Page 1
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
Chapter 12 Complex Numbers and Functions
複數 (complex numbers) 與複變數 (complex variables)
複數 (complex numbers) : z = a + i b , 其中 a 與 b 均為實數 , . 1i
複變數 (complex variables) : z = x + i y , 其中 x 與 y 均為實變數 , . 1i
複數運算規則 :
相等 (equality) : z1 = z2 x1 = x2 , y1 = y2
加法 (addition) : z1 + z2 = (x1 , y2) + (x2 , y2) = (x1+ x2 , y1+y2)
相乘 (multiplication) :
z1 z2 = (x1 , y2) · (x2 , y2) = (x1 x2 - y1y2 , x1 y2 + x2y1)
相除 (division) :
22
22
211222
22
2121
22
21
2
1
yx
yxyxi
yx
yyxx
zz
zz
z
z
Page 2
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複數平面化
(x,y)
x
y
Oθ
r
x
y
複數平面
z = x + i y
x = r cosθ y = r sinθ z = r (cosθ + i sinθ)
z = r eiθ
Polar representation
r : the modulus or magnitude of z
θ : the argument or phase of z
挪威人 – Caspar Wessel
相乘 (multiplication) :
相除 (division) :
)]sin(i)[cos(rr
)]sini(cosr)][sini(cosr[zz
212121
22211121
)]sin(i)[cos(r
r
)sini(cosr
)sini(cosr
z
z2121
2
1
222
111
2
1
2121zzzz
2121zargzarg)zzarg(
2
1
2
1
z
z
z
z 212
1 zargzarg)z
zarg(
Euler’s Formula : zsinizcoseiz
Chapter 12 Complex Numbers and Functions
Page 3
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
De Moivre’s ( 隸美弗 ) Formula
nin )sini(cose n)sini(cosnsinincos
二項式定理展開
...sincos)4
n(
sincos)2
n(cosncos
44n
22nn
...sincos)5
n(
sincos)3
n(sincos)
1
n(nsin
55n
33n1n
)nsinin(cosrerz ninnn Q : 試證明 cos3cos43cos 3
3sini3cos)sini(cos 3 A :
...sincos)4
n(
sincos)2
n(cosncos
44n
22nn
cos3cos4
)cos1(cos3cos
sincos3cos3cos
3
23
23
Chapter 12 Complex Numbers and Functions
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Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
Q : 試解 1. 2. 3.
A :
1.
2.
3.
3 1 i1 0z)1z( 33
n2i3 e1z 3/n2iez 1z0n
3/2iez1n 3/4iez2n
)n24/(i2 e2i1z 2/)n24/(i4/1 e2z
8/i4/1 e2z0n
8/9i4/1 e2z1n
0z)1z( 33 0)1zz)(1z2( 2
2/1z 2/)3i1(z
Chapter 12 Complex Numbers and Functions
Page 5
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
x
y
z
θθ
(x,y)
(x,-y)
*z
共軛複數 (Complex Conjugation)
iyxz iyxz*
irez i* rez
複變函數 (complex functions)
Complex function w(z) = u(x,y) + iv(x,y) where u(x,y) and v(x,y) are pure real
w(z) = z2 = (x + iy)2 = (x2 - y2) + i 2xy
22 yx)y,x(u)z(w
xy2)y,x(v)z(w x
y
z - plane
1 2
u
v
w - plane
1 2mapping
For example :
Chapter 12 Complex Numbers and Functions
Page 6
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
Q : 某一楔形金屬板 , 其兩面之溫度固定為恆溫 ( 如圖所 示 ), 試求其中之溫度分佈 .
0oC
100oC
π/3
x
π/3
y
u
v
π/3
0oC
100oC
在 u-v 平面上的解為 : , 又v3/
100)v,u(T
)x/y(tanv 1
在 x-y 平面上的解為 : )x/y(tan3/
100)y,x(T 1
A :
Chapter 12 Complex Numbers and Functions
Page 7
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
)n2(irlnirlnrelnzlnw i
主值 (Principle value)
複變對數函數
z = r eiθ
θ : 主幅角 (the principle argument)
w : 多值函數
x
y
z - plane
u = lnr
v = θ + 2nπ
w - plane
mapping
n = 1
n = 0
n = -1
n = -2
Q : 試計算 之值 )i31ln(
A : 假設 3/2ie2)3
2sini
3
2(cos23i1z
)n23/2(i2ln)3i1ln(zlnw
3/2,2r
通解
3/2i2lnw 主值
Chapter 12 Complex Numbers and Functions
Page 8
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變冪函數
azw 其中 a 亦為複數zlnaa ez )n2(irlnzln 為多值函數 azw 亦為多值函數
)]n2(ir[lnaa ezw
Q : 試計算下列之值 i)i1(
)n2i4/i2(lni)i1ln(ii ee)i1(
)2lnsini2ln(cose
ee)n24/1(
n24/2lni
通解
主值)2lnsini2ln(cose)i1( 4/1i
A :
Chapter 12 Complex Numbers and Functions
Page 9
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變指數函數
zsinhzcoshz!3
1z
!2
1z1e 32z zsinizcoseiz
)zz(zz 2121 eee zi2z ee 複變指數函數具有虛週期 i2
複變三角函數
i2
eez
!5
1z
!3
1zzsin
iziz53
2
eez
!4
1z
!2
11zcos
iziz42
122121 zcoszsinzcoszsin)zzsin( 212121 zsinzsinzcoszcos)zzcos(
複變三角函數仍具有週期 , 但為無界函數2複變雙曲函數
2
eez
!5
1z
!3
1zzsinh
zz53
2
eez
!4
1z
!2
11zcosh
zz42
122121 zcoshzsinhzcoshzsinh)zzsinh( 212121 zsinhzsinhzcoshzcosh)zzcosh(
Chapter 12 Complex Numbers and Functions
Page 10
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
Q : 1. 試解 2. 求 之值
A : 1.
2.
5zsin )icos(
ysinhxcosiycoshxsiniysinxcosiycosxsin)iyxsin(zsin
0ysinhxcos 0y or 2
)1n2(x
1ycosh 5xsin
n)1(xsin ycosh 恆為正n 為偶數
2
)1n4(x
5ycosh
)5(coshi2
)1n4(iyxz 1
2
ee
2
ee2
eeee
2
ee)icos(
11
ii1)i(i)i(i
Chapter 12 Complex Numbers and Functions
Page 11
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
映射轉換 (mapping transformation)平移 (translation)
x
y
u
v
x0
y0
z-plane w-plane 0zzw
)yy(i)xx(ivu 00
0xxu 0yyv
旋轉 (rotation)
x
y
u
v
θ0
z-plane w-plane
i0 ezzw
irez 0i00 erz
)(i0
i 0erre
0rr 0
Chapter 12 Complex Numbers and Functions
Page 12
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
映射轉換 (mapping transformation)
放大 (enlargement)
x
y
u
vz-plane w-plane
innn erzw
ncosru n
)nsinr(i)ncosr(ivu nn
nsinrv n
反轉 (inversion)
x
yz-plane
v
u
w-plane
ii
er
1
re
1
z
1w
)sinr
1(i)cos
r
1(ivu
cosr
1u sin
r
1v
Chapter 12 Complex Numbers and Functions
Page 13
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
映射轉換 (mapping transformation)
x
yz-plane
y = c1 1 2 3 4
iyx
1ivu
反轉 (inversion)
z
1)z(f 22 yx
xu
22 yx
yv
22 vu
ux
22 vu
vy
line circle
122c
vu
vy
222 ryx 22
22
r
1vu
)v2
vu(
c2
1 22
1
v2
uv)
v2
vu(v
c2
1v
2222
1
2
1
2
1
2 )c2
1()
c2
1v(u
w-plane
u
v
)c2
1,0(
1
1
2
34
Chapter 12 Complex Numbers and Functions
Page 14
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
映射轉換 (mapping transformation)
2i22 erzw
two-to-one correspondence
非線性轉換 : 係數平方 , 幅角變兩倍
Upper half-plane of z, 0 θ < π whole plane of w, 0 φ < 2π
Lower half-plane of z, π θ < 2π whole plane of w, 0 φ < 2π Cover by two times
0z 0i
0 zez ,
20zw For example : two-to-one correspondence
w(z) = z2 = (x + iy)2 = (x2 - y2) + i 2xy22 yx)y,x(u xy2)y,x(v
122 cyx
2cxy2
x
yz-plane
u
v1cu
2cv
w-plane
Chapter 12 Complex Numbers and Functions
Page 15
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
映射轉換 (mapping transformation)
i2/i2/12/1 eerzw
one-to-two correspondence
, two-to-one correspondence
2
w-plane z-plane
How to make the function of w a singled-values function ?
2,
z-plane w-plane
, 在 z 平面上某一點映射到 w 平面時 , 可以有兩個值 .
one-to-one correspondence
x
yz-plane
cut line 20
限制 z 的幅角branch point singularities
z = 0 的點除外
Chapter 12 Complex Numbers and Functions
Page 16
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
映射轉換 (mapping transformation)
iiyxiyxz eeeeew y,ex
many-to-one correspondence
z-plane w-plane
)n2y(ix any points the same point
ivu)n2(irln)reln(zlnw )n2(i also
If 1r
x
y
The Riemann surface for ln z
cut line
x
y
Chapter 12 Complex Numbers and Functions
Page 17
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變函數的微分—直角座標
)z(fdz
)z(df
z
)z(flim
z)zz(
)z(f)zz(flim '
0z0z
yixz viuf yix
viu
z
f
x
y
z0δx0
δy = 0 δx = 0δy0
δx0
δy = 0
x
vi
x
u)
x
vi
x
u(lim
z
)z(flim
0x0z
δx = 0
δy0
First approach
Secondapproach y
v
y
ui)
y
v
y
ui(lim
z
)z(flim
0y0z
y
v
x
u
x
v
y
u
Cauchy-Riemann conditions : if exists, then ,
dz
)z(df
f (z) is analytic at z = z0
if does not exist at z = z0, then z0 is labeled a singular point .dz
)z(df
holomorphic
regular
Chapter 12 Complex Numbers and Functions
Page 18
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變函數的微分—極座標
),r(iv),r(u)z(f irez
r
vi
r
u
dz
dfe
r
z
dz
df
r
)z(f i
v
iu
dz
dfire
z
dz
df)z(f i
實部對實部 , 虛部對虛部
f
r
1
r
fi
v
r
1
r
u
u
r
1
r
v
Cauchy-Riemann conditions
)r
vi
r
u(e
dz
df i
極座標 )
x
vi
x
u(
dz
df
直角座標
Chapter 12 Complex Numbers and Functions
Page 19
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變函數的微分性質
複變函數的微分性質與實變函數相同
)z(fc)z(fc)]z(fc)z(fc[ '22
'11
'2211
)z(f)z(f)z(f)z(f)]z(f)z(f[ '212
'1
'21
1.
2.
)z(f
)z(f)z(f)z(f)z(f]
)z(f
)z(f[ 2
2
'212
'1'
2
1 3.
dz
)z(dg
)z(dg
)]z(g[df
dz
)]z(g[df4.
zz eedz
d
zsinzcosdz
d zcoszsindz
d
zcoshzsinhdz
d zsinhzcoshdz
d
z
1zln
dz
d
For any points
1aa azzdz
d
Except for branch points and cut lines
Chapter 12 Complex Numbers and Functions
Page 20
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變解析 (analytic) 函數
定義 : 若複變函數 f (z) 在 z0 處以及包圍 z0 點的小封閉曲線範圍內均可微分 , 則稱 f (z) 為複變解析函數 .
複變全 (entire) 函數
定義 : 若複變函數 f (z) 在整個複數平面均可微分 , 則稱 f (z) 為複變全(entire) 函數 .
奇異點 (singular point)
有理函數除分母為零的位置外 , 均為解析函數 .
對數函數與冪函數除了在分支點與分支切割外 , 均為解析函數 .
複變多項式函數以及 , , , , , 等均為全函數
ze zcos zsin zcosh zsinh
若複變函數 f (z) 在 z0 處不可微分 , 則稱 z0 處為奇異點 .
Chapter 12 Complex Numbers and Functions
Page 21
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
Q : 試問下列函數在何處解析 ?
z)z(f 2z)z(f 2z)z(f *z)z(f z/1)z(f
ireiyxz y
v
x
u
x
v
y
u
Cauchy-Riemann conditions :
A :
v
r
1
r
u
u
r
1
r
v
1. z)z(f
ivu)z(f
r),r(u 0),r(v
均不成立 任何地方皆不解析
v
r
1
r
u
u
r
1
r
v
1. 2
z)z(f 2r),r(u 0),r(v
僅原點成立 原點處解析
v
r
1
r
u
u
r
1
r
v
Chapter 12 Complex Numbers and Functions
Page 22
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
A :
4. x)y,x(u y)y,x(v
均不成立 任何地方皆不解析
v
r
1
r
u
u
r
1
r
v
5. r
cos),r(u
除原點不成立 除原點外均解析
3. 22 yx)y,x(u xy2)y,x(v
均成立 任何地方皆解析
2z)z(f
y
v
x
u
x
v
y
u
*z)z(f
y
v
x
u
x
v
y
u
r
sin),r(v
z/1)z(f
Chapter 12 Complex Numbers and Functions
Page 23
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
共軛座標系 (conjugate coordinates)
複數 z = x + iy 為以實數 x 與 y 為變數的函數
iyxz iyxz* 2
zzx
*
i2
zzy
*
使用 z 及 z* 為變數的座標系 (z , z*) 稱為共軛座標系
複變函數 f (z) = u + iv 在共軛座標系的表示方式時
)i2
zz,
2
zz(iv)
i2
zz,
2
zz(u)y,x(iv)y,x(u)z(f
****
此時 Cauchy-Riemann conditions 為
0z
f*
Chapter 12 Complex Numbers and Functions
Page 24
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
共軛座標系與直角座標系在偏微分的關係
)y
ix
(2
1
yi2
1
x2
1
yz
y
xz
x
z
iyxz iyxz* 2
zzx
*
i2
zzy
*
)y
ix
(2
1
yi2
1
x2
1
yz
y
xz
x
z ***
*zzx
)
zz(i
y *
Chapter 12 Complex Numbers and Functions
Page 25
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
Q : 試將拉式運算子 以共軛座標表示之
2
2
2
22
yx
A: yyxxyx 2
2
2
22
*zzx
)zz
(iy *
*
2
*
2
2
2
2
2
*
2
2
2
2
2
****2
2
2
22
zz4
)zz
2*zz
()zz
2*zz
(
)zz
(i)zz
(i)zz
)(zz
(yyxxyx
Chapter 12 Complex Numbers and Functions
Page 26
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
諧和函數 (harmonic functions)
對於任何實變函數 u (x , y), 若其滿足 Laplace’s equation :
則函數 u (x , y) 稱之為諧和函數 .
0y
u
x
uu
2
2
2
22
若複變函數 f (z) = u (x , y) + i v (x , y) 在某區域內為解析函數 , 則實變函數 u (x , y)
以及 v (x , y) 在此區域內必為諧和函數 , 但反之未必然 .
此證明利用下列定理 :
若複變函數 f (z) = u (x , y) + i v (x , y) 在某區域內為解析函數 , 則複變函數 f (z) 的
各階導數均存在且仍為解析函數 .
Chapter 12 Complex Numbers and Functions
Page 27
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
線積分 (Contour Integrals)
積分路徑
x
y
z2
z0
z1
z0’=zn
ζ1
ζ2
n
1j1jjjn )zz)((fS Consider the sum :
Let n with for all j 0zz 1jj
If the sum exists and is independent of the
details of choosing the points zj and j .
nn
Slim
then
'0
0
z
z
n
1j1jjj
nn
ndz)z(f)zz)((flimSlim
f (z) 沿著特定路徑 C ( 由 z = z0 到 z = z0’) 的線
積分
Chapter 12 Complex Numbers and Functions
Page 28
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
線積分 (Contour Integrals)
22
11
22
11
22
11
'0
0
y,x
y,x
y,x
y,x
y,x
y,x
z
z
]dy)y,x(udx)y,x(v[i]dy)y,x(vdx)y,x(u[
]idydx)][y,x(iv)y,x(u[dz)z(f線積分定義
將複變積分簡化為實變積分的複數和
舉例 : Cndzz 此處 C 是以 z = 0 為中心 , 半徑 r 的圓
我們以極座標來處理 irez diredz i
當 n -1 0e)1n(
rdeirdireerdzz
2
0
)1n(i1n
2
0
)1n(i1n2
0
iinn
C
n
在極座標下另一種做法
當 n = -1 i2didireerdzz
1 2
0
2
0
ii1
C 與 r 無關 !
Chapter 12 Complex Numbers and Functions
Page 29
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分定理 (Cauchy’s Integral Theorem)
當一複變函數 f (z) 在某封閉區域中為可解析 (analytic), 且其微分仍為連續的 ,
則對於在此封閉區域的任一封閉路徑 C, f (z) 的線積分為零 .
0dz)z(fdz)z(fCC
記得上一例子中 :
i2dzz
1dz
z
1CC
? 此乃因 f (z) = 1/z 在 z = 0 處為不解析
i2dzz
1dz
z
1dz
z
13z2z1z
0dzz
1dz
z
1dz
z
113z
2
11z12z
含原點
不含原點
Chapter 12 Complex Numbers and Functions
Page 30
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分定理 (Cauchy’s Integral Theorem) 證明
利用 Stokes’s theorem 與 Cauchy-Riemann condition 可證明 Cauchy’s Integral Theorem
CCC
)udyvdx(i)vdyudx(dz)z(f
yx VyVxV
Stokes’s theorem : dxdy)y
V
x
V()dyVdxV( xy
C yx
Let u = Vx and v = -Vy
0dxdy)y
u
x
v(dxdy)
y
V
x
V()dyVdxV()vdyudx( xy
C yxC
對於第一項
對於第二項 Let v = Vx and u = Vy
0dxdy)y
v
x
u(dxdy)
y
V
x
V()dyVdxV()udyvdx( xy
C yxC
Cauchy-Riemann condition
If f (z) is analytic
y
v
x
u
x
v
y
u
Chapter 12 Complex Numbers and Functions
Page 31
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分定理 (Cauchy’s Integral Theorem)
當一複變函數 f (z) 在某封閉區域中為可解析 (analytic), 且其微分仍為連續的 ,
則對於在此封閉區域的任一封閉路徑 C, f (z) 的線積分為零 . 0dz)z(fdz)z(fCC
C1
C2
0dz)z(fdz)z(f21 CC
C1
C2
C3
0dz)z(fdz)z(fdz)z(fCCC 31
0dz)z(fdz)z(fdz)z(fCCC 32
21 CC
dz)z(fdz)z(f 與路徑無關
Chapter 12 Complex Numbers and Functions
Page 32
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
全函數之積分 2
1
z
zdz)z(f
假設 )z(f)z(F'
)z(F)z(F)z(dFdz)z(Fdz)z(f 12
z
z
z
z
'z
z
2
1
2
1
2
1
77 交大控制計算 ?dz)1z(
C
2 }20:t)2
tsin
2
t(it{:C
1z)z(f 2 解 為全函數 只考慮端點 0 , 2 + i
3
i148)z
3
z(dz)1z(
i2
0
3
C
2
Chapter 12 Complex Numbers and Functions
Page 33
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分公式 (Cauchy’s Integral Formula)
當一複變函數 f (z) 在某封閉區域中為可解析 (analytic), 且其微分仍為連續的 ,
則對於在此封閉區域的任一封閉路徑 C, 則
)z(fdzzz
)z(f
i2
10C
0
其中 z0 位於封閉路徑 C 內
部因為 f (z) 可解析 , 在 z = z0 處不是解析的 , 除非 f (z0) = 0 0zz
)z(f
C1
z0
C2
contour line
0dzzz
)z(fdz
zz
)z(fdz
zz
)z(f21 C
0C
0C
0
22 C
ii
i0
C0
direre
)rez(fdz
zz
)z(f
As r 0 )z(if2d)z(ifdzzz
)z(f0C0C
022
)z(fdzzz
)z(f
i2
10C
0
Chapter 12 Complex Numbers and Functions
Page 34
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分公式 (Cauchy’s Integral Formula)
0
)z(f{dz
zz
)z(f
i2
1 0
C0
z0 interior
z0 exterior
f(z) 的微分可利用歌西積分公式來表示
)dzzz
)z(fdz
zzz
)z(f(
zi2
1
z
)z(f)zz(fC
0C
0000
000
C 20
C000
0
00z
0' dz
)zz(
)z(f
i2
1dz
)zz)(zzz(
)z(fz
zi2
1lim)z(f
0
同理
C 3
00
)2( dz)zz(
)z(f
i2
2)z(f
C 1n0
0)n( dz
)zz(
)z(f
i2
!n)z(f
Chapter 12 Complex Numbers and Functions
Page 35
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分公式 (Cauchy’s Integral Formula)
計算 之值 2z 2dz
1z
zsin
)iz
1
iz
1(
i2
1
1z
12
]dziz
zsindz
iz
zsin[
i2
1dz
1z
zsin2z2z2z 2
取 z0 = i , f (z) = sinz
0
)z(f{dz
zz
)z(f
i2
1 0
C0
z0 interior
z0 exterior
1sinh21sinhi
1i2isini2dz
iz
zsin2z
取 z0 = -i , f (z) = sinz
1sinh21sinhi
1i2)isin(i2dz
iz
zsin2z
1sinhi2]dziz
zsindz
iz
zsin[
i2
1dz
1z
zsin2z2z2z 2
Chapter 12 Complex Numbers and Functions
Page 36
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分公式 (Cauchy’s Integral Formula)
計算 之值 , 其中 C 為單位圓C 3
z
dzz
e
取 z0 = 0 , f (z) = ez
0
)z(f{dz
zz
)z(f
i2
1 0
C0
z0 interior
z0 exterior
C 30
0)2( dz
)zz(
)z(f
i2
2)z(f
iie)0(ifdz)0z(
edz
z
e 0)2(
C 3
z
C 3
z
Chapter 12 Complex Numbers and Functions
Page 37
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變序列 (Complex sequences)
A complex sequence : z1, z2, z3 , z4,…. zn,…
zN
zN+1
zN+3
Lε
A complex sequence z1, z2,… is said to converge to the number L if ,given ε > 0, there is some positive integer N such that whenever n N.
zN+2
Lzn
Lzlim nn
Theorem :
Let zn = xn +iyn. Then, zn A + iB if and only if xn A and yn B
Cauchy sequence
nlim
nlim
nlim
Chapter 12 Complex Numbers and Functions
Page 38
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變級數 (Complex series)
Given a complex sequence : z1, z2, z3 , z4,…. zn,…
The complex series : ...zzz 211n
n
The sum : m21
m
1nnm z...zzzS
Theorem :
Let zn = xn +iyn. Then, if and only if andiBAz1n
n
Ax
1nn
By
1nn
Theorem :
If converges, then
1nnz 0zlim n
n
Chapter 12 Complex Numbers and Functions
Page 39
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變級數收斂 (Complex series convergence)
絕對收斂 (absolute convergence) :
1nnz convergence and
條件收斂 (conditional convergence) :
1nnz divergence but
1nnz convergence
要判斷收斂 , 可以利用比例測試法 (ratio test) :
1nnz
1nnz convergence
取)z(z
)z(zlim)z(r
n
1n
n
1. 對滿足 的 z 而言 , 此級數為絕對收斂
2. 對滿足 的 z 而言 , 此級數為發散
3. 對滿足 的 z 而言 , 無法判定收斂性
1)z(r0
1)z(r
1)z(r
Chapter 12 Complex Numbers and Functions
Page 40
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變冪級數 (Complex power series)
複變冪級數 ...)zz(a....)zz(a)zz(aa)zz(a n0n
202010
0n
n0n
利用比例測試法 (ratio test) 判斷收斂 :
0
n
1n
n0n
0n
1n01n
n
zz
a
alimzz
)zz(a
)zz(alim)z(r
where
1n
n
n a
alim
對滿足 的 z 而言 , 此冪級數為絕對收斂1)z(r0
0zzz0
ρ
展開中心
收斂區域
1n
n
n a
alim
收斂半徑
Chapter 12 Complex Numbers and Functions
Page 41
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變冪級數 (Complex power series)
複變冪級數在其收斂範圍內 可以逐項微分及積分 , 且其收斂範圍不會因積分或微分而改變 .
0zz
...)zz(a....)zz(aa)zz(a)z(f n0n010
0n
n0n
0zz
...)zz(na....)zz(a2a)zz(na)z(f 1n0n021
0n
1n0n
'
0zz
常數
0n
1n0
n )zz(1n
adz)z(f 0zz
Chapter 12 Complex Numbers and Functions
Page 42
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變冪級數 (Complex power series)
定理 : 若 f(z) 為連續 , 且在區域 D 中存在 R+, 而有 , 則在區域 D 中1)z(f
...)z(f....)z(f)z(f1)z(f1
1 n2
證明 : 對於複變冪級數 ...)z(f....)z(f)z(f1 n2 )z(f)z(r
取 )z(f....)z(f)z(f1)z(S 1n2n
)z(f1
1)z(g
)z(f1
)z(f1)z(S
n
n
)z(f1
)z(f)z(S)z(g
n
n
因為 f(z) 在區域 D 中為收斂且小於 1 0)z(flim n
n
0)z(f1
)z(flim)z(S)z(glim
n
nnn
)z(f1
1)z(g)z(Slim n
n
Chapter 12 Complex Numbers and Functions
Page 43
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變冪級數 (Complex power series)
試求複變冪級數 之收斂區域及其和....)1z
1z(
2
1)
1z
1z(
2
1
1z
1z
2
1 33
22
11z
1z
2
1lim)z(rn
區域內為絕對收斂 1z21z
...)z(f....)z(f)z(f1)z(f1
1 n2 利用
)z(f1
)z(f1
)z(f1
1....)
1z
1z(
2
1)
1z
1z(
2
1
1z
1z
2
1 33
22
3z
1z
2z23z2z2
1z
)1z1z
(21
1
)1z1z
(21
Chapter 12 Complex Numbers and Functions
Page 44
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變泰勒級數 (Complex Taylor series)
複變函數在解析點的無限級數展開形式
考慮單連封閉曲線 C 為圓 , 函數 f(z) 在 C 上及其內部均為解析 ,z 為 C 內部的一點 .
0zz
z0
ρ
C
歌西積分公式 (Cauchy’s Integral Formula)
C0
0C
dszs
zs
zs
)s(f
i2
1ds
zs
)s(f
i2
1)z(f
C
0
00C
0
0
ds
zszz
1
1
zs
)s(f
i2
1ds
zszs
1
zs
)s(f
i2
1
z
z 為 C 內部的一點 , 而 s 在圓上 1zs
zz
0
0
Chapter 12 Complex Numbers and Functions
Page 45
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變泰勒級數 (Complex Taylor series)
...)z(f....)z(f)z(f1)z(f1
1 n2
...)zs
zz(....)
zs
zz(
zs
zz1
zszz
1
1 n
0
02
0
0
0
0
0
0
C
n
0
02
0
0
0
0
0C
0
00
ds...])zs
zz(...)
zs
zz(
zs
zz1[
zs
)s(f
i2
1ds
zszz
1
1
zs
)s(f
i2
1)z(f
...)z(f!n
)zz(....)z(f
!2
)zz()z(f)zz()z(f 0
)n(n
00
''2
00
'00
f(z) 在 z0 點處的複變泰勒級數 (Complex Taylor series) 冪級數
Chapter 12 Complex Numbers and Functions
Page 46
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變泰勒級數 (Complex Taylor series)
ze)z(f
...)z(f!n
)zz(....)z(f
!2
)zz()z(f)zz()z(f)z(f 0
)n(n
00
''2
00
'00
Let and 0z0 z)n( e)z(f
0n
n
0n0
)n(n
0z
!n
z)z(f
!n
)zz(e For all z 得到驗證
試求 在 0 之泰勒展開級數z1
1)z(f
展開中心在 z = 0, f(z) 的奇異點在 z = -1, 因此收斂半徑 ρ = 1
0n
nn
0n
n )z()1()z()z(1
1
z1
1if 10z
Chapter 12 Complex Numbers and Functions
Page 47
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變泰勒級數 (Complex Taylor series)
試求 在 展開之泰勒級數 , 並求收斂半徑 (74 台大化工 )2z
1)z(f
1z
展開中心在 z = 1, f(z) 的奇異點在 z = 2, 因此收斂半徑 ρ = 1
...])1z(...)1z()1z(1[1)1z(1
1
2z
1)z(f n2
...)z(g....)z(g)z(g1)z(g1
1 n2
試求 在 之泰勒展開級數1z
1)z(f
i2z
展開中心在 z = -2i, f(z) 的奇異點在 z = -1, 因此收斂半徑 ρ =5
0n
nn )i21
i2z()1(
)i21(
1]
)i21i2z
(1
1[
)i21(
1
)i2z()i21(
1
1z
1)z(f
0n
n1n
n )i2z()i21(
1)1(
Chapter 12 Complex Numbers and Functions
Page 48
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變勞倫級數 (Complex Laurent series)
C2
C1
r2
z0 r1
z
f(z) 在 區域中為解析201 rzzr
C
][i2
1ds
zs
)s(f
i2
1)z(f
downCupCC
12
取 0
]dszs
)s(fds
zs
)s(f[
i2
1)z(f
12 CC
222 C
0C
0
00C
ds1
1
zs
)s(f
i2
1ds
zszz
1
1
zs
)s(f
i2
1ds
zs
)s(f
i2
1
1zs
zz
0
0
0n
n0n )zz(a
2C 1n
0n ds
)zs(
)s(f
i2
1a
f(z) 在 C2 中並非都解析
Chapter 12 Complex Numbers and Functions
Page 49
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變勞倫級數 (Complex Laurent series)
1111 C
0C
0
00CC
ds1
1
zz
)s(f
i2
1ds
zzzs
1
1
zz
)s(f
i2
1ds
sz
)s(f
i2
1ds
zs
)s(f
i2
1
1zz
zs
0
0
0hC 1h
0
h0
0hC
h
0
0
011 )zz(
1]ds)zs)(s(f
i2
1[ds)
zz
zs(
zz
)s(f
i2
1
1mm
0
m
)zz(
b
1C
1m0m ds)zs)(s(f
i2
1b
1m
m0m
0n
n0nCC
)zz(b)zz(a]dszs
)s(fds
zs
)s(f[
i2
1)z(f
12
2C 1n0
n ds)zs(
)s(f
i2
1a
1C
1m0m ds)zs)(s(f
i2
1b
Chapter 12 Complex Numbers and Functions
Page 50
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變勞倫級數 (Complex Laurent series)
1m
m0m
0n
n0nCC
)zz(b)zz(a]dszs
)s(fds
zs
)s(f[
i2
1)z(f
12
2C 1n0
n ds)zs(
)s(f
i2
1a
1C
1m0m ds)zs)(s(f
i2
1b
此時 z0 是不解析的
m = -n
1C
1m0m ds)zs)(s(f
i2
1b
1C 1n
0n ds
)zs(
)s(f
i2
1b
n
n0n
1n
n0n
0n
n0n )zz(a)zz(b)zz(a)z(f
C 1n0
n ds)zs(
)s(f
i2
1a
201 rzzr C : 區域
Chapter 12 Complex Numbers and Functions
Page 51
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變勞倫級數 (Complex Laurent series)
泰勒級數與勞倫級數主要的差別在於 : 泰勒級數之有效範圍必是一個圓的內部 , 而勞倫級數的有效範圍必定是一個環狀區域 .
複變泰勒級數 (Complex Taylor series)
C
n
0
02
0
0
0
0
0C
0
00
ds...])zs
zz(...)
zs
zz(
zs
zz1[
zs
)s(f
i2
1ds
zszz
1
1
zs
)s(f
i2
1)z(f
...)z(f!n
)zz(....)z(f
!2
)zz()z(f)zz()z(f 0
)n(n
00
''2
00
'00 n
00n
C 1n0
)zz(]ds)zs(
)s(f
i2
1[
n
n0n
1n
n0n
0n
n0n )zz(a)zz(b)zz(a)z(f
C 1n0
n ds)zs(
)s(f
i2
1a
201 rzzr C : 區域
泰勒級數必定是以解析點為展開中心 , 而勞倫級數的展開中心可以是解析點 ,也可以是不解析點 .
Chapter 12 Complex Numbers and Functions
Page 52
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變勞倫級數 (Complex Laurent series)
試求函數 在原點展開之所有的勞倫級數 (89 清大動機 )
54 zz
1)z(f
對 f(z) 而言 , 不解析點 : z = 0, z = 1
在 1z0
...)zz1(z
1
z1
1
z
1
zz
1)z(f 2
4454
...
z
1
z
1
z
1234
在 1z
...)z
1
z
11(
z
1
z/11
1
z
1
z1
1
z
1
zz
1)z(f
255454
....z
1
z
1
z
1765
Chapter 12 Complex Numbers and Functions
Page 53
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變勞倫級數 (Complex Laurent series)
試求函數 在原點展開之所有的勞倫級數 (89 清大動機 )
54 zz
1)z(f
n
n0n )zz(a)z(f
C 1n
0n ds
)zs(
)s(f
i2
1a
1z0 C : 區域
C 6n5nC 1nC 1n0
n ds])s()s(
1[
i2
1ds
)s(
)s(f
i2
1ds
)zs(
)s(f
i2
1a
0mC m5n
0mC 5n
m
C 5nds
s
1
i2
1ds
s
s
i2
1ds]
s1
1[
s
1
i2
1
4n0
4n1{i2
i2
1
0m1,m5n
0m1,m5n
...zzz)zz(a)z(f 234
n
n0n
Chapter 12 Complex Numbers and Functions
Page 54
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
複變勞倫級數 (Complex Laurent series)
試求函數 在 z = -1 展開之勞倫級數 )i3z)(1z(
1)z(f
)i3z
1)(
10
i31()
1z
1)(
10
i31(
)i3z)(1z(
1)z(f
i3z
1
在 -1 處是可解析的 ,故可用泰勒展開式 i3z
1)z(f
0n1n
n
)i31(
)1z()z(f
2'
)i3z(
1)z(f
3''
)i3z(
2)z(f
i31
1
i31
1)1(f
2'
)i31(
1)1(f
33''
)i31(
2
)i31(
2)1(f
for 101z
1z
1
已是在 -1 處的勞倫展開式 for 01z
)10
i31()
1z
1)(
10
i31(
)i3z)(1z(
1)z(f
0n
1n
n
)i31(
)1z(
for 101z0
Chapter 12 Complex Numbers and Functions
Page 55
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
奇異點 (Singularities)
當 f(z) 在 區域中為解析函數 , 但在 z0 處不為解析函數時 , 我們稱 f(z) 有一孤立奇異點 (isolated singularity) z0..
我們將 f(z) 對 z0 做勞倫級數展開
rzz0 0
1n
n0n
0n
n0n )zz(b)zz(a)z(f rzz0 0
1. 當展開式中沒有 (z-z0) 的負冪次項 , 則稱 z0 為可移除奇異點 (removable singularity).
2. 當展開式中有無窮多個 (z-z0) 的負冪次項 , 則稱 z0 為本質奇異點 (essential singularity).
3. 當展開式中有 (z-z0) 的負冪次項一個以上 , 則稱 z0 為極點 (pole).
4. 當展開式中 (z-z0) 的負冪次項部分稱為主要部份 (principal part).
5. 當展開式中 (z-z0) 的負冪次項部分只到第 k 項 , 則稱 z0 為 k 階極點 (kth order pole).
6. 當展開式中 沒有主要部分 , 且 則稱 z0 為 k 階零點 (kth order zero).0a..aa 1k10
7. b1 稱為 f(z) 在 z0 的殘 ( 留 ) 數(residue).
Chapter 12 Complex Numbers and Functions
Page 56
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
8. 當 z0 為 f(z) 的一個 k 階零點 , 則 z0 為 1/ f(z) 的一個 k 階極點 ,反之亦然 .
當 z0 為 f(z) 的 k 階極點 , 則 f(z) 在 z0 的殘數記為 R(z0):
)]z(f)zz[(dz
dlim
)!1k(
1)z(R k
01k
1k
zz0
0
9. 當展開式中 (z-z0) 的負冪次項只一個 , 則稱 z0 為簡單極點 (simple pole).
若 f(z) 為有理函數 : , 而 z0 是一個簡單極點 :
)z(g
)z(h)z(f 0)z(h,0)z(g 00
)z(g
)z(h]
)zz/()z(g)z(g
)z(h[lim
])zz/()z(g
)z(h[lim]
)z(g
)z(h)zz[(lim)z(R
0'
0
00zz
0zz
0zz
0
0
00
Chapter 12 Complex Numbers and Functions
Page 57
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
找出 以及 在 z = 0 之極點階數與殘數)1e(z
1)z(f
z
z
1sinz)z(g
)z(hz
1
...!3
z!2
z1
1
z
1
...!3
z!2
zz
1
z
1
)1e(z
1)z(f
22232z
h(z) 在 z = 0 處解析 , 且 h(0) 0, 故 f(z) 在 z = 0 處為一個二階極點
2
1...)
!3
z
!2
z1(
dz
dlim)
...!3
z!2
z1
1(
dz
dlim]
1e
z[
dz
dlim)]z(f)z[(
dz
dlim)0(R
2
0z20zz0z
2
0z
...z
1
!5
1
z
1
!3
11...)
z
1
!5
1
z
1
!3
1
z
1(z
z
1sinz)z(g
4253 為本質奇異點
0b)0(R 1
Chapter 12 Complex Numbers and Functions
Page 58
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
z0
r
C當 f(z) 在 區域中為解析函數 , 但在 z0 處不為解析函數時 , 我們稱 f(z) 有一孤立奇異點(isolated singularity) z0.我們將 f(z) 對 z0 做勞倫級數展開
rzz0 0
1n
n0n
0n
n0n )zz(b)zz(a)z(f rzz0 0
...)zz(a)zz(aa)zz(
b
)zz(
b
)zz(
b... 2
020100
12
0
23
0
3
由歌西積分公式 : )z(gdzzz
)z(g
i2
10C
0
1C
bdz)z(fi2
1
)z(iR2ib2dz)z(f 01C
殘數定理
f(z) 在 區域中為解析函數 , 但除了 z0, z1,…, zn 點為不為解析 , 當 C 為包含上述奇異點的單連封閉曲線時 , 則
rzz0 0
])z(R[i2dz)z(fn
0jjC
Chapter 12 Complex Numbers and Functions
Page 59
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
計算 之值 ,C 為包含 z = 1 之單連封閉曲線 . (89 清華動機 )
C 3
2
dz)1z(
2z3z5
3
2
)1z(
2z3z5)z(f
z = 1 處為三階極點
3
2
3
2
)1z(
4)1z(7)1z(5
)1z(
2z3z5)z(f
5)1(R
)z(iR2ib2dz)z(f 01C
i10)1(iR2dz)1z(
2z3z5C 3
2
Chapter 12 Complex Numbers and Functions
Page 60
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
計算 之值 , 曲線 C 為以原點為中心且半徑為 3 的圓
C 22dz
)4z(z
zsin
z = 0 處為簡單極點 , z = 2i, -2i 處為簡單極點)4z(z
zsin)z(f
22
)z(g
)z(h)z(R
0'
00
)z(g
)z(h
)4z(z
zsin)z(f
22
16
)i2sin(i
i16
)i2sin(
)i2(8)i2(4
)i2sin()i2(R
3
16
)i2sin(i
i16
)i2sin(
)i2(8)i2(4
)i2sin()i2(R
3
)z(g
)z(h
)4z(z
...z!3
11
)4z(z
zsin)z(f
2
2
22
4
1
)0(g
)0(h)0(R
'
z = 2i, -2i 處為簡單極點
z = 0 處為簡單極點
)]i2sin(i2[4
)]i2(R)i2(R)0(R[i2dz)4z(z
zsinC 22
Chapter 12 Complex Numbers and Functions
Page 61
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
2
0
d]cos,[sinfI
殘數定理之應用 -- 三角函數的積分
Where f is both finite and single-valued for all values of
Let iez diedz i
z
dzid
i2
zzsin
1
2
zzcos
1
z
dz)
2
zz,
i2
zz(fid]cos,[sinfI
112
0
The path of integration is the unit circle
By the residue theorem
circleunitthewithinz
)z(fofresiduesi2)i(I
circleunitthewithinz
)z(fofresidues2
Chapter 12 Complex Numbers and Functions
Page 62
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
計算 1cos1
dI
2
0
z
dz)
2
zz,
i2
zz(fid]cos,[sinfI
112
0
2
zzcos
1
1z)/2(z
dz2i
z
dz
2)zz(
1
1i
21
分母有兩個根 2111
z
21
11z
單位圓之外 單位圓之內
2zz
'12/2z2
1
)z(g
)z(h)z(R
22 1
2
12
4)z(iR2
2iI
很重要 !
Chapter 12 Complex Numbers and Functions
Page 63
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
計算 之值 , 其中 a > 0 (91 交大物理 )
2/
02sina
dI
0
2/
0
2/
02 cos1a2
d
2cos1a2
d2
sina
dI
1z2
1z1
2
0
dz1z)1a2(2z
1i
iz
dz
2/)zz(1a2
1
2
1
cos1a2
d
2
1
2
zzcos
1
z
dzid
分母有兩個根 aa2)1a2(z 2 aa2)1a2(z 2
單位圓之外單位圓之內
aa4
1
2a4z2
1
)z(g
)z(h)z(R
2zz
'
aa2aa4
12)z(iR2iI
22
Chapter 12 Complex Numbers and Functions
Page 64
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
RCCR
]dz)z(q
)z(pdz
)z(q
)z(p[limdx
)x(q
)x(pdx)x(fI
殘數定理之應用 -- 有理函數的積分
CR
RC
Rdz
)z(q
)z(plim]
)z(q
)z(p[i2 在上半平面之殘數和函數
在 CR
上iRez
0
zC
Rd
)z(q
)z(zpilimdz
)z(q
)z(plim
R
izddReidz i
取 01m
m aza....za)z(p 01n
n bzb....zb)z(q
n0
n11n
m0
m11m
1mnz01
nn
02
11m
m
zz zbzb...b
zaza...a
z
1lim
bzb...zb
zaza...zalim
)z(q
)z(zplim
當 n-m-1 > 0 時 , 上式之極限值將趨近於零
當 q(z) 之次方數 p(z) 之次方數 +2 時 0d)z(q
)z(zpilimdz
)z(q
)z(plim
0z
CR
R
])z(q
)z(p[i2dx)x(fI 在上半平面之殘數和函數
Chapter 12 Complex Numbers and Functions
Page 65
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
2x1
dxI計算 之值
dx)x(q
)x(pdx
x1
1
x1
dxI
22 q(z) 之次方數 p(z) 之次方數 +2
])z(q
)z(p[i2dx)x(fI 在上半平面之殘數和函數
iz
1
iz
1
1z
12
分母有兩個根 iz iz
上半面下半面]
)z(q
)z(p[i2dx)x(fI 在上半平面之殘數和函數
i2
1i2)
z2
1(i2
izz
q(z) 沒有實根 , 以避免函數 p(z)/q(z) 在實軸上出現極點
Chapter 12 Complex Numbers and Functions
Page 66
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分主值 (The Cauchy Principal Value : PV)
當線積分路徑上出現極點時
CR
CxC
x
semicircleiniteinfdx)x(fdz)z(fdx)x(fdz)z(f00x
0
Infinite radius semicircle
residuesenclosedi2
當走下面箭頭之路徑時 , 小半球 Cx0 包含x0 點
i0 exz deidz i
)x(Riidie
dei
xz
dz0
2
i
i
0
)x(Riidie
dei
xz
dz0
0
i
i
0
counterclockwise
clockwise
0xC
dz)z(f )x(iR o residuesenclosedi2 )x(iR2 onet
)x(iR o
Chapter 12 Complex Numbers and Functions
Page 67
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西積分主值 (The Cauchy Principal Value : PV)
CR
CxC
x
semicircleiniteinfdx)x(fdz)z(fdx)x(fdz)z(f00x
0
Infinite radius semicircle residuesenclosedi2
當走上面箭頭之路徑時 , 小半球 Cx0排除x0 點
0xC
dz)z(f )x(iR o residuesenclosedi2
net
)x(iR o0
dx)x(fPV}dx)x(fdx)x(f{lim
0
0
x
x
0
歌西積分主值
])z(f[i])z(f[i2 在實數軸上之殘數和在上半平面之殘數和
Chapter 12 Complex Numbers and Functions
Page 68
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
dx)axcos()x(f and
dx)axsin()x(f常出現在傅氏轉換(Fourier Transform)
Consider
dxe)x(fdx)axsin()x(fidx)axcos()x(fI iax
CR
RC
iaz
C
iaz
R
iax ]dze)z(q
)z(pdze
)z(q
)z(p[limdxe
)x(q
)x(p
RC
iaz
R
iaz dze)z(q
)z(plim]e
)z(q
)z(p[i2 在上半平面之殘數和函數
在 CR
上iRez izddReidz i
0
cosiaRsinaR
zC
iaz
Rdee]
)z(q
)z(zp[ilimdze
)z(q
)z(plim
R
收斂至零與否取決於 : 在 0 到之間 sin恆為正實數sinaRe a > 0即可
)z(q
)z(zp: q(z) 之次方數 p(z) 之次方數 +1即可Jordan’s Lemma
Chapter 12 Complex Numbers and Functions
Page 69
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
0
dxx
xsinI計算 之值
0
dxx
xsinI sinx 與 x 均為奇函數
dxx
xsin
2
1dx
x
xsinI
0
線積分路徑上出現極點 x = 0 利用歌西積分主值公式
dx)x(gPV ])z(g[i])z(g[i2 在實數軸上之殘數和在上半平面之殘數和
dxx
eix
0
z
e)z(g
iz
1
idxx
eI
ix'
dx)x(gdxx
edxe)x(fI
ixix' ]IIm[
2
1I '
2]IIm[
2
1I '
Chapter 12 Complex Numbers and Functions
Page 70
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
歌西不等式 (Cauchy’s Inequality)
當 f(z) 在 區域中為解析函數時 , 若 M 為 在 上之極大值 , 則
恆有 :
rzz 0 )z(f rzz 0
n0)n(
r
M!n)z(f
rzz:C 0 取
C1n
00
)n( dz)zz(
)z(f
i2
!n)z(f
2
0
i)1n(i1n
i0
C1n
00
)n( direer
)rez(f
i2
!ndz
)zz(
)z(f
i2
!n)z(f
2
0n
2
0
i0n
2
0
ini0n
Mdr2
!nd)rez(f
r2
!nde)rez(f
r2
!n
n
2
0n r
M!nd
r2
M!n
圓
證明
Chapter 12 Complex Numbers and Functions
Page 71
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
理奧維爾定理 (Liouville Theorem)
任何有界 (bounded) 的全函數 , 必然是一個常數函數 .
當 f(z) 為全函數時 ,f(z) 在 區域當然為解析函數 , 假設 m 為 在圓上的極大值 , 則
rzz 0 )z(f
r
m)z(f 0
'
若 f(z) 為有界 , 則必定存在一正實數 M 為 在整個複數平面上的絕對極大值 :
)z(f
r
M
r
m)z(f 0
'
此不等式與 r 的大小無關 , 因此當 r 趨近於無窮大時
0r
Mlimr
0)z(flim 0'
r
0)z(f 0
' f(z) 是一個常數函數
證明
Chapter 12 Complex Numbers and Functions
Page 72
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
最大模數定理 (Maximum Modulus Theorem)
若 f(z) 在單連封閉曲線 C 上及內部為解析函數 , 則 在 C 上及內部的極大值 , 必發生在解析區域的邊界 C 上 , 而不在 C 內部 .
)z(f
dZ0
C 單連封閉曲線 C 的長度為 L, z0距離 C 之最短距離為 d假設 M 為 在曲線 C 上的極大值 .)z(f
C1n
00
)n( dz)zz(
)z(f
i2
!n)z(f
Ld
M
2
1dz
)zz(
)z(f
2
1dz
)zz(
)z(f
i2
1)z(f
n
C 0
n
C 0
nn
0
C 00 dz
)zz(
)z(f
i2
1)z(f
C 0
n
0n dz
)zz(
)z(f
i2
1)z(f
n/10 )
d2
L(M)z(f
當 時n M)z(f 0
z0 為 C 內之任一點 , 因此得證 !
Chapter 12 Complex Numbers and Functions
Page 73
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
最小模數定理 (Minimum Modulus Theorem)
若 f(z) 在單連封閉曲線 C 上及內部為解析函數 , 且 , 則 在 C 上及內部的極小值 , 必發生在解析區域的邊界 C 上 , 而不在 C 內部 .
)z(f0)z(f
試求 在 上之極大與極小模數
zcos)z(f 1z
2
eez
!4
1z
!2
11zcos
iziz42
在 為解析且恆不為零1z Sol.
的極大值與極小值均發生在 上1z sinicosz 222
)sinh(sin)sin(cosi)cosh(sin)cos(cos)sinicos(cos)z(f
)z(f
)(coscos)(sinsinh)(sinsinh)(cossin)(sincosh)(coscos 222222 2
)z(f 的極值發生在 0d
)z(fd2
0sin)sin(cos)cos(cos2cos)cosh(sin)sinh(sin2
2
n
若 n 為偶數 m 1cos)z(f
若 n 為奇數2
n 1cosh)icos()z(f
1cosh)z(f1cos
Example
Chapter 12 Complex Numbers and Functions
Page 74
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
幅角原理 (The Argument Principle)
在區域 D 中 , 除了幾個極點外 ,f(z) 是可解析的 . 假設單連封閉曲線 C 在區域 D 中沒有通過任何極點或零點時 , 則
C
'
dz)z(f
)z(f
i2
1
曲線 C 中 f(z) 之零點數目 - 曲線 C 中 f(z) 之極點數目
CC
dzzsin
zcoszdzcot zsin)z(f
2 3--2-3 0
在曲線 C 中 ,sinz 包含五個零點 , 且沒有極點
i
-ii10)05(i2dzzsin
zcoszdzcot
CC
Chapter 12 Complex Numbers and Functions