國 立 交 通 大 學
應用數學系
碩 士 論 文
N相黎曼空間的理論與應用
Theory and Applications of Riemann Surfaces of
genus N
研 究 生：凃偉隆
指導教授：李榮耀 教授
中 華 民 國 九 十 九 年 九 月
N相黎曼空間的理論與應用
Theory and Applications of Riemann Surfaces
of genus N
研 究 生：凃偉隆 Student：Wei-Long Tu
指導教授：李榮耀 Advisor：Jong-Eao Lee
國 立 交 通 大 學
應 用 數 學 系
碩 士 論 文
A ThesisSubmitted to Department of Applied Mathematics
College of Science
National Chiao Tung University
in Partial Fulfillment of the Requirements
for the Degree of
Master
In
Applied Mathematics September 2010
誌 謝
論文的完成首先要感謝辛勞的指導教授－李榮耀老師，在交大的這些日子裡 曾經迷失了來交大念研究所的初衷，經過李榮耀老師的提點，漸漸的埋首於論文 的研究，重新找回了做研究的熱忱。更感謝李榮耀老師一次又一次的與我一起修 正論文的內容，使得這篇論文能夠趨於完備。 在口試期間，要感謝余啟哲教授與李志豪教授費心的審閱我的論文，同時也 給學生在論文上與生涯規劃上寶貴的意見，學生銘記在心。 同時感謝在求學的路途上對我人生有著極大影響的梁忠三老師，楊連祥老師 與黃豐國老師，承蒙你們一路上的教導與提攜，使學生在求學的路途上能夠更順 遂。也感謝我的好友們暐翎、喻琳、美玲、冠良、文煥、永霖、富源、承鴻... 等等好友，在我投入研究論文的同時給我的鼓勵與陪伴，讓我又更有動力再重新 投入論文的完成。 最後要感謝我的父母，全力的支持，讓我追求自己理想和完成夢想的藍圖， 在我遭遇挫折和迷惘的同時，一直給予我無比的力量陪我完成學業。感謝交大所 有老師辛勤的教學，一路走來學生獲益良多。N相黎曼空間的理論與應用
研究生：凃偉隆 指導老師：李榮耀 教授
國 立 交 通 大 學
應 用 數 學 系
摘要
我們利用代數與幾何分析的方法建構多值函數(開方函數)的黎曼空間使得 一個定義在複數平面上是多值的函數在黎曼空間上是唯一值且可解析的函數。在 黎曼空間上對封閉曲線 a，b cycles 的積分可以解決許多微分方程上的問題， 而且可以找到 a，b cycles 之等價路徑，再經由 Cauchy Integral Theorem 可得知 a，b， cycles 之積分值與它們的等價路徑積分值會相等。藉由這樣的方 法，當我們執行黎曼空間的積分時，無論是數值上或是理論上，我們都可以解決 問題進而求得解答。Theory and Applications of Riemann Surfaces
of genus N
Student：Wei-Long Tu Advisor：Jong-Eao Lee
Department of Applied Mathematics
National Chiao Tung University
Abstract
We use algebraic and geometric analysis to develop two-sheet Riemann surface R of genus N such that muti-valued function on the complex plane
C become single-valued and analytic on R. The integrals over a,b cycles
on R can solve many problems in Differential Equations. By Cauchy Integral Theorem, we can find equivalent paths of a,b cycles such that their integrals are equal. When we do the integral on the Riemann surface ,no matter what on theoretically or in value , by the principle ,we could solves the problem and get the solution.
Contents
Abstract (in Chinese) Ⅰ
Abstract (in English) Ⅱ
Contents
Ⅲ
1 Introduction the Riemann Surface 1
1.1 Construct the corresponding Riemann Surface ……….. 1
1.2 The a,b cycles and its equivalent paths ……… 11
1.3 The homotopic………...………14
1.4 Conclusion of Riemann Surface ………..… 15
2. The integrals of
1 ( ) f zover a,b cycles for horizontal cut 16
3. The integrals of
1 ( ) f zover a,b cycles for vertical cut 44
4. The integrals of
1 ( ) f zover a,b cycles for slant cut 81
5. Applications of Differential Equations 125
I.1 Construct the corresponding Riemann Surface
2 1 2 , , - , , , - , , ( ) ( ) k k i i n i n k k Where z C and z z z e z e n z z e n
First, we take a example f z( ) z
,f C: C
1 2 2 1 2 1 2 2 ( 2 ) 1 2 2 ( ) - i i n i e if n even e if n odd z f z z z e z
Where
i i( 2n ) z z e z e n Z
Because f f is a two-valued function.
We will take f z( ) becomes a single valued function, so that we
modify its domain C to develop the corresponding Riemann Surface such that
f becomes a single-valued on Riemann Surface .
( 2 ) i i zre re r e12 i2 1 2 i2 r e
Now define that
1 2 2 1 2 2 -3( )
* ** i i e ez
f z
z
Where (*) called sheet-I and (**) called sheet-II
The each sheet has two edges, the starting edge with (-) and terminal edge with (+).Next we will use the stereographic sphere help us to consider the two sheets work process.
Sheet-Ⅰ ＋ － － ＋ 0 Sheet-Ⅱ － 3 ＋ － ＋ 0
Example 1: Construct the Riemann Surface of 7 7 1 1 ( ) ( - k) ( - k) k k f z z z z z
where k=1.2…...7Figure 5: The cut from zk
We have crossing one cut need to change the sign by ＂-1＂.
So that when crosse even cuts we will not change the sing and crosse odd cuts will change sign.
Figure 6：The cut plane
＋ ＋ ＋ ＋ 7 Z Z6 Z5 Z4 Z3 Z2 Z1
－
－
－
－
7 Z Z6 Z5 Z4 Z3 Z2 Z1 7 1 6 1 5 1 4 1 3 1 2 1 1
There are branch cuts in
z z1, 2
, z z3, 4
, z z5, 6
, z7,
and then using same idea to construct the corresponding Riemann Surface.Figure 7: Placing the cuts open
z 7 z 6 z 5 z 4 z 3 z 2 z 1
z 7 z 6 z 5 z 4 z 3 z 2 z 1
Figure 8：Take the two sheet together with (+) edge to (－) edge
z
7z
6z
5z
4z
3z
2z
1Example 2：Construct the Riemann Surface of 8 8 1 1 ( ) ( - k) ( - k) k k f z z z z z
where k=1.2…...8Figure 10 : The cut from zk
There are branch cuts in
z z1, 2
, z z3, 4
, z z5, 6
, z z . 7, 8
Figure 11 : The cut plane
Figure 11 : Placing the cuts open z8 z 7 z6 z5 z4 z3 z2 z1 ＋ ＋ ＋ ＋ 8 Z Z7 Z6 Z5 Z4 Z3 Z2 Z1
－
－
－
－
8 Z Z7 Z6 Z5 Z4 Z3 Z2 Z1 8 1 17 16 15 14 13 12 18
z z 7 z 6 z 5 z 4 z 3 z 2 z 1 8
z z 7 z 6 z 5 z 4 z 3 z 2 z 1
Figure 12：Take the two sheet together with (+) edge to (－) edge
z 8 z 7 z 6 z 5 z 4 z 3 z 2 z 1
In general 1 1 ( ) ( - ) ( - ) n n k k k k f z z z z z
, where k=1.2…...nFigure 10 : cut plane zk where n = odd or even
Figure 14: cut plane zk
Case 1. if the nodd
There are branch cuts in
z z1, 2
, z z3, 4
,,
z2N3,z2N2
, z2N1,
Figure 15: The cut plane of n=2N-1
2N 3 z ……
z2N1z2N2 z 4 z 3 z 2 z 1
Figure 16: Placing cuts open in both sheets ＋ ＋ ＋ ＋ 2N 1 Z Z2N2 Z2N3 …… Z4 Z3 Z2 Z1
－
－
－
－
n Z Zn1 Zn2 Z3 Z2 Z1 1n 1n1 1n2 13 12 1 z2N1 ……z 4 z 3 z 2 z 1
z2N1 ……z 4 z 3 z 2 z 1
Figure 17: Together two sheets
Figure 18: n=2N-1 and N-1 holes
2N1
Case 2. if the n even
There are branch cuts in
z z1, 2
, z z3, 4
,,
z2N3,z2N2
, z2N1,z2N
Figure 19: The cut of n=2N
z2N3 2 N
z z2N1 z2N2 z 4 z 3 z 2 z 1
Figure 20: Placing cuts open in both sheets
2 N
z z2N1 ……z 4 z 3 z 2 z 1
2 N
z z2N1 ……z 4 z 3 z 2 z 1
Figure 21: Together two sheets
＋ ＋ ＋ ＋ 2 N
Z Z2N1 Z2N3 Z2N4 …… Z4 Z3 Z2 Z1
z2 N z2N1 ……. z 2 z 1
Figure 22: n=2N and n-1 holes
So the conclusion , no matter the n=2N or n=2N-1 ,there are N cuts and N-1 holes called Riemann surface of genus N-1.
b a
1.2 The a,b cycles
Example f z( ) z
z1
z2
z3
－ － 0 ＋ 1 2 ＋ 3
Figure 23: a,b cycle of f z( ) z
z1
z2
z3
Figure 24 : The cut construct of sheet Ⅰ
b
a
0 1 2 3
Figure 25 : The cut construct of sheet Ⅱ
Figure 26 : The geometric structure In general If n=2N-1 ＋ bN1 ＋ ＋ ＋ Z2N1 Z2N2 Z2N3 …… Z4 Z3 Z2 Z1
－ －
aN1 a2－ －
1 a 2 b 1 b 0 1 2 3 b a 0 ＋ 1 2 ＋ 3 0 1 2 3 － － ＋ ＋ － － b bIf n=2N
Figure 27 :a,b cycles on complex plane
Figure 28: The a,b cycles on Riemann Surface
＋ bN1 ＋ ＋ ＋ 2 N Z Z2N1 Z2N2 Z2N3 …… Z4 Z3 Z2 Z1
－ －
aN1 a2－ －
a1 2 b 1 b 1 Z 2 Z Z 3 2N 1 Z 1 a 1 b Na
N b1 r 2 r
1.3 homotopic
C C 1 C 2Figure 29 : Homotopic path
In the homotopic case ,C is homotopic to C and 1 C . 2
So that 1 2 1 1 1 = ( ) ( ) ( ) C f z dz C f z dz C f z dz
,and we compress the curveC ,finally the equivalent paths C homotopic to r1 r2
1 2 1 1 1 ( ) ( ) ( ) C f z dz r f z dz r f z dz
.1.4 The conclusion of Riemann Surface
For any branch points 2N-1 or 2N we have I. There are N branch cuts in complex plane.
II. There are N-1 holes and that called Riemann Surface of genus N-1. III.There are N-1 a-cycles and N-1 b-cycles
2. The integrals of
1( )
f z
over a,b cycles for horizontal cut
First we discuss the value different in sheet-I of theory and on Mathematica .
1
in sheet-I, 1 i but 1in Mathematica, 1 i.
We will find the mistake in this.
We know that
,
in MathematicaFigure 30: Domain and range in Mathematica
Lemma Ⅰ:
,
in sheet-I for horizontal cut
( ) | , ( ) | ( ) | MATH sheet MATH f z f z f z =-Proof: Theory : 1 e i 1 e 2i i Mathematica : 1 e i 1 e2i i Thus ( ) ( ) math f z f z where in Mathematica. in Mathematica 2 2 ＋
Now we take a example to test and verify the Lemma Ⅰ.
To evaluate 1 , ( ) ( -1)( - 2) ( )dz where f z z z z f z
－ 0 1 r － 2 2 r 1 ＋Figure 31: cut plane of f z( ) z z( -1)( - 2)z
Proof : 1. z r1 (1) theory: 1 1 2 1 2 1 1 2 2 2 1 1 1 2 2 2 2 1 0 1 0 1 2 0 2 2 2 2 1 1 2 ( ) i i r z z z z z z z e z e i z i z z z f z
ｚ ｚ-1 ｚ-2 (2)mathematica: 1 1 2 1 2 1 1 2 2 2 1 1 1 2 2 2 2 1 0 1 0 1 2 0 2 2 2 2 1 1 2 ( ) i i r z z z z z z z e z e i z i z z z f z
ｚ ｚ-1 ｚ-22. z r2 (1) theory: 2 1 2 1 2 1 1 2 2 2 1 1 1 1 2 2 2 2 0 1 0 1 2 0 2 2 2 2 1 1 2 ( ) i i r z z z z z z z e z e i z i z z z f z
ｚ ｚ-1 ｚ-2 (2) mathematica: 2 1 2 1 2 1 1 2 2 2 1 1 1 1 2 2 2 2 0 1 0 1 2 0 2 2 2 2 1 1 2 ( ) i i r z z z z z z z e z e i z i z z z f z
ｚ ｚ-1 ｚ-2The value of (1) and (2) are the same.
So that 1 1 1 2 2 2 2 1 1 2 1 2 ( ) 0 0 5.24412 0 i z z z in theory dz f z in mathematica i in theory in mathematica
When = we see that have a problem between theory and
mathematica,they difference a ｀ －＇sign.
Now we get the result Lemma Ⅰ,we can use the result to do some example. Example 1. 1. z r1 . . . 0 arg( ) 0 1 0 arg( 1) 0 1 1 2 0 arg( 2) 2 2 Math Math Math z z z z z z z z z z z z 1 1 1 1 . 2 2 2 2 1 1 1 2 ( ) math r f z z z z
2. z r2 . . . 0 arg( ) 0 1 0 arg( 1) 0 1 1 2 0 arg( 2) 2 2 Math Math Math z z z z z z z z z z z z 2 1 1 1 . 2 2 2 2 1 1 1 2 ( ) math r f z z z z
So that we have 1 1 1 . 2 2 2 2 1 1 2 1 2 0 5.24412 ( ) math r f z z z z i
Example 2. Evaluate 1 ( )dz f z
and over a,bcycle.where f z( ) (z3)(z1)(z1)(z3)(z4)(z6)(z9) a 3 a 2 a 1 － －a3 －a2 －a1 ＋ -3 -1 ＋ 1 3 ＋ 4 6 ＋ 9
Figure 32: a-cycle and homotopic path a
Proof:
Take z19,z2 6,z34,z4 3,z51,z6 1,z7 3
1.z
6, 9Let a be a cycle center at 1 15
2 with r=2 and
15
2
2
iz
e
. (1) 1 7 1 191
2
( )
15
(
)
2
1.0842 10
0.0776642
i a i k kie
dz
d
f z
e
z
i
(by Cauchy Thm)
(2)theory:
z
a
1* (Ⅰ)z
6 → 9 1 1 2 2 2 1 2 1 9 0 9 9 9 9 9 1 , 2, 3, 4,5, 6, 7 i i k k k k z z z e z e i z z z z z z z z k z z 9 7 1 2 6 9 1 6 1 ( ) k k dz i z z dz f z
(Ⅱ)z
6 ← 9 1 1 2 2 2 1 2 1 9 0 9 9 9 9 9 1 , 2, 3, 4,5, 6, 7 i i k k k k z z z e z e i z z z z z z z z k z z 6 7 1 2 9 6 1 9 1 ( ) ( ) k k dz i z z dz f z
by(Ⅰ) (Ⅱ) we know that in theory
1 6 7 1 2 1 9 1 ( 2 ) ( ) k a k dz i z z dz f z
= 0.0776642i (3) mathematic :z
a
1* (Ⅰ)z
6 → 9 1 1 2 2 2 1 2 1 9 0 9 9 9 9 9 1 , 2, 3, 4,5, 6, 7 i i k k k k z z z e z e i z z z z z z z z k z z 9 7 1 2 6 9 1 6 1 ( ) ( ) k k dz i z z dz f z
(Ⅱ)
z
6 ← 9 1 1 2 2 2 1 2 1 9 0 9 9 9 9 9 1 , 2, 3, 4,5, 6, 7 i i k k k k z z z e z e i z z z z z z z z k z z 6 7 1 2 6 9 1 9 1 ( ) ( ) k k dz i z z dz f z
In mathematic the value of
1 1 0 ( ) a f z dz
(4)by the Lemma (Ⅰ)
z
a
1*(Ⅰ)
z
6 → 9 . 1 1 1 . . arg( ) arg( ) 0 2...7 ( ) ( ) math math k k k math z z z z z z z z z z z z k f z f z (Ⅱ)z
6 ← 9 . 1 1 1 . . arg( ) arg( ) 0 2...7 ( ) ( ) math math k k k math z z z z z z z z z z z z k f z f z So that 1 9 6 1 1 2 0.0776642 ( ) ( ) a f z dz f z dz i
2.z
3, 4(1)Let a be a cycle center at 2 7
2 with r=1 and
7
2
iz
e
2 7 11
( )
7
(
)
2
0 0.200969
i a i k kie
dz
d
f z
e
z
i
(by Cauchy Thm) (2)theory z a2 (Ⅰ)z 3 4 1 1 2 2 2 1 2 1 0 1, 2, 3 1 , 4,5, 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 4 7 1 3 2 3 4 1 3 1 ( ) k k dz i z z dz f z
(Ⅱ) z 3 4 1 1 2 2 2 1 2 1 0 1, 2, 3 1 , 4,5, 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 3 7 1 3 2 3 4 1 4 1 ( ) ( ) k k dz i z z dz f z
by(Ⅰ) (Ⅱ) we know that in theory
2 3 7 1 3 2 1 4 1 2 ( ) ( ) k a k dz i z z dz f z
= 0-0.200969i(3)mathematic z a2 (Ⅰ)z 3 4 1 1 2 2 2 1 2 1 0 1, 2, 3 1 , 4,5, 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 4 7 1 3 2 3 4 1 3 1 ( ) ( ) k k dz i z z dz f z
(Ⅱ) z 3 4 1 1 2 2 2 1 2 1 0 1, 2, 3 1 , 4,5, 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 3 7 1 3 2 3 4 1 4 1 ( ) ( ) k k dz i z z dz f z
By (Ⅰ)(Ⅱ) 2 1 0 ( ) a f z dz
in mathematic. (4) by Lemma (Ⅰ) (Ⅰ)z 3 4 . . . arg( ) 1, 2, 3 arg( ) 0 4,5, 6, 7 ( ) ( ) math k k k math k k k math z z z z z z k z z z z z z k f z f z (Ⅱ) z 3 4 . . . arg( ) 1, 2, 3 arg( ) 0 4,5, 6, 7 ( ) ( ) math k k k math k k k math z z z z z z k z z z z z z k f z f z So that 2 4 3 1 1 2 ( ) ( ) a f z dz f z dz
= 0-0.200969i 3.z
1,1
Let a is a cycle center at 0 with r = 23 z 2ei
(1) 3 7 1 18
1
2
( )
(2
)
3.46945 10
0.151409
i a i k kie
dz
d
f z
e
z
i
(by Cauchy thm) (2)theory :z a3 (Ⅰ) z 1 1 1 1 2 2 2 1 2 1 0 1, 2, 3, 4,5 1 , 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 1 7 1 5 2 1 1 1 1 1 ( ) k k dz i z z dz f z
(Ⅱ) z 1 1 1 1 2 2 2 1 2 1 0 1, 2, 3, 4,5 1 , 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 1 7 1 5 1 ( ) dz i z z dz
3 1 7 1 5 2 1 1 1 2 0.0151409 ( ) k a k dz i z z dz i f z
(3)mathematic: z a3 (Ⅰ) z 1 1 1 1 2 2 2 1 2 1 0 1, 2, 3, 4,5 1 , 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 1 7 1 5 2 1 1 1 1 1 ( ) ( ) k k dz i z z dz f z
(Ⅱ) z 1 1 1 1 2 2 2 1 2 1 0 1, 2, 3, 4,5 1 , 6, 7 i i k k k k k k k k k k z z z z z z e z z e i z z k z z z z z z z z k z z 1 7 1 5 2 1 1 1 1 1 ( ) ( ) k k dz i z z dz f z
By (Ⅰ)(Ⅱ) 3 1 0 ( ) a f z dz
in mathematic.(4)by the Lemma (Ⅰ) (Ⅰ)z 1 1 . . . arg( ) 1, 2, 3, 4,5 arg( ) 0 6, 7 ( ) ( ) math k k k math k k k math z z z z z z k z z z z z z k f z f z (Ⅱ) z 1 1 . . . arg( ) 1, 2, 3, 4,5 arg( ) 0 6, 7 ( ) ( ) math k k k math k k k math z z z z z z k z z z z z z k f z f z So that 3 1 1 1 1 2 0.151409 ( ) ( ) a f z dz f z dz i
b 1 b 2 b 3 － － － － ＋ -3 -1 ＋ 1 3 4 6 ＋ 9 Figure 33:b-cycles 4. － b3 － － － ＋ -3 -1 ＋ 1 3 ＋ 4 6 ＋ 9
Figure 34:b s equivalent path b3 3
(1)theory z b3 (Ⅰ)z 3 1 1 2 7 7 7 1 1 2 2 2 1 0 1 0 , 1, 2, 3, 4,5, 6 i i k k k k k k z z z z z z z z z z z z e z z e i z z k z z
7 1 1 6 2 3 1 3 1 1 ( ) k k dz i z z dz f z
(Ⅱ)z 3 1 1 2 7 7 7 1 1 2 2 2 1 0 1 0 , 1, 2, 3, 4,5, 6 i i k k k k k k z z z z z z z z z z z z e z z e i z z k z z 7 1 3 6 2 3 1 3 1 1 1 1 1 ( ) ( ) k k dz dz i z z dz f z f z
In the theory we have
3 7 1 7 1 3 6 3 2 2 1 1 1 1 1 2 2 0.0765026 ( ) k k b k k dz i z z dz z z dz f z
(2) mathematic z b3 (Ⅰ)z 3 1 1 2 7 7 7 1 1 2 2 2 1 0 1 0 , 1, 2, 3, 4,5, 6 i i k k k k k k z z z z z z z z z z z z e z z e i z z k z z 7 1 7 1 1 6 1 2 2 3 1 3 3 1 1 1 ( ) ( ) k k k k dz i z z dz z z dz f z
(Ⅱ)z 3 1 1 2 7 7 7 1 1 2 2 2 1 0 1 0 , 1, 2, 3, 4,5, 6 i i k k k k k k z z z z z z z z z z z z e z z e i z z k z z 7 1 3 2 3 1 1 1 1 ( ) k k dz z z dz f z
In mathematic3 1 0 ( ) b f z dz
(3)by the Lemma (Ⅰ)
(Ⅰ) z 3 1 7 7 7 arg( ) 0 arg( ) ( ) ( ) math math k k k math z z z z z z z z z z z z f z f z (Ⅱ)z 3 1 Because f z( ) f z( )
Consider the z 3 1 in sheet-Ⅰ
7 7 7 1 --- 3 1 3 arg( ) 0 arg( ) 1...6 ( ) ( ) ( ) ( ) ( ) math math k k k math math z z z z z z z z z z z z k f z f z f z f z f z 3 1 3 1 1 2 0.0765026 ( ) ( ) b f z dz f z dz
－ b2 － － －
＋ -3 -1 ＋ 1 3 ＋ 4 6 ＋ 9
Figure 35: b s equivalent path b2 2
(1)theory * 2 z b (Ⅰ) z 1 1 1 2 7 1 1 2 2 2 1 0 , 6, 7 1 0 , 1, 2, 3, 4,5 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 1 1 5 2 1 1 1 1 ( ) k k dz i z z dz f z
(Ⅱ) z-1 1 Because 1 1 1 1 1 1 ( ) ( ) f z f z
Consider z 1 1 1 2 7 1 1 2 2 2 1 0 , 6, 7 1 0 , 1, 2, 3, 4,5 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 1 1 5 2 1 1 1 1 1 1 1 ( ) ( ) i z zk dz f z f z
(Ⅲ) z 1 3 1 2 7 1 1 2 2 2 1 0 , 6, 7,5 1 0 , 1, 2, 3, 4 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 3 4 2 1 3 1 1 1 ( ) k k i z z dz f z
(Ⅳ) z1 3 1 3 1 3 1 1 ( ) ( ) f z f z
1 2 7 1 1 2 2 2 1 0 , 6, 7,5 1 0 , 1, 2, 3, 4 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 1 4 2 1 3 1 3 3 1 1 1 ( ) ( ) k k i z z dz f z f z
By (Ⅰ)(Ⅱ)(Ⅲ)(Ⅳ) * 2 7 1 7 1 1 5 3 4 2 2 1 1 1 1 1 2 2 ( ) k k b k k i z z dz i z z dz f z
=0.157328 (2)mathematic * 2 z b (Ⅰ) z 1 1 1 2 7 1 1 2 2 2 1 0 , 6, 7 1 0 , 1, 2, 3, 4,5 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 1 1 5 2 1 1 1 1 ( ) ( ) k k dz i z z dz f z
(Ⅱ) z-1 1 1 2 7 1 1 2 2 2 1 0 , 6, 7 1 0 , 1, 2, 3, 4,5 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 1 5 2 1 1 1 1 1 ( ) ( ) k k i z z dz f z
(Ⅲ) z 1 3 1 2 7 1 1 2 2 2 1 0 , 6, 7,5 1 0 , 1, 2, 3, 4 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 3 4 2 1 3 1 1 1 ( ) ( ) k k i z z dz f z
(Ⅳ) z1 3 1 2 7 1 1 2 2 2 1 0 , 6, 7,5 1 0 , 1, 2, 3, 4 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 1 4 2 1 3 3 1 1 ( ) ( ) k k i z z dz f z
By (Ⅰ)(Ⅱ)(Ⅲ)(Ⅳ) * 2 1 0 ( ) b f z
(3)by Lemma (Ⅰ) (Ⅰ) z 1 1 7 7 arg( ) 0 , 6, 7 arg( ) , 1, 2, 3, 4,5 ( ) ( ) math k math k k k math z z z z z z k z z z z z z k f z f z (Ⅱ) z-1 1 7 7 arg( ) 0 , 6, 7 arg( ) , 1, 2, 3, 4,5 ( ) ( ) math k math k k k math z z z z z z k z z z z z z k f z f z (Ⅲ) z 1 3 7 7 arg( ) 0 , 5, 6, 7 arg( ) , 1, 2, 3, 4 ( ) ( ) math k math k k k math z z z z z z k z z z z z z k f z f z (Ⅳ) z1 3 1 3 1 3 1 1 ( ) ( ) f z f z
Consider z 1 3 7 7 arg( ) 0 , 5, 6, 7 arg( ) , 1, 2, 3, 4 math k math k k k z z z z z z k z z z z z z k 1 3 ( ) ( ) math f z f z 1 3 1 3 ( ) ( ) ( ) math f z f z f z * 2 7 1 7 1 1 3 2 2 1 1 1 1 1 2 2 ( ) k k b k k z z dz z z dz f z
=0.157328 b1 － － － ＋ -3 -1 ＋ 1 3 ＋ 4 6 ＋ 9Figure 36: b s equivalent path b1 1
(1) theory z b1 (Ⅰ)z 3 4 1 2 7 1 1 2 2 2 1 0 , 4,5, 6, 7 1 0 , 1, 2, 3, k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 4 3 2 3 4 3 1 1 ( ) k k i z z dz f z
(Ⅱ) z 3 4 3 4 1 2 7 1 1 2 2 2 1 0 , 4,5, 6, 7 1 0 , 1, 2, 3, k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 3 3 2 3 -- 4 4 1 1 ( ) k k i z z dz f z
(Ⅲ)z 4 6 1 2 7 1 1 2 2 2 1 0 , 3, 4,5, 6, 7 1 0 , 1, 2 k k i i z z z z k z z z z z z z z e z z e i z z k 7 1 6 2 2 4 6 4 1 1 ( ) k k i z z dz f z
(Ⅳ) z 4 -- 6 4 -- 6 4 6 1 1 ( ) ( ) f z f z
1 2 7 1 1 2 2 2 1 0 , 3, 4,5, 6, 7 1 0 , 1, 2 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 4 2 2 4 -- 6 4 6 6 1 1 1 ( ) ( ) k k i z z dz f z f z
* 1 7 1 7 1 7 1 3 6 3 4 6 2 2 2 2 1 1 4 1 1 1 1 2 2 2 ( ) k k k b k k k i z z dz i z z dz i z z dz f z
(2)mathematic z b1 (Ⅰ)z 3 4 1 2 7 1 1 2 2 2 1 0 , 4,5, 6, 7 1 0 , 1, 2, 3, k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 4 3 2 3 4 3 1 1 (- ) ( ) k k i z z dz f z
(Ⅱ) z 3 4 3 4 1 2 7 1 1 2 2 2 1 0 , 4,5, 6, 7 1 0 , 1, 2, 3, k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 3 3 2 3 -- 4 4 1 1 ( ) ( ) k k i z z dz f z
(Ⅲ)z 4 6 1 2 7 1 1 2 2 2 1 0 , 3, 4,5, 6, 7 1 0 , 1, 2 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 6 2 2 4 6 4 1 1 ( ) ( ) k k i z z dz f z
(Ⅳ) z 4 -- 6 1 2 7 1 1 2 2 2 1 0 , 3, 4,5, 6, 7 1 0 , 1, 2 k k i i k k k k k k z z z z k z z z z z z z z e z z e i z z k z z 7 1 4 2 2 4 -- 6 6 1 1 ( ) ( ) k k i z z dz f z
(3)by the Lemma (Ⅰ)
(Ⅰ)z 3 4 7 7 arg( ) 0 , 4,5, 6, 7 arg( ) , 1, 2, 3, ( ) ( ) math k math k k k math z z z z z z k z z z z z z k f z f z (Ⅱ) z 3 4 3 4 7 7 arg( ) 0 , 4,5, 6, 7 arg( ) , 1, 2, 3, ( ) ( ) math k math k k k math z z z z z z k z z z z z z k f z f z
(Ⅲ)z 4 6 7 7 arg( ) 0 , 3, 4,5, 6, 7 arg( ) , 1, 2, ( ) ( ) math k math k k k math z z z z z z k z z z z z z k f z f z (Ⅳ) z 4 -- 6 ( ) ( ) f z f z Consider z 4 6 7 7 arg( ) 0 , 3, 4,5, 6, 7 arg( ) , 1, 2, ( ) ( ) math k math k k k math z z z z z z k z z z z z z k f z f z So ( )4 --6 ( )4 6 ( ) math f z f z f z By (1) (2) (3) * 1 7 1 7 1 7 1 3 3 6 2 2 2 1 1 4 1 1 1 1 2 2 2 ( ) k k k b k k k dz z z dz z z dz z z dz f z
Now we discuss in general situation:
Compute 1
( )dz
f z
over a,b cycles for horizontal cut where1 ( ) ( ) m k k f z z z
, 1 2 1... , ... k k m where z z z 1.a-cycle aN1 a j a 2 a 1 _ _ 1 N a _ aj _ 2 a _ 1 a ＋ Z2N1 Z2N2＋ Z2N3… Z ＋2j Z2j1 … Z ＋4 Z 3 Z ＋2 Z 1
Figure 37: a-cycles for 2N-1 points
a N aN1 a j a 2 a 1 _ aN _ 1 N a _ aj _ 2 a _ 1 a Z2 N＋ Z2N1 Z2N2＋Z2N3… Z ＋2j Z2j1 … Z ＋4 Z 3 Z ＋2 Z 1
Figure 38:a-cycles for 2N points
By Cauchy theorem, we can get that
* 1 1 ( ) ( ) j j a f z dz a f z dz
Now we consider the path by the Lemma Ⅰ
2 2 1 2 2 1 2 2 1 2 2 1 : ( ) + : ( ) j j j j j j j j
z z the path z z on edge z z the path z z on edge
(1) consider z2jz2j1 ( ) on edge
arg( ) 0 , 2 , 2 1,..., arg( ) , 1, 2,..., 2 1 math k k k math k k k z z z z z z where k j j m z z z z z z where k j 2 1 ( ) ( 1) ( ) ( ) math j f z f z f z So that 2 1 2 2 2 1 1 1 ( ) ( ) j j math z j j f z z f z
(2) consider z2jz2j1 ( ) on edge arg( ) 0 , 2 , 2 1,..., arg( ) , 1, 2,..., 2 1 math k k k math k k k z z z z z z where k j j m z z z z z z where k j ( ) ( ) math f z f z So that 2 2 1 2 2 1 1 1 ( ) ( ) j j math z j j f z z f z
So we have the conclusion
2 1 * 2 1 1 2 ( ) ( ) j j j math z a f z z f z
2.b-cycle(1) Give b is a circle centered at x with radius r and enclosed j
b j
－ － － － － ＋ Z2N1 Z2N2＋ Z2N3… Z ＋2 j Z2j1 … Z ＋4 Z 3 Z ＋2 Z 1
Figure 39: b -cycle for 2N-1 points j
(2) Give b is a circle centered at x with radius r and enclosed j
theZ2N1,Z2j and intersect at the points on Z2j1,Z2j and
Z2N1,Z2N
.b j
- - _ _ _
2 N
Z ＋ Z2N1 Z2N2＋ Z2N3… Z ＋2 j Z2j1 … Z ＋4 Z 3 Z ＋2 Z 1
Figure 40: b -cycle for 2N points j
j i , 0 2 , 3 . If zb and z x re where Note that f z( ) f z( ) 0 3 2 1 1 0 0 1 ( ) = j i i m m b i i k k k k i i m m i i k k rie rie dz d d f z x re z x re z rie rie d d x re z x re z
3.Now we consider the equivalent path bj bj － － － － － ＋ Z2N1 Z2N2＋ Z2N3… Z ＋2 j Z2j1 … Z ＋4 Z 3 Z ＋2 Z 1 Figure 41: j
b-cycle for 2N-1 points
bj － － － － － 2 N Z ＋ Z2N1 Z2N2＋ Z2N3… Z ＋2 j Z2j1 … Z ＋4 Z 3 Z ＋2 Z 1 Figure 42: j
b-cycle for 2N points
By Cauchy Theorem,we know that
1 1
( ) ( )
j j
b f z b f z
The path we must have two part to discuss. (1) path on the cut is
2 2 2 1 2 2 2 1 , , 1,..., 2 ( ) , , 1,..., 2 ( ) s s s s
z z where s j j N on the edge in sheet
z z where s j j N on the edge in sheet
