe )1(0 0 ==⋅≠ − eeee zzz

3) xexf =)( is one-to-one function *** zezf =)( is a periodic function.

Its period is 2πi *** ]sin[cos yiyee xz +=

( 2 )

2

2

[cos( 2 ) sin( 2 )]

,

x

x i y n

x i y n i

z n i

e y n i y ne eee n I

π

π

π

π π+

+ +

+

= + + +

=

=

= ∈

iT π2= (imaginary period)，故知 ze 為週期函數。 ♣ MATLAB Commands for plotting ze , ( )Re ze , and ( )Im ze :

% Magnitude of exp(z) x=[-1:0.05:1]; y=[-6:0.05:6]; [X,Y]=meshgrid(x,y); Z=X+i*Y; w=exp(Z); wm=abs(w); meshz(X,Y,wm);hold on title('Magnitude of exp(z)') % Real part of exp(z) x=[-1:0.05:1]; y=[-6:0.05:6]; [X,Y]=meshgrid(x,y); Z=X+i*Y; w=exp(Z); wm=real(w); meshz(X,Y,wm);hold on surf(X,Y,wm) title('Real Part of exp(z)') % Imaginary part of exp(z) x=[-1:0.05:1]; y=[-6:0.05:6]; [X,Y]=meshgrid(x,y); Z=X+i*Y;

40

w=exp(Z); wm=imag(w); meshz(X,Y,wm);hold on title('Imaginary Part of exp(z)')

41

5. For iyxz += , we have

1) 0≠ze

2) 1=yie and xz ee = (The magnitude of ze is determined by the real part of z.)

3) If ikze z π21 =⇔= , k = constant.

4) If ikzzee zz π22121 =−⇔= where k is an integer.

1) From 0 1z ze e e−⋅ = = , since the product is never zero, neither factor can be zero. Therefore, Cze z ∈∀≠ 0 .

2) 1sincos 22 =+= yye yi

Since ]sin[cos yiyee xz += ,

then yiyee xz sincos +=

1xx

ee

= ⋅

=

3) Let 1]sin[cos =+= yiyee xz

thus cos 1

sin 0

x

x

e y

e y

⎧ =⎪⎨

=⎪⎩

Since Rxe x ∈∀≠ 0 , we see that 0sin =y . Hence Inny ∈= ,π

Because ( )不合11cos0 −=⇒> ye x ⇒ Ikky ∈= ,2 π Thus, we seek that kn 2= . If we take 01 =⇒= xe x 因此， iyxz +=

02

ink i

ππ

= += ←此為必要條件

現在，尚須證明其充分條件亦成立，故 Suppose that ikz π2= , where k I∈ . Then, we obtain that

2

cos2 sin 21

z k ie ek i k

π

π π== +=

4) Since 21 zz ee = 121 =⇔ −zze Hence from the property 3), we have ikzz π221 =− , where k I∈ . The theorem is thus established. Since Cze z ∈∀≠ 0 , and 1=− zzee We have that

zz

ee 1=−

6. Let us show that z ze e=

z x i y x i ye e e+ −= =

(cos sin )

x i y

xe ee y i y

−=

= −

42

(cos sin ) (cos sin )z x xe e y i y e y i y= + = −

Thus, we see that z ze e= ♣ Some useful identities:

1) 0 / 2 3 / 2 / 21, , 1,i i i i ie e i e e i eπ π π π−= = = − = − =

2) 1 0ie π + =

3) DeMoivre’s Theorem: (cos sin ) cos sinni n i nθ θ θ θ+ = + or ( )ni ine eθ θ= . 4) Complex functions of a real variable: ( ) ( ) ( )f t u t iv t= + , where ( )u t and ( )v t are real.

⇒ ( ) ( ) ( )f t u t iv t′ ′ ′= +

Example 1 Find 7 2

7

cos(2 )td e t

dt

⎡ ⎤⎣ ⎦ .

1. Direct differentiation ⇒ laborious ! 2. Note that ( )2 2(1 )cos(2 ) Ret i te t e += .

Let 2(1 )( ) i tf t e += . Then, we have (7) 7 7 2(1 ) 7 2(1 )( ) 2 (1 ) 2 ( 8 )(1 )i t i tf t i e i i e+ += + = × − +

Thus,

( ) ( ){ }7 2

7 2(1 ) 10 27

10 2

cos(2 )Re 2 ( 8 )(1 ) Re 2 sin 2 cos2 sin 2 cos2

2 (sin 2 cos2 )

ti t t

t

d e ti i e e t t i t t

dte t t

+⎡ ⎤⎣ ⎦ ⎡ ⎤= × − + = + + −⎡ ⎤⎣ ⎦⎣ ⎦

= +

H.W. 1 (a) Suppose we want the n-th derivative, with respect to t, of ( )2( ) / 1f t t t= + . Notice that

1( ) Ref tt i

⎛ ⎞= ⎜ ⎟−⎝ ⎠ and that the n-th derivative of the function in the brackets is exactly taken. Using

the method of Example 1, as well as the binomial theorem (which perhaps should be reviewed), show that

( 1) / 2 1 2( )

2 10

( 1) !( 1)! ( 1)( )(2 )!( 1 2 )!( 1)

n k n kn

nk

n n tf tk n kt

+ + −

+=

− + −=

+ −+∑ , for n odd,

1 2/ 2( )

2 10

!( 1)! ( 1)( )(2 )!( 1 2 )!( 1)

k n knn

nk

n n tf tk n kt

+ −

+=

+ −=

+ −+∑ , for n even.

(b) Using the method of part (a), find similar expressions for the nth derivative of ( )2( ) 1/ 1f t t= + . Note that this function is identical to ( )Im 1/( )t i− .

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3.1, Problem 24, Pearson Education, Inc., 2005.】 H.W. 2 The absolute magnitude of the expression

12 ( 1)

01

Ni i i N in

nP e e e eψ ψ ψ ψ

−−

== + + + + = ∑

is of interest in many problems involving radiation from N identical physical elements (e.g., antennas, loudspeakers). Here ψ is a real quantity that depends on the separation of the elements and the position of an observer of the radiation. P can tell us the strength of the radiation observed. (a) Using the formula for the sum of a finite geometric series (see H.W. 8 in page 6, lecture note of

ch_01), show that

43

sin / 2( )sin / 2

NP ψψψ

=

(b) Find 0lim ( )Pψ ψ→ .

(c) Use a calculator or a simple computer program to plot ( )P ψ for 0 2ψ π≤ ≤ when N = 3. 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3.1, Problem 25, Pearson Education, Inc., 2005.】

(a) Let iz e ψ= , then 1 1

2 1 1 111 1

N NN z zP z z z

z z

− +− − −= + + + + = =

− −.

⇒ ( ) ( )( ) ( )

/ 2 / 2 / 2

/ 2 / 2 / 2

/ 2

/ 2

/ 2

/ 2

1 1( )1 1

cos( / 2) sin( / 2) cos( / 2) sin( / 2)cos( / 2) sin( / 2) cos( / 2) sin( / 2)

2 sin( / 2)2 sin(

iN iN iN iN iN

i i i i i

iN

i

iN

i

e e e e ePe e e e e

N i N N i Nei ie

e i Nie

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ

ψ

ψ

ψ

ψ

ψ ψ ψ ψψ ψ ψ ψ

ψψ

−

−

⎡ ⎤− − −= = = ⎢ ⎥

− − −⎢ ⎥⎣ ⎦⎡ ⎤+ − −

= ⎢ ⎥+ − +⎣ ⎦

=/ 2)

Thus, we have / 2

/ 22 sin( / 2) sin( / 2)( )2 sin( / 2) sin( / 2)

iN

ie i N NP

ie

ψ

ψψ ψψψ ψ

= = ------------ Q.E.D.

(b) Consider

0

sin( / 2) / 2 cos( / 2)limsin( / 2) 1/ 2 cos( / 2)

N N N Nψ

ψ ψψ ψ→

= =

Thus, we have

0lim ( )P Nψ

ψ→

=

(c) MATLAB commands: % H.W.1 part (c) N=3; phi=[0:0.1:2*pi]; y=sin(N*phi)./sin(phi); Y=abs(y); plot(phi,Y)

0 1 2 3 4 5 6 70

0.5

1

1.5

2

2.5

3

44

§3-2 The Trigonometric Function

1. Since xixe xi sincos +=

xixe xi sincos −=− then we see that

sin2

cos2

i x i x

i x i x

e exi

e ex

−

−

⎧ −=⎪⎪

⎨+⎪ =⎪⎩

⇒ 屬於實數

and sinh

2

cosh2

x x

x x

e ex

e ex

−

−

⎧ −=⎪⎪

⎨+⎪ =⎪⎩

⇒ 屬於實數

1) Given any complex number z , we define

sin2

cos2

i z i z

i z i z

e ezi

e ez

−

−

−=

+=

2) The functions zsin and zcos are analytic for all values of z. Moreover,

sin2

2cos

i z i z

i z i z

d i e i ezdz i

e e

z

−

−

+=

+=

=

cos2

( )2

sin

i z i z

i z i z

d i e i ezdz

e ei

z

−

−

−=

− −=

= −

2. 底下是一些三角函數和 Hyperbolic Function 的性質：

1) 1sincos 22 =+ xx , 1sinhcosh 22 =− xx

2) cxdxx

+=−

−∫ 12 sin11 ,

cxdxx

+=+

−∫ 12 sinh11

3) ℒ 2 2{sin }t sωωω

=+

,

ℒ 2 2{sinh }t sωωω

=−

4) ℒ 2 2{cos }st

sω

ω=

+,

ℒ 2 2{cosh }st

sω

ω=

−

3. If iyxz += , then

1) yxiyxz sinhcoscoshsinsin += 2) yxiyxz sinhsincoshcoscos −=

45

1) ieez

zizi

2sin

−−=

( ) ( )

2

2[cos sin ] [cos sin ]

2cos ( ) sin ( )

2

cos sin2 2

sin cosh cos sinh

i x i y i x i y

y i x y i x

y y

y y y y

y y y y

e ei

e ei

e x i x e x i xi

x e e i x e ei

e e e ei x

x y i x y

+ − +

− + − −

− −

− −

−=

−=

+ − −=

− + +=

− += +

= +

2) 同理可證。在此另舉一證法：由 Taylor's Series，知

3 5 2 1

03 5 2 1

0

( 1)sin3! 5! (2 1)!

sinh3! 5! (2 1)!

n n

nn

n

z z zz zn

z z zz zn

+∞

=+∞

=

⎧ −= − + − + =⎪

+⎪⎨⎪ = + + + =⎪ +⎩

∑

∑

比較上列二式可得 sinh sinsin sinh

iz i ziz i z

=⎧⎨ =⎩

同樣地，由下二式 2 4 2

02 4 2

0

( 1)cos 12! 4! (2 )!

cosh2! 4! (2 )!

n n

nn

n

z z zzn

z z zz zn

∞

=∞

=

⎧ −= − + − + =⎪

⎪⎨⎪ = + + + =⎪⎩

∑

∑

比較可得 cosh coscos cosh

iz ziz z

=⎧⎨ =⎩

由上述結果我們便可證明 2) 之結果： Since )cos(cos yixz +=

cos cos sin sincos cosh sin sinh

x i y x i yx y i x y

= −= −

4. 1sin ≤x ，在實變函數中成立；但在複變函數中， 1sin ≤z ，不一定成立。

Since yxiyxz sinhcoscoshsinsin += ，

thus yxyxz 22222 sinhcoscoshsinsin +=

yx

yxyx22

2222

sinhsin

sinh)sin1()sinh1(sin

+=

−++=

If we take 2π

=x ， 1=y ，then obtain

1sin 2 =x ， 0sinh 2 >y

⇒ 1sin 2 >z ⇒ 1sin ≤z ，不一定成立

5. If Innzz ∈=⇒= ，π0sin .

46

From 02

sin =−=−

ieez

zizi

⇒ 12 =⇒= − zizzi eee i

⇒ 1)22( =+ yixie ⇒ 1]2sin2[cos2 =+− xixe y ⇒ 02sin =x Since 02 >ye ，when 02sin =x , we know 12cos ±=x But in this case we must take 12cos +=x （如此 02 >ye 才能配合題意之要求）. ⇒ πnx 22 = ⇒ πnx = 而因 12cos =x ，故取 12 =− ye 故知 0=y ⇒ iyxz +=

0,

n in n Iππ

= += ∈

6. If 1cos 02

z z n n Iπ⎛ ⎞= ⇒ = + ∈⎜ ⎟⎝ ⎠

， .

From 02

cos =+=− zizi eez , we see that zzi ee i−= .

⇒ 12 −=zie ⇒ 1]2sin2[cos2 −=+− xixe y Then, we have sin 2 0x = + .

Since 02 >− ye ，when 02sin =x , we know that 12cos ±=x . But in this case we must take 12cos −=x （such that they satisfy 02 >− ye ， 12cos2 −=− xe y ）. ⇒ π)12(2 += nx

⇒ 12

x n π⎛ ⎞= +⎜ ⎟⎝ ⎠

又因 12cos −=x ，故取 12 =− ye 。 ⇒ 02 =− y ⇒ 0=y 故 iyxz +=

1 0212

n i

n

π

π

⎛ ⎞= + +⎜ ⎟⎝ ⎠⎛ ⎞= +⎜ ⎟⎝ ⎠

7. Given the complex number z , we define

1) sin 1tancos 2

zz z nz

π⎛ ⎞= ≠ +⎜ ⎟⎝ ⎠

，

2) coscotsin

zz z nz

π= ≠，

3) 1seccos 2

z z nz

π π= ≠ +，

4) 1cscsin

z z nz

π= ≠，

where In∈ in all above cases. 5) ztan and zcot both have a fundamental period of π. 6) zsec and zcsc both have a fundamental period of 2π.

8. ztan ， zcot ， zsec and zcsc are analytic functions of z except for the above mentioned

y

x 0 π 2π 3π -3π -2π -π

y

x0

π2

3− π

2

1− π

2

1π

2

3 π

2

5

47

limiting values. We can define that

1) ππ nzzzdzd

+≠=2

sec)(tan 2 ，

2) πnzzzdzd

≠−= ，2csc)(cot

3) ππ nzzzzdzd

+≠=2

tansec)(sec ，

4) πnzzzzdzd

≠−= ，cotcsc)(csc

where in all cases , In∈ .

9. If iyxz += , then

1) sin sinz z= ， cos cosz z=

sin sin( ) sin( )z x i y x i y= + = −

yxiyxyixyix

sinhcoscoshsinsincoscossin

−=−=

⇒ sin sinz z= yxiyxz sinhcoscoshsinsin +=

sin sin cosh cos sinhz x y i x y= −

同理 cos cosz z=

2) yxz 222 sinhsin|sin| +=

3) yxz 222 sinhcos|cos| +=

4) 1cossin 22 =+ zz 5) 212121 sincoscossin)sin( zzzzzz ±=± 6) 212121 sinsincoscos)cos( zzzzzz ∓=±

7) sin cos2

z zπ⎛ ⎞− =⎜ ⎟⎝ ⎠

8) zzz cossin22sin = 9) zzz 22 sincos2cos −=

以上各式之證明均非常簡單，留待給讀者證明。

10. Let us show that

21

2121 tantan1

tantan)tan(zzzzzz

−+

=+

where 1 212

z z n π⎛ ⎞+ ≠ +⎜ ⎟⎝ ⎠

, In∈ .

)cos()sin()tan(

21

2121 zz

zzzz++

=+

21

21

2121

2121

tantan1tantan

sinsincoscossincoscossin

zzzz

zzzzzzzz

−+

=

−+

=

♣ MATLAB Commands for plotting cos( )z :

% Magnitude of cos(z) x=[-6:0.05:6]; y=[-1:0.05:1]; [X,Y]=meshgrid(x,y); Z=X+i*Y; w=cos(Z);

48

wm=abs(w); meshz(X,Y,wm);hold on title('Magnitude of cos(z)')

H.W.1 Let ( ) sin(1/ )f z z= .

(a) Express this function in the form ( , ) ( , )u x y iv x y+ . Where in the complex plane is this function analytic?

(b) What is the derivative of ( )f z ? Where in the complex plane ( )f z′ is analytic? 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3.2, Problem 22, Pearson Education, Inc., 2005.】 (a) Since ( ) sin(1/ )f z z= , where 1/ z is analytic except for z = 0. But, sinz is analytic for all z. Therefore,

we have sin(1/ )z is analytic for all 0z ≠ . Since z x iy= + , we have

2 2 2 2 2 2

2 2 2 2 2 2 2 2

sin(1/ ) sin sin

sin cosh cos sinh

x iy x yz ix y x y x y

x y x yix y x y x y x y

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

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

where

2 2 2 2( , ) sin coshx yu x y

x y x y

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

and 2 2 2 2( , ) cos sinhx yv x y

x y x y

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

(b) 21sin(1/ ) cos(1/ )d z z

dz z⎛ ⎞= −⎜ ⎟⎝ ⎠

----------- analytic for all 0z ≠ .

H.W. 2 Show that sin(2 ) sinh(2 )tancos(2 ) cosh(2 )

x i yzx y+

=+

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3.2, Problem 28, Pearson Education, Inc., 2005.】

Since

2 2 2 2

sin sin cosh cos sinhtancos cos cosh sin sinh(sin cosh cos sinh )(cos cosh sin sinh )

cos cosh sin sinh

z x y i x yzz x y i x yx y i x y x y i x y N

Dx y x y

+= =

−+ −

= ≡+

Consider the denominator D:

49

( ) ( )

[ ]

2 2 2 2 2 2 2 2

2 2

cos cosh sin sinh cosh 1 sin sin cosh 1

cosh sin(1 cosh 2 ) [1 cos(2 )]

21 cosh(2 ) cos(2 )2

x y x y y x x y

y xy x

y x D

+ = − + −

= −+ − −

=

= + =

Now, consider the numerator N: The real part of N is

( )2 2

2 2

cosh sin cos sinh sin cos1sin cos cosh sinh sin cos sin(2 )2

y x x y x x

x x y y x x x

−

= − = =

Then, consider the imaginary part of numerator:

( )2 2

2 2

cos cosh sinh sin cosh sinh1sinh cosh cos sin sinh cosh sinh(2 )2

x y y x y y

y y x x y y y

+

= + = =

Thus, we have

( )

1 12 2

12

sin(2 ) sinh(2 )

cosh 2 cos2

x i yND y x

+=

+

This implies that sin(2 ) sinh(2 )tancos(2 ) cosh(2 )

x i yzx y+

=+

---------- Q.E.D.

H.W. 3 (a) Since yxiyxz sinhcoscoshsinsin += and sinh coshy y≤ , show that

sinh sin coshy z y≤ ≤ .

(b) Derive a comparable double inequality for cos z . 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3.2, Problem 30, Pearson Education, Inc., 2005.】 (a) Since yxiyxz sinhcoscoshsinsin += , we have

( )

2 2 2 2

2 2 2 2

2 2 2

2

sin sin cosh cos sinh

sin cosh cos sinh

sin cosh cos cosh sinh cosh

cosh sin cos

cosh cosh cosh (since cosh is positive!)

z x y i x y

x y x y

x y x y y y

y x x

y y y y

= +

= +

≤ + ≤

= +

= = =

∵

∴ sin coshz y≤ ----------- (1) On the other hand, we have

( )

2 2 2 2

2 2 2 2

2 2 2

2

sin sin cosh cos sinh

sin sinh cos sinh sinh cosh

sinh sin cos

sinh sinh

z x y x y

x y x y y y

y x x

y y

= +

≥ + ≤

= +

= =

∵

⇒ sinh siny z≤ ------------ (2) From Eqs.(1) and (2), we conclude that

sinh sin coshy z y≤ ≤ ------------ Q.E.D. (b) Since cos cos cosh sin sinhz x y i x y= − , we have

50

( )

2 2 2 2

2 2 2 2

2 2 2

2

cos cos cosh sin sinh

cos cosh sin sinh

cos cosh sin cosh sinh cosh

cosh sin cos

cosh cosh cosh (since cosh is positive!)

z x y i x y

x y x y

x y x y y y

y x x

y y y y

= −

= +

≤ + ≤

= +

= = =

∵

∴ cos coshz y≤ ----------- (3) Similarly, we have

( )

2 2 2 2

2 2 2 2

2 2 2

2

cos cos cosh sin sinh

sin sinh cos sinh sinh cosh

sinh sin cos

sinh sinh

z x y x y

x y x y y y

y x x

y y

= +

≥ + ≤

= +

= =

∵

⇒ sinh cosy z≤ ------------ (4) From Eqs.(3) and (4), we conclude that

sinh cos coshy z y≤ ≤ ------------ Q.E.D. H.W. 4 Find all roots of the equation sin cosh 4z = .

( )2 2 4 , 0, 1, 2,z n i nπ π= + ± = ± ± 【本題摘自：James Ward Brown and Ruel V. Churchill, Complex Variables and Applications, 6rd ed., Exercise 24, Problem 16, McGraw-Hill, Inc., 2005.】 H.W. 5 Find all roots of the equation cos 2z = . ( )12 cosh 2 or 2 ln 2 3 , 0, 1, 2,z n i n i nπ π−= + ± + = ± ± 【本題摘自：James Ward Brown and Ruel V. Churchill, Complex Variables and Applications, 6rd ed., Exercise 24, Problem 17, McGraw-Hill, Inc., 2005.】

51

§3-3 The Hyperbolic Functions

1. Recall that when we study in the real variable, the hyperbolic functions have been defined as following:

2sinh

xx eex−−

= , its graph is as shown in Fig. (a).

and 2

coshxx eex

−+= , its graph is as shown in Fig. (b).

2. Given any complex number z , we define

2

sinhzz eez

−−= ，

ieez

zizi

2sin

−−=

2

coshzz eez

−+= ，

2cos

zizi eez−+

=

1) 若取 z 為實數，則上述定義與實變函數中的雙曲線函數之定義可相符合。 2) zsinh and zcosh both have a fundamental period of 2π i . 3) zz sinh)sinh( −=−

zz cosh)cosh( =−

4) 1sinhcosh 22 =− zz

5) sinh2

z zd d e ezdz dz

−⎛ ⎞−= ⎜ ⎟

⎝ ⎠

zeezz

cosh2

)(=

−−=

−

)2

(coshzz ee

dzdz

dzd −+

=

zeezz

sinh2

=−

=−

6) If Ininzz ∈=⇒= ,0sinh π . Since 0sinh =z

⇒ 02

z ze e−⎛ ⎞−=⎜ ⎟⎜ ⎟

⎝ ⎠

⇒ zz ee −= Then, we have

122 =+ yixe This means that

y

x 0

3π i

2π i

π i

−2π i

−π i

2sinh

xx eexy−−

==

-5

-60

-40

-20

20

40

60

y

0 5x

Fig. (a)

2cosh

xx eexy−+

==

-5

-60

-40

-20

20

40

60

y

0 5x

Fig. (b)

52

1)2sin2(cos2 =+ yiye x 利用實部=實部，虛部=虛部，我們可得知 02sin =y ----------

and 12cos2 =⋅ ye x ----------

Since 02 >xe ⇒ 12cos =y （ 1− 不合） From , we can obtain that πny 22 = ⇒ πny = Again, we need

12 =xe ⇒ 0=x So, if sinh 0z = , we conclude that

iyxz += inin ππ =+= 0

7) If 1cosh 0 ,2

z z n i n Iπ⎛ ⎞= ⇒ = + ∈⎜ ⎟⎝ ⎠

Since 02

cosh =+=− zz eez ,

⇒ zz ee −−= ⇒ 12 −=ze ⇒ 1]2sin2[cos2 −=+− yiye x ⇒ 02sin =y

Since 02 >xe ⇒ 12cos −=y ⇒ π)12(2 += ny

故 1 ,2

y n n Iπ⎛ ⎞= + ∈⎜ ⎟⎝ ⎠

.

Again we need 12 −=xe ⇒ 0=x So, if sinh 0z = , we conclude that

iyxz += 12

n iπ⎛ ⎞= +⎜ ⎟⎝ ⎠

* 由(6)及(7)知，只有純虛數才能使 sinh z 和 cosh z 等於 0。

3. Given the complex number z, we define the other hyperbolic functions as followings:

1) sinh 1tanhcosh 2

zz z n iz

π⎛ ⎞= ≠ +⎜ ⎟⎝ ⎠

， .

2) coshcothsinh

zz z n iz

π= ≠， .

3) 1 1seccosh 2

h z z n iz

π⎛ ⎞= ≠ +⎜ ⎟⎝ ⎠

， .

4) 1cscsinh

h z z n iz

π= ≠， .

where In∈ in all cases. 5) Both ztanh and zcoth have a fundamental period of π i . 6) Both zhsec and zhcsc have a fundamental period of 2π i .

4. ztanh , zcoth , zhsec and zhcsc are analytic functions of z，其中 z 值不可為上述重點 3

所限制的各不允許值，則

1) 2 1(tanh ) sech2

d z z z n idz

π⎛ ⎞= ≠ +⎜ ⎟⎝ ⎠

，

2) 2(coth ) cschd z z z n idz

π= − ≠，

y

x0

3π i

2π i

π i

−π i

−2π i

iπ23

−

iπ21

−

iπ21

iπ23

iπ25

53

3) 1(sech ) sech tanh2

d z z z z n idz

π⎛ ⎞= ≠ +⎜ ⎟⎝ ⎠

，

4) (csch ) csch cothd z z z z n idz

π= − ≠，

where In∈ in all cases.

5. If iyxz += , then 1) xyixyz coshsinsinhcossinh += 2) xyiyyz sinhsincoshcoscosh +=

1) 2

sinhzz eez

−−=

( ) ( )

21 [ (cos sin ) (cos sin )]2

cos sin2 2

cos sinh sin cosh

x i y x i y

x x

x x x x

e e

e x i y e y i y

e e e ey i y

y x i y x

+ − +

−

− −

−=

= + − −

− += +

= +

2) zcosh 同理可證，亦可仿 §3-2 3.之 2) 證之。

6. If iyxz += , then a) zizi sin)(sinh = ， zizi sinh)(sin = b) zzi cos)(cosh = ， zzi cosh)(cos =

c) sinh sinhz z= ， cosh coshz z=

d) xyz 222 sinhsin|sinh| +=

xyz 222 sinhcos|cosh| += H.W. 1 (a) Where on the line x y= is the equation sin sinh 0z i z+ = satisfied?

(b) Using MATLAB, obtain a three-dimensional plot of sin sinhz i z+ and verify that the surface obtained has zero height at points in Part (a). Include 0z = and at least one other solution, on the line, of the given equation.

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3.3, Problem 20, Pearson Education, Inc., 2005.】 (a) Since sin sinh 0z i z+ = , we have

sin cosh cos sinh [sinh cos cosh sin ] 0x y i x y i x y i x y+ + + = ---------- (A) Using Re = Re and Im =Im, gives

sin cosh cosh sin 0x y x y− = Put x y= in the preceding equation, yields

sin cosh cosh sinx x x x= ---------- satisfied! Equating the imaginary part in Eq.(A), gives

cos sinh sinh cos 0x y x y+ = Put x y= in the preceding equation, yields

2sinh cos 0x x =

⇒ 0x = or 2

, 0, 1, 2, 3,x n nππ⎡ ⎤= ± + =⎢ ⎥⎣ ⎦. Also, y x= in this case.

(b) MATLAB commands:

x=[0:0.05:2]; y=[0:0.05:2] ; [X,Y]=meshgrid(x,y); Z=X+i*Y;

54

w=sin(Z)+i*sinh(Z); wm=abs(w); meshz(X,Y,wm);view(150,70)

H.W. 2 Find all roots of the equation (a) 1

2cosh z = , (b) sinh z i= , and (c) cosh 2z = − .

【本題摘自：James Ward Brown and Ruel V. Churchill, Complex Variables and Applications, 6rd ed., Exercise 25, Problem 14, Pearson Education, Inc., 2005.】

(a) ( )132 , 0, 1, 2,n i nπ± = ± ± (b) ( )122 , 0, 1, 2,n i nπ+ = ± ± (c) ( )ln 2 3 (2 1) , 0, 1, 2,n i nπ± + + + = ± ±

55

§3-4 The Logarithmic Function

1. For any complex number 0≠z , there exists complex numbers w such that 0≠= zew . In particular, one of such w’s is the complex number as shown below:

θizw += ||ln

Let θierz = , where rz =|| . Hence, we have

ln| |

ln

ln

w z i

r i

r i

i

e eee er ez

θ

θ

θ

θ

+

+=

=

= ⋅

==

Thus, we can define that

zizizzarg||ln

||lnln+=+= θ

1) Define πθπθ ≤

56

2 2 2 2

2 21

#

x yix y x yx i y z

z z zx y

= −+ +−

= = =⋅+

ii) Polar coordinate method： Let )(zfw = and (cos sin )z x iy r iθ θ= + = + , then

⎩⎨⎧

==

θθ

sincos

ryrx

In Chapter 2, we have derived the formula as shown below:

rwi

zdwd

∂∂

−= )sin(cos θθ

Note that θizzzfw +=== ||lnLn)(

θir += ln

⇒ rwi

zdwd

∂∂

−= )sin(cos θθ

1(cos sin )

1(cos sin )

1

ir

r i

z

θ θ

θ θ

= − ⋅

=⋅ +

=

----------- Q.E.D.

2. In the real variable analysis, we know that if 01 >x and 02 >x , then

2121 lnlnln xxxx +=⋅ However, in the complex variable analysis, the following 2121 LnLnLn zzzz +=⋅ may be not true. For example: Let iziz +−== 1, 21 then izz −−=⋅ 121 First, we have

iz LnLn 1 =

ln | |2

2

i i

i

π

π

= +

=

and )1Ln(Ln 2 iz +−= 3ln | 1 |4

3ln 24

i i

i

π

π

= − + +

= +

But 1 2Ln( ) Ln ( 1 )z z i⋅ = − − 3ln | 1 | ( )4

3ln 24

i i

i

π

π

= − − + −

= −

Hence, iizz ππ432ln

2LnLn 21 ++=+

iπ452ln +=

57

1 2Ln( ) #z z≠ ⋅

♣ But, the following theorems are still satisfied.

3. If the complex numbers 1z , 2z , 3z are different from zero, then the principal values of the arguments and logarithms of the product, quotient, and powers among these complex numbers are given by

1) )(),(2argarg)(Arg 2111121 azznzzzz －π++=⋅ )(),(2LnLn)(Log 2112121 bzznizzzz －π++=⋅

where 1n is assumed to be the values of 1, 0, 1− as following：

⎪⎪⎩

⎪⎪⎨

⎧

−≤+

58

where 2n is assumed to be the values of 1, 0, 1− as followings：

⎪⎪⎩

⎪⎪⎨

⎧

−≤−

59

Example 3 Find )22(Ln i+ .

Ln ( 2 2)ln | 2 2 | arg( 2 2)

ln 24

ii i i

i π

+= + + +

= +

Example 4 Find the principal value of i

ieπ3

)31(2 ⎥⎦

⎤⎢⎣⎡ −− .

3

3

33

1 3 1 33 ln arg2 2

( 1 3)2

1 3( )2

1 32

[cos3 sin 3 ]

i

i

ii

i ii i

e i

ie

ie

i e

π

π

ππ

ππ π

⎡ ⎤⎛ ⎞− − − −+⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤− −⎢ ⎥⎣ ⎦

⎡ ⎤− −= ⎢ ⎥⎣ ⎦

⎛ ⎞− −= ⎜ ⎟

⎝ ⎠

= +

2

23 ln13

233

2

( 1)i i

i i

e

e

e

ππ

ππ

π

⎡ ⎤⎛ ⎞+ −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞−⎜ ⎟⎝ ⎠

= − ⋅

= −

= −

5. If 0≠z , w and λ are any complex number,

then kizwz w π2Ln)(Ln += ----------- (1)

kiww ezz λπλλ 2)( ⋅= ----------- (2) where k is the integer given by bracket function

⎥⎦⎤

⎢⎣⎡ +−=

π2Arg)Re(||ln)Im(

21 zwzwk

Denote wz by α , zarg by θ , and let viuw += , then

zww ez Ln==α

]||ln[||ln

]||][ln[

θθ

θ

uzvivzu

izviu

ee

e

+−

++

⋅=

=

Thus,

⎪⎩

⎪⎨⎧

++=

= −

kuzvz

e vzu

πθ

α θ

2||lnarg

|| ||ln ------------ (3)

where k is the integer such that ππθπ ≤++

60

⎥⎦⎤

⎢⎣⎡ +−=

⎥⎦⎤

⎢⎣⎡ +−=+=

π

πθ

2arg)Re(||Ln)Im(

21

2||Ln

21]1[

zwzw

uzvtk

Now, using Eq. (3), we see that

kizw

kiizviu

kuzvivzu

kuzvie

i

z

vzu

w

π

πθ

πθθ

πθ

αα

α

θ

2Ln

2)||)(ln(

]2||ln[||ln

]2||ln[][Ln

arg||ln

Ln)(Ln

||ln

+=

+++=

+++−=

+++=

+=

=

−

故(1)式已被證得。同理，可證得(2)式。

H.W. 1 Consider the identity ln lnnz n z= , where n is an integer, which is valid for appropriate choice of

the logarithms on each side of the equation. Let 1z i= + and 5n = . (a) Find the values of ln nz and ln z that satisfy ln lnnz n z= . (b) For the given z and n, is ln lnnz n z= satisfied? (c) Suppose n = 2 and z is unchanged. Is ln lnnz n z= then satisfied?

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3.4, Problem 26, Pearson Education, Inc., 2005.】

(a) 54

5ln ln 5 Ln 2z z i π⎡ ⎤= = +⎢ ⎥⎣ ⎦

(b) No. (c) Yes.

6. Definition (Branch)：

A branch of a multivalued function is a single-valued function analytic in some domain. At every point of the domain, the single-valued function must assume exactly one of the various possible values that the multivalued function can assume.

Definition (Branch cut) A line used to create a domain of analyticity is called a branch line or branch cut.

Definition (Branch point) Any point that must lie on a branch cut ⎯ no matter what branch used ⎯ is called a branch point of a multivalued function.

Ex. (1) Ln z ≡ principal value of ln z, which is defined for all z except z = 0. (2) Ln z is also used to denote the principal branch of the logarithm function, which is defined for all

z except z = 0 and values of z on the negative real axis. (3) ( ) Lnf z r iθ= + , where 3 / 2 / 2π θ π− < ≤ .

⇒ It is discontinuous at the origin and at all points on the positive imaginary axis. (4) ( ) Lnf z r iθ= + , where 3 / 2 2 / 2 2 , 0, 1, 2,k k kπ π θ π π− + < ≤ + = ± ± , are, for each k,

analytic branches, provided z is confined to the domain D1. Example 5 (a) Find the largest domain of analyticity of ( ) Ln[ (3 4 )]f z z i= − + .

(b) Find the numerical value of (0)f .

(a) For the given function, the non-analytic points are located at：

61

Im 0, Re 0w w= ≤

If (3 4 )w z i= − + , these two conditions can be rewritten as

[ ]Im ( ) (3 4 ) 0 4x iy i y+ − + = ⇒ = [ ]Re ( ) (3 4 ) 0 3x iy i x+ − + ≤ ⇒ ≤

So, the full domain of analyticity is shown in the following Fig. (a).

(b) (0) Ln( 3 4 ) Ln(5) arg( 3 4 )f i i i= − − = + − − , where arg( 3 4 )iπ π− < − − < .

⇒ (0) Ln(5) 2.214f i= − ⇒ Fig. (b).

H.W. 2 (a) Show that ( )Ln Ln( )z is analytic in the domain consisting of the z-plane with a branch cut along

the line 0, 1y x= ≤ . (See below) (b) Find ( )Ln Ln( )d z dz within the domain of analyticity found in part (a). (c) What branch cut should be used to create the maximum domain of analyticity for

( )Ln Ln Ln( )z⎡ ⎤⎣ ⎦ ? 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3-5, Problem 15, Pearson Education, Inc., 2005.】

H.W. 3 The complex electrostatic potential ( , ) Ln(1/ )x y i zφ ψΦ = + = , where 0z ≠ , can be created by an

electric line charge located at 0z = and lying perpendicular to the xy-plane. (a) Sketch the streamlines for this potential. (b) Sketch the equipotentials for 1, 0, 1, 2φ = − .

jy

x

− 4i

− 3

− 2.214 (approx.)

jy

x

Branch cut y = 4, x ≤ 3

3 + 4i

Fig. (a)Fig. (b)

jy

x.

Domain of analytic of Ln z (shaded)

jy

x

Branch cut Domain D1

jy

x

Branch cut

1

Fig. HW-2

62

(c) Find the components of the electric field at an arbitrary point (x, y). 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3-5, Problem 16,

Pearson Education, Inc., 2005.】 (a) Since ( , ) Ln(1/ ) Lnx y i z zφ ψΦ = + = = −

⇒ 2 2( , ) Lnx y x yφ ⎛ ⎞= − +⎜ ⎟⎝ ⎠

, ( )1( , ) tan yxx yψ −= − , where ψ ’s are ramp emanating from origin. (b) Since 2 21 Ln 1x y φ⎛ ⎞− = − + = = −⎜ ⎟

⎝ ⎠, therefore

2 2x y e+ = if 1φ = −

Also, when 2 20 Ln 0x y φ⎛ ⎞= − + = =⎜ ⎟⎝ ⎠

, therefore

2 2 1x y+ = if 0φ =

When 2 21 Ln 1x y φ⎛ ⎞= − + = =⎜ ⎟⎝ ⎠

, therefore

2 2 1/x y e+ = if 1φ = The electric field is

x ydE iEdzΦ⎛ ⎞+ = −⎜ ⎟

⎝ ⎠

⇒ 2 2xxE

x y=

+ and 2 2y

yEx y

=+

(c) MATLAB commands:

% HW (c) x=[-2.5:0.02+0.001:2.5]; y=x; [X,Y]=meshgrid(x,y); Z=X+i*Y; w=log(Z-1-1*i); %wm=real(w); % for real part wm=imag(w); % for imag part Meshz(X,Y,wm); % for imag part and real View(45,45);

H.W. 4 Let

3( )( ) 10 zf z = . This function is evaluated such that ( )f z′ is real when 1z = . Find (1 )f i′ + . Where in the complex plane is ( )f z analytic?

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3-6, Problem 24, Pearson Education, Inc., 2005.】

ψ = π/4

ψ = −π/4

ψ = −3π/4

x

y

x

yφ = 1

φ = 0

φ = − 1

63

(1 ) 0.137 0.0148f i i′ + = − ; ( )f z is an entire function.

(a) real part

(b) imaginary part

H.W. 5 Let ( )( ) 10

zef z = . This function is evaluated such that 2( / 2)f i e ππ −= . Find ( )f z′ and ( / 2)f iπ′ .

【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3-6, Problem 24, Pearson Education, Inc., 2005.】

[ ][Ln(10) 2 ]( ) Ln(10) 2zi e zf z e i eπ π+′ = + and ( / 2) 0.00464 0.01f i iπ′ = −

64

§3-5 Inverse Trigonometric and Hyperbolic Functions

1. 1) Let sinz w= , then ( )1/ 21 2sin ln 1z i zi z− ⎡ ⎤= − + −⎢ ⎥⎣ ⎦

Since 2

iw iwe ezi

−−= , assume that iwp e= and 1/ iwp e−= , then we have

1/2

p pzi

−=

⇒ 22 1izp p= −

⇒ 2 2 1 0p izp− − = Solve this equation for p:

( )1/ 221p zi z= + − or ( )1/ 221iwe zi z= + − ⇒ ( )1/ 221 ln 1w zi zi

⎡ ⎤= + −⎢ ⎥⎣ ⎦

Since 1sinw z−= , we have

( )1/ 21 21sin ln 1z zi zi− ⎡ ⎤= + −⎢ ⎥⎣ ⎦

2) Other inverse trigonometric functions:

( )1/ 21 21cos ln 1z z zi− ⎡ ⎤= + −⎢ ⎥⎣ ⎦

1tan ln2i i zz

i z− +⎛ ⎞= ⎜ ⎟−⎝ ⎠

2. Inverse of Hyperbolic function

1) ( )1/ 21 2sinh ln 1z z z− ⎡ ⎤= + +⎢ ⎥⎣ ⎦

2) ( )1/ 21 2cosh ln 1z z z− ⎡ ⎤= + −⎢ ⎥⎣ ⎦

3) 1 1 1tanh ln2 1

zzz

− +⎛ ⎞= ⎜ ⎟−⎝ ⎠

3. Derivative of inverse trigonometric functions:

1) ( )( )

1/ 21 21/ 22

1 1sin ln 11

d dz zi zdz dz i z

− ⎧ ⎫⎡ ⎤= + − =⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭ −

2) ( )

11/ 22

1cos1

d zdz z

− −=−

3) ( )

12

1tan1

d zdz z

− =+

4. Derivative of inverse hyperbolic functions:

1) ( )

11/ 22

1sinh1

d zdz z

− =+

2) ( )

11/ 22

1cosh1

d zdz z

− =−

65

3) ( )

12

1tanh1

d zdz z

− =−

Example 1 Find 1sin (1/ 2)− .

1/ 21 1 3sin (1/ 2) ln

2 4i

i− ⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

Taking ( )1/ 23/ 4 3 / 2= , the above expression becomes

( )1 1 3sin (1/ 2) ln ln 1 ( / 6) 2 , 0, 1, 2,2 2 6i i k k

iππ π−

⎡ ⎤= + = − ∠ = + = ± ±⎢ ⎥

⎣ ⎦

Taking ( )1/ 23/ 4 3 / 2= − , the above expression becomes

( )1 1 3 5sin (1/ 2) ln ln 1 (5 / 6) 2 , 0, 1, 2,2 2 6i i k k

iππ π−

⎡ ⎤= − = − ∠ = + = ± ±⎢ ⎥

⎣ ⎦

Example 2 Find all the numbers whose sine is 2.

This question is to solve sin 2z = . Thus, we have

( )1/ 21 1sin (2) ln 2 3ii

− ⎡ ⎤= + −⎢ ⎥⎣ ⎦

Taking ( )1/ 23 3i− = + , the above expression becomes

( ) ( )( )

12

2

1sin (2) ln 2 3 Ln 2 3 2

2 1.317, 0, 1, 2,

i i i i ki

k i k

π

π

π

π

− ⎡ ⎤⎡ ⎤= + = − + + +⎣ ⎦ ⎢ ⎥⎣ ⎦

= + − = ± ±

Taking ( )1/ 23 3i− = − , the above expression becomes

( ) ( )( )

12

2

1sin (2) ln 2 3 Ln 2 3 2

2 1.317, 0, 1, 2,

i i i i ki

k i k

π

π

π

π

− ⎡ ⎤⎡ ⎤= − = − − + +⎣ ⎦ ⎢ ⎥⎣ ⎦

= + + = ± ±

Example 3 Find all the numbers whose sine is i.

This question is to solve sin z i= . Thus, we have

( )1/ 21 21sin ( ) ln 2 ln(1 2)i i ii

− ⎡ ⎤= + = − ±⎢ ⎥⎣ ⎦

Taking ( )1/ 23 3i− = + , the above expression becomes

( ) ( )( ) ( ) ( )

1Ln 2 1 2 2 Ln 2 1

sin ( ) ln 1 2Ln 2 1 2 2 Ln 2 1

i i k k ii i

i i k k i

π π

π π π π−

⎧ ⎡ ⎤− − + = − −⎪ ⎣ ⎦⎡ ⎤= − − ± = ⎨⎣ ⎦ ⎡ ⎤− + + + = + − +⎪ ⎣ ⎦⎩

where 0, 1, 2,k = ± ± . H.W. 1 Show that ( )1 1sin ( ) ( 1) ln 1 2ni n iπ− += + − + . 【本題摘自：James Ward Brown and Ruel V. Churchill, Complex Variables and Applications, 6rd ed., Section 29, Example 1, McGraw-Hill, Inc., 2005.】 H.W. 2 Show that ( ) ( )1tanh (1/ 2) ln cot / 2ie iθ θ− = ⎡ ⎤⎣ ⎦ . 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Exercise 3-7, Problem 16,

66

Pearson Education, Inc., 2005.】

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