黎曼空間的理論和其在微分方程上的應用

國

立

交

通

大

學

應用數學系

碩

士

論

文

黎曼空間的理論和其在微分方程上的應用

Theory of Riemann Surfaces and Its Applications

to Differential Equations

研 究 生：吳昀庭

指導教授：李榮耀 教授

黎曼空間的理論和其在微分方程上的應用

Theory of Riemann Surfaces and Its Applications

to Differential Equations

研 究 生：吳昀庭 Student：Yun-Ting Wu

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 Thesis

Submitted 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 April 2010

Hsinchu, Taiwan, Republic of China

誌

謝

在交大研究所的日子裡，經歷了許多心境的轉折，曾經徬徨迷失不知所措過， 忘記當初想來交大的目的。幸好在做研究寫論文的日子裡，不只找回了目的，更 找回能吃苦肯努力且認真有目標的態度面對人生的自己。接下來的人生也將會全 力以赴。 非常感謝李榮耀教授，對我的指導和鼓勵，不論是做學問或是做研究的 方法，都給了學生許多教導，讓我對做研究產生了興趣和自信。不僅僅課業上， 還有許多各方面，像是生涯規劃、做學問的態度等都給了學生許多教誨。謝謝林 松山教授，題點了學生許多人生的道理。感謝李志豪教授，對於我的疑惑都耐心 解答。 感謝父母對於我人生的夢想圖給予很大的尊重和支持，不只讓我專心的 完成學業。並在我失敗，遇到挫折和錯誤時鼓勵我，點醒我。還要感謝弟不少的 幫忙，教我如何使用繪圖軟體。謝謝朋友們的陪伴，讓我在交大的日子多了很多 精彩。最後感謝交通大學的老師、系辦的人員及學長姐們，在我交大求學階段帶 給我ㄧ個很棒的經歷，學到了很多，成長了很多。真的由衷地感謝。

i

黎曼空間的理論和其在微分方程上的應用

研究生：吳昀庭 指導老師：李榮耀 教授

國 立 交 通 大 學

應 用 數 學 系

摘 要

此篇文章主要在探討擁有

0 ) ( 2 2  P u dt u d N

形式的非線性二階微分方程，其解的函數理論，其中

PN(u)

是 2N

或 2N-1 次多項式。此方程的解存在於 N-1 相黎曼空間上。我們要利

用正確的代數結構來建構這些黎曼空間。以此為基準，無論是理論或

數值上我們可以在黎曼空間執行路徑的積分，並在此原則上獲得其解。

其中

PN(u)

的根扮演了重要的角色，而複數分析是我們主要的工具。

中 華 民 國 九 十 九 年 四 月

Theory of Riemann Surfaces and Its Applications to

Differential Equations

Student：Yun-Ting Wu Advisor：Jong-Eao Lee

Department of Applied Mathematics

National Chiao Tung University

Abstract

In this paper, we study the function theory of the solutions of

the nonlinear second-order equations which have the

following forms,

0 ) ( 2 2  P u dt u d N

where

PN(u)

is a polynomial of degree 2N-1 or 2N. Solutions

of such equations reside on Riemann surfaces of genus N-1.

We construct those Riemann surfaces with the correct

algebraic structures. From which, we are able to perform

path integrals on the Riemann surfaces theoretically and

numerically, and, in principle, solutions can be derived. The

roots of

P_N(u)

play the essential roles in every aspects, and

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 curve in algebraic and geometric structure . . . 10

1.3 The a,b cycles and its equivalent paths . . . 11

1.4 Conclusion of Riemann Surface . . . 15

2 The integrals of _{f (z)}1 over a,b cycles for horizontal cut 17

3 The integrals of _{f (z)}1 over a,b cycles for vertical cut 46

4 The integrals of 1

f (z) over a,b cycles for slant cut 95

5 Applications of differential equations 164

6 Conclusion 177

1

Introduction the Riemann Surface

u00+ PN(u) = 0 is a nonlinear second-order equation where PN(u) is a polynomial of

u with degree 2N-1 or 2N. From

u00+ PN(u) = 0, we have u00u0+ PN(u)u0 = 0 and 1 2(u 0 )2+ PN +1(u) = E

where E is the integration constant. It is a conservation law. So the function theory of solutions u of the equation involve

s

N +1

Q

k=1

(u − uk), we must investigate the space where u

reside. Indeed, f (z) = s n Q k=1

(z − zk) is a two-valued function of z on complex plane C. We

use algebra and analysis to develop a new surface such that f becomes a single-valued and analytic function on this surface, namely, a Riemann Surface.

1.1

Construct the corresponding Riemann Surface

When w, z ∈ C and wm = z, we use polar form to find the solution

wm = z = |z|eiθ = |z|eiθ+2nπ_{, θ ∈ [−π, π), n ∈ Z} then w = |z|m1e

(θ+2nπ)i

m , θ ∈ [−π, π), n ∈ Z

First, take f (z) = √_{z for example, f : C → C. Using polar form, let z = |z|e}iθ = |z|ei(θ+2nπ) , n ∈ Z f (z) = √z = |z|12e θ+2nπ 2 i = ( |z|12e θ 2i if n:even, −|z|12e θ 2i if n:odd. (1) is a two-valued function. Now we want to let f (z) becomes a single valued function, so we modify its domain C to develop the corresponding Riemann Surface such that f becomes

a single-valued and analytic function on this surface.

Figure 1: The idea of two sheets

Starting at z = reiθ_{, we have f (z) =} √_{z =} √_reθ₂i_{, r 6= 0. Fixing r and continuing}

along a closed path once around the origin so that θ increases by 2π, f (z) comes to the value √reθ+2π2 i = −

√

reθ2i which is just the negative of its original value. Continuing

above way then θ increases by 2π and f (z) comes to original value. First, image two sheets lying over the complex plane and cut the plane along negative real axis (i.e. from 0 to infinite) and restrict ourselves so as never to continue f (z) over this cuts, we get single-valued branches of f (z). Define that

f (z) = |z|12e iθ 2, −π ≤ θ < π (2) f (z) = |z|12e iθ 2, π ≤ θ < 3π (3)

called sheet-I and sheet-II, respectively. The cut on each sheet has two edges, label the edge of starting edge with − and the edge of terminal edge with +. (Show in Figure 1.) Moreover, we cross the cut, we pass from one sheet to another. Second we extend the plane of complex numbers with one additional point at infinity constitute a number system known as the extended complex numbers. Use stereographic projection, we can consider the two sheets to be a spheres.

Figure 2: complex plane and extended complex plane

Next, image that the spheres are made of rubber and stretch each cut into circular holes.

Figure 3: place the cuts open

Rotate the spheres until the holes face each other, and paste two cuts together (+)edge of sheet-I with (−)edge of sheet-II and (−)edge of sheet-I with (+)edge of sheet-II. We can derive a sphere. We called this sphere, Riemann surface of genus 0 , denoted R0.

Show in Figure 4. Notice that in Riemann Surface (+)edge of sheet-I is equivalent to (−)edge of sheet-II and (−)edge of sheet-I is equivalent to (+)edge of sheet-II.

We could using similar way to develop the corresponding Riemann surface for f (z) = s

n

Q

k=1

Figure 4: construct R0

Example 1: There are 7 roots of f(z). Construct the Riemann Surface of f (z) = s 7 Q k=1 (z − zk) = 7 Q k=1

p(z − zk), zk ∈ Z , z1 > z2 > . . . > z7, we cut plane starts from zk

to −∞, k = 1, . . . , 7.

Figure 5: Cut plane start from zk to −∞

Since cross one cut, we pass from one sheet to another, the argument of z increases by 2π then argument of f(z) increases by π which is just the negative of its original value. We have crossing one cut need to change the sign, using −1 represent that. So crossing odd times will change sign and even times will no change.

There are branch cuts in [z1, z2], [z3, z4], [z5, z6], [z7, −∞) and then using same idea to

construct the corresponding Riemann Surface.

Figure 7: Placing the cuts open

Figure 8: Geometric graph of R3

Finally, to place the cuts open and put two sheet together with the rule (+)edge with (−)edge and then we obtain corresponding Riemann Surface of genus 3.

Example 2: Construct the Riemann Surface of f (z) = s 8 Q k=1 (z − zk) = 8 Q k=1 p(z − zk),

zk∈ R , z1 > z2 > ... > z8. Similarly we cut plane start from zk to −∞, k = 1, ..., 8.

Figure 9: cut start from zk to −∞

As same as example 1, so there are branch cuts in [z1, z2], [z3, z4], [z5, z6], [z7, z8].

Figure 10: The cut plane of example 2

Figure 12: R3

We found f (z) of 7 or 8 roots have different algebraic structures but same geometric graph with 3 holes. That is no matter 7 or 8 points, we can construct corresponding Riemann Surface of genus 3.

In general situation, using same idea to construct Riemann surface of f (z) where f (z) = s n Q k=1 (z − zk) = n Q k=1

p(z − zk), zk ∈ R and z1 > z2 > ... > zn for horizontal cut.

First, we cut plane starts from zk to −∞. If the curve cross even cuts it will no change

that is becomes no cut. If the curve cross odd cuts, it will has a branch cut.

So the cuts plane structure is

Case1. If n=2N-1. There are cuts along [z1, z2], [z3, z4], · · · , [z2j−1, z2j], · · · , [z2N −1, −∞)

Figure 14: n = 2N − 1

Case2. If n=2N. There are cuts along [z1, z2], [z3, z4], · · · , [z2j−1, z2j], · · · , [z2N −1, z2N]

Figure 15: n = 2N

We use same idea to construct the corresponding Riemann Surface: (1) n=2N-1

Figure 16: Placing cuts open in both sheets

Figure 18: N − 1 holes for n = 2N − 1

becomes Riemann Surface with N-1 holes, that is RN −1 (2) n=2N

Figure 19: do this in two sheets

Figure 21: N − 1 holes for n = 2N So f (z) = v u u t 2N −1 Y k=1 (z − zk) or = v u u t 2N Y k=1 (z − zk)

will make N cuts and construct Riemann surface of genus N-1. (There is N-1 holes of geometric graph.)

1.2

The curve in algebraic and geometric structure

For convenience, we use algebraic to discuss and compute the integrals later. We already know the relation of algebraic and geometric structure with f (z) =

s

n

Q

k=1

(z − zk)

and how to create the Riemann surface. Here give some examples to show that the curve in algebraic structure and its corresponding in geometric structure.

We defined something as follow:

1. The curve in sheet-I is solid line and the curve in sheet-II is dash line in algebraic structure.

2. The curve in overhead Riemann Surface is solid line and the curve in ventral Rie-mann Surface is dash line in geometric structure.

Example 1.

r1 is the curve from a point at (I,+) to (I,−) in sheet-I and r2 is the curve from a point

Figure 22: Example 1.

Example 2. The curve r is start from point A in sheet-I and cross the cut to point B on sheet-II

Figure 23: Another example

1.3

The a,b cycles and its equivalent paths

We know every closed curve on Riemann Surface RN can be deformed into an integral

combination of the loop-cut ai and bi, i=1,2,...,N. So in this paper, we will consider

the integrals of f (z) over a, b-cycles help us to obtain the integrals easier. Example f (z) =pz(z − 1)(z − 2)(z − 3)

Figure 24: Cut plane and a,b cycle of f (z) =pz(z − 1)(z − 2)(z − 3)

If f(z) has four roots and then construct two cuts and one a, b cycle. Notice a, b cycles have the same amount.

Figure 25: Step 1

Figure 27: Step 3 and Step 4

Now if f (z) has 2N-1 or 2N roots, there are loop-cuts ai and bi, i=1,2,...,N-1.

Figure 29: a,b cycles on Riemann Surface

Each a cycles are non-overlapping and each b cycles are non-overlapping. Also a, b cycles have the same number.

Sometimes the curves are difficult to write out their parameters, but straight lines are ease to write out their parameters. It could help us quicker and easier to obtain the integrals over the curves. So now using homotopic of curves to find the equivalent paths of curves. Take an example to explain.

From C is homotopic to C1, denotes C ≈ C1. We have Z C f (z)−1dz = Z C1 f (z)−1dz + Z Γ2 f (z)−1dz

In figure 30, C ≈ C1 ≈ C2 ≈ C3 and finally we compression the curve C until we find the

equivalent paths of curves C ≈ Γ1S Γ2. So

Z C f (z)−1dz = Z Γ1 f (z)−1dz + Z Γ2 f (z)−1dz

We will use this tool in hole paper.

1.4

Conclusion of Riemann Surface

The above statement and result all in horizontal cut, but the way is the same in any cut. Take w2 _{= a(z − z}

1)(z − z2)(z − z3), where z1, z2, z3 are distinct for example. We let

f (z) =p(z − z1)p(z − z2)p(z − z3) and discuss f (z). Cause

√

a does not influence the cuts. For f (z) the factor p(z − zk) change the sign when when arg (z − zk) changes by

2π. We cut complex plane from z1 to z2 and from z3 to ∞. Label left of cut with (+)edge

and right of cut with (−)edge

Figure 31: The cut-plane and a, b cycle in sheets

Using similarly way to construct the Riemann Surface. Image this two sheets are made of rubber, and together the (+)edges of sheet-I with the (−)edges of sheet-II. We

get correspond Riemann Surface R1. The curve a,b correspond to the meridian curve a

and latitude curve b on Riemann Surface R1, respectively.

Figure 32: Corresponding Riemann Surface

For arbitrary cut, if f(z) has 2N-1 or 2N roots, then 1. There are N cuts in complex plane.

2. It’s geometric graph has N-1 holes, that is construct corresponding Riemann Surface of genus N-1.

2

The integrals of

_{f (z)}1

over a,b cycles for horizontal

cut

We will use Mathematica help us to obtain the values of integrals of _{f (z)}1 over a,b cycles. First, discuss the values in sheet-I, sheet-II and Mathematica for horizontal cuts. f (z) =

s

n

Q

k=1

(z − zk), using polar form n

Q

k=1

(z − zk) = reθi. Let θ1 denotes θ in sheet-I and

θ2 denotes in sheet-II. So θ2 = θ1+ 2π We have f (z)|(II)= √ reθ22 i =√reθ1+2π2 i =√reθ12 ieπi = −√reθ12 i = −f (z)|_(I) (4)

where f (z)|(I) denote the value of f (z) with z in sheet-I and f (z)|(II) means z in sheet-II.

Because the difference of argument between z in sheet-I and in sheet-II is 2π , that is the difference between f (z)|(I) and f (z)|(II) is π . So f (z)|(I) = −f (z)|(II).

Now discuss the difference in sheet-I of theory and in Mathematica. First,√−1. From the definition of argument in sheet-I, √−1 = −i, but we compute √−1 in Mathematica obtain√−1M ath.= i. Why? We found that θ ∈ (−π, π] of reiθ in Mathematica, actually.

For any other θ of reiθ _{which does not belong to (−π, π], Mathematica will conversion}

reiθ _{into re}iθ∗_{, θ}∗ _{∈ (−π, π] where re}iθ _{= re}iθ∗_.

Compare the value of f (z) with z in sheet-I and in Mathematica, we discover that Lemma 1. If

n

Q

k=1

(z − zk) = reiθ in sheet-I for horizontal cut

f (z)|(I) =

 f (z)|M athematica if θ ∈ (−π, π),

−f (z)|M athematica if θ = −π

Proof.

Since−π does not in (−π, π], Mathematica will conversion re−πi into reπi and re−πi= reπi but f (z) will different.

In theory: − 1 = e−πi

√

→√−1 = e−πi2 = −i.

In Mathematica: − 1 = e−πi M ath.= eπi

√

→√−1 = eπi2 = i

So f (z)M ath.= −f (z) if θ = −π in Mathematica.

In hole paper, f (z) M ath.= −f (z) denotes the polynomial f (z) in front of M ath.= is the value of f (z) in theory and the polynomial f(z) behind the M ath.= is the value of f (z) in Mathematica.

After we known the state above, we must modify the computation when we want to use Mathematica to calculate the value. Take example to explain: evaluate R

r 1

f (z)dz where

f (z) =pz(z − 1)(z − 2), z ∈ R and r = r1S r2 where r1 = the path on a horizontal cut

from 1 to 2 with (+)edge of sheet-I and r2 = the path on a horizontal cut from 2 to 1

with (−)edge of sheet-I.

Figure 34: cuts in complex plane of f (z) =pz(z − 1)(z − 2)

1. If z ∈ r1.

(1) In theory: z ≥ 0 then√z = |z|12 and z − 1 ≥ 0 then

√

z − 1 = |z − 1|12

z − 2 < 0 then z − 2 = |z − 2|e−πi, so √z − 2 = |z − 2|12e− π 2i = −i|z − 2| 1 2 Z r1 1 f (z)dz = i Z 2 1 |z|−12|z − 1|− 1 2|z − 2|− 1 2dz

(2) In Mathematica: z ≥ 0 then √z = |z|12 , z − 1 ≥ 0 then

√

z − 1 = |z − 1|12

z − 2 < 0 then z − 2 = |z − 2|eπi. We have√z − 2 = |z − 2|12e π 2i = i|z − 2| 1 2 Z r1 1 f (z)dz = −i Z 2 1 |z|−12|z − 1|− 1 2|z − 2|− 1 2dz

Compare (1) and (2), we found there a difference of a minus sign with the value in sheet-I and in Mathematica.

2. z ∈ r2

(1) In theory: z ≥ 0 then√z = |z|12 , z − 1 ≥ 0 ⇒

√

z − 1 = |z − 1|12

z − 2 < 0 then z − 2 = |z − 2|eπi _then √_{z − 2 = |z − 2|}1_2eπ₂i _{= i|r − 2|}1 2 Z r2 1 f (z)dz = −i Z 1 2 |z|−12|z − 1|− 1 2|z − 2|− 1 2dz

(2) In Mathematica: z ≥ 0 then √z = |z|12 , z − 1 ≥ 0 then

√

z − 1 = |z − 1|12

z − 2 < 0 then z − 2 = |z − 2|eπi _then √_{z − 2 = |z − 2|}1_2eπ₂i _{= i|r − 2|}1 2 Z r2 1 f (z)dz = −i Z 1 2 |z|−12|z − 1|− 1 2|z − 2|− 1 2dz

Compare (1) and (2), it is the same. By 1,2 we have Z r 1 f (z)dz =  2i R2 1 |z| −1 2|z − 1|− 1 2|z − 2|− 1 2dz in sheet-I , 0 in Mathematica =  0. + 5.24412i in sheet-I , 0 in Mathematica

Clearly, there is a mistake when θ = −π. When we use Mathematica to get the value of integration we want, we need modify some range where the value will wrong.

Determine the difference of sign(f) (same or negative) and then modify the computation of Mathematica to get right value. Because sometimes the form of integration is complex, if we could simplify the way about modify the difference of sigh(f), it will help us to get right value easier.

Example: Same f(z) as the example before, using lemma1 to modify. 1. If z ∈ r1, z : 1 → 2

z ≥ 0 then arg(z) = 0 then √z M ath.= √z

z − 1 ≥ 0 then arg(z − 1) = 0 then √z − 1M ath.= √z − 1 z − 2 < 0 then arg(z) = −π then √z − 2M ath.= −√z − 2

Z r1 1 f (z)dz M ath. = − Z 2 1 1 √ z 1 √ z − 1 1 √ z − 2dz 2. If z ∈ r2, z : 2 → 1

z ≥ 0 then arg(z) = 0 then √z M ath.= √z

z − 1 ≥ 0 then arg(z − 1) = 0 then √z − 1M ath.= √z − 1 z − 2 < 0 then arg(z) = π then √z − 2M ath.= √z − 2

Z r2 1 f (z)dz M ath. = Z 1 2 1 √ z 1 √ z − 1 1 √ z − 2dz By 1,2 we have Z r 1 f (z)dz M ath. = −2 Z 2 1 1 √ z 1 √ z − 1 1 √ z − 2dz = 0. + 5.24412i

Take another example to consider how to modify computation in Mathematica such that numeral result is right for horizontal cuts. And discuss the difference between the value in theory and in Mathematica.

Example 2 : Evaluate R _{f (z)}1 dz over a1, a2, a3, b1, b2 and b3 cycles. where

Figure 35: a-cycles and their equivalent path a∗

Mathematica and in theory to compare the result and using the result of angle to modify the computation to get value. Let z1 = 9, z2 = 6, z3 = 4, z4 = 3, z5 = 1, z6 = −1, z7 = −3

Solution:

1. Let a1 is a cycle center at 15₂ with radius 2 and enclosed the cut [6, 9]. So let

z = 15₂ + 2eiθ_{, we have} Z a1 1 f (z)dz = Z π −π 2ieiθ 7 Q k=1 q 15 2 + 2e iθ_{− z} k dθ = 1.0842 × 10−19+ 0.0776642i

By Cauchy Theorem. Since ak cycle is simple connected, we can use some equivalent

paths, say a∗_k, to easily compute the integrals for ak cycle.

(1) If z ∈ a∗₁ of theory in sheet-I where a∗₁ = 6 → 9+ S 6← 9−

(a) 6→ 9 : the path along x-axis from 6 to 9 on (+)edge of sheet-I.+

z − 9 = −|z − 9| = |z − 9|e−πi then √ 1

z − 9 = |z − 9| −1 2e π 2i = |z − 9|− 1 2i z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 2, 3, 4, 5, 6, 7 Z 6→9+ 1 f (z)dz = Z 9 6 i 7 Y k=1 |z − zk|− 1 2dz

(b) 6← 9 : the path along x-axis from 9 to 6 on sheet-I with (−)edge.−

z − 9 = −|z − 9| = |z − 9|eπi then √ 1

z − 9 = |z − 9| −1 2e− π 2i = −i|z − 9|− 1 2 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 2, 3, 4, 5, 6, 7

Z 6←9− 1 f (z)dz = Z 6 9 (−i) 7 Y k=1 |z − zk|− 1 2dz

So by (a) and (b) we have Z a∗1 1 f (z)dz = −2i Z 6 9 7 Y k=1 |z − zk|− 1 2dz = 0.0776642i

(2) Analysis the integral over a∗₁ in Mathematica

(a) 6→ 9 : the path along x-axis from 6 to 9 on sheet-I with (+)edge+

z − 9 = −|z − 9| = |z − 9|eπi then √ 1

z − 9 = |z − 9| −1₂ e−π2i = −i|z − 9|− 1 2 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 2, 3, 4, 5, 6, 7 Z 6→9+ 1 f (z)dz = Z 9 6 (−i) 7 Y k=1 |z − zk|− 1 2dz

A difference of a minus sign with in sheet-I

(b) 6← 9 : the path along x-axis from 9 to 6 of sheet-I with (−)edge−

z − 9 = −|z − 9| = |z − 9|eπi then √ 1

z − 9 = |z − 9| −1 2e− π 2i = −i|z − 9|− 1 2 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 2, 3, 4, 5, 6, 7 Z 6←9− 1 f (z)dz = Z 6 9 (−i) 7 Y k=1 |z − zk|− 1 2dz as same as in sheet-I. But in Mathematica Z a∗ 1 1 f (z)dz = 0 (3) Using the results before to modify

(a) 6→ 9: the path along x-axis from 6 to 9 on sheet-I with (+)edge+ arg(z − z1) = −π then √ z − z1 M ath. = −√z − z1 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 2, . . . , 7 So f (z)M ath.= −f (z)

(b) 6← 9: the path along x-axis from 9 to 6 on sheet-I with (−)edge− arg(z − z1) = π then √ z − z1 M ath. = √z − z1 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 2, . . . , 7 So f (z)M ath.= f (z) We have Z a∗₁ 1 f (z)dz M ath. = −2 Z 9 6 1 f (z)dz = 0.0776642i

2. Let a2is a cycle center at 7₂ with radius 1 and enclosed the cut [3, 4]. So let z = 7₂+eiθ,

we have Z a2 1 f (z)dz = Z π −π ieiθ 7 Q k=1 q 7 2 + eiθ − zk dθ = 0. − 0.200969i

Same as a1 , by Cauchy Theorem to compute equivalent path a∗2 where a∗2 = 3 +

→ 4S 3← 4−

(1) Analysis the integral of a∗₂ in sheet-I

(a) 3→ 4: the path along x-axis from 3 to 4 on sheet-I with (+)edge+

z − zk= −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z k|− 1 2i, k = 1, 2, 3 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7 Z 3→4+ 1 f (z)dz = Z 4 3 i3 7 Y k=1 |z − zk|− 1 2dz

(b) 3← 4 : the path along x-axis from 4 to 3 on sheet-I with (−)edge−

z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z k|− 1 2, k = 1, 2, 3 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7

Z 3←4− 1 f (z)dz = Z 3 4 (−i)3 7 Y k=1 |z − zk|− 1 2dz So we have Z a∗ 2 1 f (z)dz = −2 Z 3 4 (−i)3 7 Y k=1 |z − zk|− 1 2dz = 0. − 0.200969i

(2) Analysis the integral of a∗₂ in Mathematica

(a) 3→ 4 : the path along x-axis from 6 to 9 on sheet-I with (+)edge+

z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2, 3 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7 Z 3→4+ 1 f (z)dz = Z 4 3 (−i)3 7 Y k=1 |z − zk|− 1 2dz

(b) 3← 4 : the path along x-axis from 4 to 3 on sheet-I with (−)edge−

z − zk= −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z k|− 1 2, k = 1, 2, 3 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7 Z 3←4− 1 f (z)dz = Z 3 4 (−i)3 7 Y k=1 |z − zk|− 1 2dz

But by (a),(b) we obtain different value in Mathematica Z

a∗₂

1

f (z)dz = 0 (3) Using lemma 1 to modify

(a) 3→ 4 : the path along x-axis from 3 to 4 on sheet-I with (+)edge+ arg(z − zk) = −π then √ z − zk M ath. = −√z − zk , k = 1, 2, 3 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk , k = 4, 5, 6, 7 So f (z)M ath.= −f (z)

(b) 3← 4 : the path along x-axis from 4 to 3 on sheet-I with (−)edge− arg(z − zk) = π then √ z − zk M ath. = √z − zk , k = 1, 2, 3 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk , k = 4, 5, 6, 7 So f (z)M ath.= f (z) We have Z a∗₃ 1 f (z)dz M ath. = −2 Z 4 3 1 f (z)dz = 0. − 0.200969i

3. a3 : Let a3 is a cycle center at 0 with radius 2 and enclosed the cut [−1, 1]. So let

z = 2eiθ_{, we have} Z a3 1 f (z)dz = Z π −π 2ieiθ 7 Q k=1 √ 2eiθ_{− z} k dθ = 3.46945 × 10−18+ 0.151409i

Same as a1 , by Cauchy Theorem to compute equivalent path a∗3 where a ∗ 3 = −1 + → 1S 1← −1− (1) Analysis of a∗₃ in theory

(a) −1→ 1 : the path along x-axis from −1 to 1 on sheet-I with (+)edge+

z − zk = −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = i|z − z k|− 1 2, k = 1, 2, 3, 4, 5 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7 Z −1→1+ 1 f (z)dz = Z 1 −1 i5 7 Y k=1 |z − zk|− 1 2dz

(b) −1← 1 : the path along x-axis from 1 to -1 on (−)edge of sheet-I−

z − zk= −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− πi 2 = −i|z − z_k|− 1 2, k = 1, 2, 3, 4, 5 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7

Z −1←1− 1 f (z)dz = Z −1 1 (−i)5 7 Y k=1 |z − zk|− 1 2dz

By (a), (b), we obtain the value Z a∗₃ 1 f (z)dz = 2 Z 1 −1 i5 7 Y k=1 |z − zk|− 1 2dz = 2 Z 1 −1 i 7 Y k=1 |z − zk|− 1 2dz M ath. = 0.0151409i (2) Consider a∗₃ in Mathematica (a) −1→ 1 :+ z − zk= −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− πi 2 = −i|z − z_k|− 1 2, k = 1, 2, 3, 4, 5 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7 Z −1→1+ f (z)dz = Z 1 −1 (−i)5 7 Y k=1 |z − zk|− 1 2dz (b) −1← 1 :− z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− πi 2 = −i|z − z k|− 1 2, k = 1, 2, 3, 4, 5 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7 Z −1←1− 1 f (z)dz = Z 1 −1 (−i)5 7 Y k=1 |z − zk|− 1 2dz

But we obtain different value in Mathematica Z

a∗₃

1

f (z)dz = 0 (3) Using lemma 1 to modify

(a) −1→ 1 : the path along x-axis from −1 to 1 on sheet-I with (+)edge+ arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2, 3, 4, 5 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 6, 7 So f (z)M ath.= −f (z)

(b) 1← −1 : the path along x-axis from 1 to -1 on sheet-I with (−)edge− arg(z − zk) = π then √ z − zk M ath. = √z − zk, k = 1, 2, 3, 4, 5 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 6, 7 So f (z)M ath.= f (z) Z a3 1 f (z)dz = Z a∗₃ 1 f (z)dz M ath. = −2 Z 1 −1 1 f (z)dz = 0.151409i Figure 36: b-cycles

4. b3 : Let b3 is a cycle which center at −2 with radius 2. We could write down the

parameter, let z = −2 + 2eiθ _{and θ ∈ [−π, 0)S[2π, 3π). Notice that f (z)|} (II) = −f (z)|(I), so we have Z b3 1 f (z)dz = Z 0 −π 2ieiθ 7 Q k=1 √ −2 + 2eiθ_{− z} k dθ − Z π 0 2ieiθ 7 Q k=1 √ −2 + 2eiθ_{− z} k dθ = −0.0765026 + 6.93889 × 10−18i

Since bk cycle is simple connected, we can use some equivalent paths, say b∗k, such

Figure 37: b3’s equivalent path b∗3

(1) Consider b∗₃ of theory in sheet-I (a) −3 → −1 z + z3 = |z + z3| then 1 √ z + z3 = |z + z3|− 1 2 z − zk = −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z k|− 1 2i, k = 1, 2, 3, 4, 5, 6 Z −3→−1 1 f (z)dz = − Z −1 −3 i6 7 Y k=1 |z − zk|− 1 2dz

(b) −1 99K −3 : the path along x-axis from -1 to -3 of sheet-II. We known that f (z)|(I) = −f (z)|(II) , so consider −1 → −3

z + z3 = |z + z3| then 1 √ z + z3 = |z + z3|− 1 2 z − zk = −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z k|− 1 2i, k = 1, 2, 3, 4, 5, 6 Z −3L99−1 1 f (z)dz = − Z −3←−1 1 f (z)dz = Z −3 −1 i6 7 Y k=1 |z − zk|− 1 2dz By (1), (2), we obtain Z b∗ 3 1 f (z)dz = 2 Z −3 −1 i6 7 Y k=1 |z − zk|− 1 2dz = −2 Z −3 −1 7 Y k=1 |z − zk|− 1 2dz = −0.0765026 (2) Consider b∗₃ in Mathematica

(a) −3 → −1 z + z3 = |z + z3| then 1 √ z + z3 = |z + z3|− 1 2 z − zk= −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z k|− 1 2, k = 1, 2, 3, 4, 5, 6 Z −3→−1 1 f (z)dz = Z −1 −3 (−i)6 7 Y k=1 |z − zk|− 1 2dz = − Z −1 −3 7 Y k=1 |z − zk|− 1 2dz

(b) −1 99K −3 : the path along x-axis from -1 to -3 on sheet-II z + z3 = |z + z3| then 1 √ z + z3 = |z + z3|− 1 2 z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2eπi = |z − z k|− 1 2i, k = 1, 2, 3, 4, 5, 6 Z −3L99−1 1 f (z)dz = − Z −3 −1 7 Y k=1 |z − zk|− 1 2dz But in Mathematica Z b∗₃ 1 f (z)dz = 0 (3) Using Lemma1 to modify

(a) −3 → −1 : the path along x-axis from -3 to -1 of sheet-I arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2, 3, 4, 5, 6 arg(z − z7) = 0 then √ z − z7 M ath. = √z − z7 So f (z)M ath.= f (z)

(b) −1 99K −3 : the path along x-axis from -1 to -3 of sheet-II We known that f (z)|(I) = −f (z)|(II), so we consider −1 → −3

arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k=1, . . . , 6 arg(z − z7) = 0 then √ z − z7 M ath. = √z − z7 From f (z)|−1→−3 M ath. = f (z). We have f (z)|_−199K−3 = −f (z)|−1→−3 M ath. = −f (z)

Z b3 1 f (z)dz = Z b∗ 3 1 f (z)dz M ath. = 2 Z −1 −3 1 f (z)dz = −0.0765026

5. b2 : Let b2 is a cycle center at 0 with radius 7₂ . So we could write down the

parameter, let z = 7₂eiθ and θ ∈ [−π, 0)S[2π, 3π). Notice that f (z)|(II)= −f (z)|(I),

so we have Z b2 1 f (z)dz = Z 0 −π 7 2ie iθ 7 Q k=1 q 7 2eiθ− zk dθ − Z π 0 7 2ie iθ 7 Q k=1 q 7 2eiθ− zk dθ = 0.157328

Using same way in (4). Consider equivalent path b∗₂ = b∗₃S −1 → 1+ S −1 L99− 1_{S 1 → 3 S 1 L99 3}

Figure 38: The equivalent path b∗₂

(1) Consider b∗₂ of theory in sheet-I (a) −1→ 1 :+ z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7 z − zk = −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = i|z − z_k|− 1 2, k = 1, 2, 3, 4, 5 Z −1→1+ 1 f (z)dz = Z 1 −1 i5 7 Y k=1 |z − zk|− 1 2dz

(b) −1_{L99 1 ≡ −1}− ← 1 i.e. the path on horizontal cut from -1 to 1 on (−)edge+ in sheet-II equals the path on horizontal cut from −1 to 1 of (+)edge in sheet-I. So consider z ∈ −1← 1+ z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7 z − zk = −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = i|z − z_k|− 1 2, k = 1, 2, 3, 4, 5 Z −1L991− 1 f (z)dz = Z −1←1+ 1 f (z)dz = Z −1 1 i5 5 Y k=1 |z − zk|− 1 2dz (c) 1 → 3: z − zk= |z − zk| ⇒ 1 √ z − zk = |z − zk|− 1 2, k = 5, 6, 7 z − zk= −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z_k|− 1 2i, k = 1, 2, 3, 4 Z 1→3 1 f (z)dz = Z 3 1 i4 7 Y k=1 |z − zk|− 1 2dz

(d) 1 L99 3 : we known that f (z)|(II) = −f (z)|(I), so we first consider 1 ← 3

z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 5, 6, 7 z − zk= −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z k|− 1 2i, k = 1, 2, 3, 4 Z 1L993 1 f (z)dz = − Z 1←3 1 f (z)dz = − Z 1 3 47 k=1|z − zk|− 1 2dz

By (a), (b), (c) and (d), we have Z b∗₂ 1 f (z)dz = 2 Z 1 −1 i5 7 Y k=1 |z − zk|− 1 2dz + 2 Z 3 1 i4 7 Y k=1 |z − zk|− 1 2dz = 0.157328

(a) −1→ 1+ z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2, 3, 4, 5 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7 Z −1→1+ 1 f (z)dz = Z 1 −1 (−i)5 7 Y k=1 |z − zk|− 1 2dz (b) −1_{L99 1}− z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z k|− 1 2, k = 1, 2, 3, 4, 5 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 6, 7 Z −1L991− 1 f (z)dz = Z −1 1 (−i)5 5 Y k=1 |z − zk|− 1 2dz (c) 1 → 3 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 5, 6, 7 z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2, 3, 4 Z 1→3 1 f (z)dz = Z 3 1 (−i)4 7 Y k=1 |z − zk|− 1 2dz (d) 1 L99 3 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 5, 6, 7 z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2, 3, 4 Z 1L993 1 f (z)dz = Z 1 3 (−i)4 7 Y k=1 |z − zk|− 1 2dz

But in Mathematica we obtain different value Z b∗ 2 1 f (z)dz = 0 (3) Using Lemma1 to modify

(a) −1→ 1+ arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, . . . , 5 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 6, 7 So f (z)M ath.= −f (z)

(b) −1 _{L99 1 ≡ −1}− ← 1 that is the path on horizontal cut from −1 to 1 of+ (−)edge in sheet-II is equal the path on horizontal cut from −1 to 1 of (+)edge in sheet-I. So consider z ∈ −1← 1+

arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, . . . , 5 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 6, 7 So f (z)M ath.= −f (z) (c) 1 → 3 arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2, 3, 4 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 5, 6, 7 So f (z)M ath.= f (z)

(d) 1 L99 3 : we known that f (z)|(II) = −f (z)|(I), so we first consider 1 ← 3

arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2, 3, 4 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 5, 6, 7 We have f (z)|1←3 M ath. = f (z) then f (z)|_1L993 = −f (z)|1←3 M ath. = −f (z) By (1), (2), (3) and Cauchy Integral Theorem

Z b2 1 f (z)dz = Z b∗₂ 1 f (z)dz M. = −2 Z 1 −1 7 Y k=1 1 f (z)dz + 2 Z 3 1 7 Y k=1 1 f (z)dz = 0.157328

6. b1: Let b1is a cycle center at 3 with radius 3 . So we could write down the parameter,

let z = 3 + 3eiθ _{and θ ∈ [−π, 0)S[2π, 3π). Notice that f (z)|}

(II) = −f (z)|(I), so we have Z b1 1 f (z)dz = Z 0 −π 3ieiθ 7 Q k=1 √ 3 + 3eiθ_{− z} k dθ − Z π 0 3ieiθ 7 Q k=1 √ 3 + 3eiθ_{− z} k dθ = 0.0565161

Consider equivalent path b∗₁ = b∗₂S b∗ 3S 3

+

→ 4S 3L99 4− S 4 → 6 S 4 L99 6

Figure 39: The equivalent path b∗₁

(1) Analysis the integration over b∗₁ in sheet-I (a) 3→ 4:+ z − zk = −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z_k|− 1 2i, k = 1, 2, 3 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7 Z 3→4+ 1 f (z)dz = Z 3 4 i3 7 Y k=1 |z − zk|− 1 2dz

(b) 3_{L99 4 ≡ 3}− ← 4 that is the path on horizontal cut from 3 to 4 of (−)edge+ in sheet-II is equal to the path on horizontal cut from 3 to 4 of (+)edge in sheet-I. So we consider z ∈ −1← 1+ z − zk = −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z k|− 1 2i, k = 1, 2, 3 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7

Z 3L994− 1 f (z)dz = Z 3←4+ 1 f (z)dz = Z 3 4 i3 7 Y k=1 |z − zk|− 1 2dz (c) 4 → 6 z − zk= −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z k|− 1 2i, k = 1, 2 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 3, 4, 5, 6, 7 Z 4→6 1 f (z)dz = Z 6 4 i2 7 Y k=1 |z − zk|− 1 2dz

(d) 4 L99 6 : we known that f (z)|(I) = −f (z)|(II), so we first consider 4 ← 6

z − zk= −|z − zk| = |z − zk|e−πi then √ 1 z − zk = |z − zk|− 1 2e π 2i = |z − z_k|− 1 2i, k = 1, 2 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 3, 4, 5, 6, 7 Z 4L996 1 f (z)dz = − Z 4←6 1 f (z)dz = − Z 4 6 i2 7 Y k=1 |z − zk|− 1 2dz Z b∗₁ 1 f (z)dz = 2 Z −3 −1 i6 7 Y k=1 |z − zk|− 1 2dz + 2 Z 3 1 i4 7 Y k=1 |z − zk|− 1 2dz − 2 Z 6 4 i2 7 Y k=1 |z − zk|− 1 2dz (2) Consider b∗₁ in Mathematica (a) 3→ 4+ z − zk= −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z k|− 1 2, k = 1, 2, 3 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7 Z −1→1+ 1 f (z)dz = Z 1 −1 (−i)3 7 Y k=1 |z − zk|− 1 2dz

(b) 3_{L99 4}− z − zk= −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2, 3 z − zk= |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 4, 5, 6, 7 Z 3L994− 1 f (z)dz = Z 3 4 (−i)4 7 Y k=1 (−i)4|z − zk|− 1 2dz (c) 4 → 6 z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 3, 4, 5, 6, 7 Z 4→6 1 f (z)dz = Z 6 4 (−i)2 7 Y k=1 |z − zk|− 1 2dz (d) 4 L99 6 z − zk = −|z − zk| = |z − zk|eπi then √ 1 z − zk = |z − zk|− 1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2 z − zk = |z − zk| then 1 √ z − zk = |z − zk|− 1 2, k = 3, 4, 5, 6, 7 Z 4L996 1 f (z)dz = Z 4 6 (−i)2 7 Y k=1 |z − zk|− 1 2dz

(3) Using Lemma1 to modify (a) 3→ 4+ arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2, 3 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 4, 5, 6, 7 So f (z)M ath.= −f (z)

(b) 3 _{L99 4 ≡ 3}− ← 4 = the path on horizontal cut from 3 to 4 of (−)edge in+ sheet-II is equal to the path on horizontal cut from 3 to 4 of (+)edge in

sheet-I. So consider z ∈ 3← 4.+ arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2, 3 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 4, 5, 6, 7 So f (z)M ath.= −f (z) (c) 4 → 6 arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 5, 6, 7 So f (z)M ath.= f (z)

(d) 4 L99 6 : we known that f (z)|(II) = −f (z)|(I) , so we first consider 4 ← 6

arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 3, 4, 5, 6, 7 From f (z)|4←6 M ath. = f (z), we have f (z)|_4L996 = −f (z)|(4←6) M ath. = −f (z) By (1), (2), (3) and Cauchy Integral Theorem

Z b1 1 f (z)dz = Z b∗₁ 1 f (z)dz M ath. = −2 Z −1 −3 1 f (z)dz + 2 Z 3 1 1 f (z)dz − 2 Z 9 6 1 f (z)dz Discuss in general situation:

ComputeR _{f (z)}1 dz over a,b cycles for horizontal cut where f (z) = s m Q k=1 (z − zk) , zk∈ R, ∀k = 1 ∼ m and z1 > z2 > ... > zm 1. a-cycle

Figure 41: a-cycles for 2N points

There are N cuts (N-1 holes), we give that aj is a cycle center at x with radius r

enclosed [z2j, z2j−1] and doesn’t intersect with other cuts.

If z ∈ aj, let z = x + reiθ where θ ∈ [−π, π)

Z aj 1 f (z)dz = Z aj 1 s m Q k=1 (z − zk) dz (5) = Z π −π rieiθ m Q k=1 √ x + reiθ_{− z} k dθ (6) 2. Consider R_a∗ j 1 f (z)dz where a ∗

j is an equivalent path for aj and it’s from z2j to z2j−1

in (+)edge and then from z2j−1 to z2j in (−)edge.

Figure 42: a∗-cycles for 2N-1 points

Figure 43: a∗-cycles for 2N points

By Cauchy theorem, we can get that Z aj 1 f (z)dz = Z a∗_j 1 f (z)dz (7)

Similarly, we use some ideas of complex number to analysis the integrations, first. (1) z2j

+

(a) Analysis in theory

z − zk > 0 then arg(z − zk) = 0, k = 2j, 2j + 1, ..., m

z − zk < 0 then arg(z − zk) = −π, k = 1, 2, ..., 2j − 1

So we have z − zk= |z − zk|eiθk where θk = arg(z − zk) then

√ z − zk = |z − zk| 1 2eiθk2 Z z2j + →z2j−1 1 f (z)dz = Z z2j−1 zj m Y k=1 |z − zk| 1 2e−(− iθk 2 )(2j−1)dz = Z z2j−1 zj m Y k=1 |z − zk| 1 2ejπi(−i)dz = (−1)j+1i Z z2j−1 zj m Y k=1 |z − zk| 1 2dz

We can use this with Mathematica to get value, and we compare with the result below to know the difference.

(b) Analysis in Mathematica: z − zk > 0 then arg(z − zk) = 0, k = 2j, 2j + 1, ..., m z − zk < 0 then arg(z − zk) = π, k = 1, 2, ..., 2j − 1 So √ z − zk = |z − zk| 1 2eiθk2 Z z2j→z+ 2j−1 1 f (z)dz = Z z2j−1 zj m Y k=1 |z − zk| 1 2e−( iθk 2 )(2j−1)dz = Z z2j−1 zj m Y k=1 |z − zk| 1 2ejπiidz = (−1)ji Z z2j−1 zj m Y k=1 |z − zk| 1 2dz

We can find that the difference of value between theory in sheet-I and Mathe-matica. is a minus and it must be pure imaginary number.

(2) z2j −

← z2j−1: That is consider the path from z2j−1 to z2j in (−) edge

(a) In theory z − zk > 0 then arg(z − zk) = 0, k = 2j, 2j + 1, ..., m z − zk < 0 then arg(z − zk) = π, k = 1, 2, ..., 2j − 1 Z z2j − ←z2j−1 1 f (z)dz = Z z2j z2j−1 m Y k=1 |z − zk|− 1 2e−(iθk2 )(2j−1)dz = Z z2j zj−1 m Y k=1 |z − zk|− 1 2ejπiidz

(b) In Mathematica, same as above

But if we can modify the computation it will more quick and easier. (3) Using Lemma 1 to modify the computation.

(a) z2j + → z2j−1 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk , k = 2j, 2j + 1, . . . , m arg(z − zk) = −π then √ z − zk M ath. = −√z − zk , k = 1, 2, . . . , 2j − 1 So f (z)M ath.= (−1)2j−1f (z) = −f (z) Z z2j + →z2j−1 1 f (z)dz M ath. = Z z2j−1 z2j 1 (−1)2j−1_{f (z)}dz = − Z z2j−1 z2j 1 f (z)dz (b) z2j − ← z2j−1 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk , k = 2j, 2j + 1, ..., m arg(z − zk) = π then √ z − zk M ath. = √z − zk , k = 1, 2j, ..., 2j − 1 So f (z) M ath.= f (z) Z z2j + →z2j−1 1 f (z)dz M ath. = Z z2j−1 z2j 1 f (z)dz Conclusion: We obtain Z a∗_j 1 f (z)dz = −2ie jπi Z z2j−1 zj m Y k=1 |z − zk|− 1 2dz (8) M ath. = −2 Z z2j−1 z2j 1 f (z)dz (9)

3. b-cycles

Figure 44: bj-cycle for 2N-1 points

Give bj is a circle centered at x with radius r and enclosed the [z2N −1, z2j] and

intersect at the points on [z2j, z2j−1] and [z2N −1, z2N].

Figure 45: bj-cycle for 2N points

Give bj is a circle centered at x with radius r and enclosed the [z2N −1, z2j] and

intersect at the points on [z2j, z2j−1] and [z2N −1, ∞). If z ∈ bj, z = x + reiθ where

θ ∈ [−π, 0)S[2π, 3π). From f (z)|(II)= −f (z)|(I) Z bj 1 f (z)dz = Z 0 −π rieiθ m Q k=1 √ x + reiθ_{− z} k dθ + Z 3π 2π rieiθ m Q k=1 √ x + reiθ_{− z} k dθ (10) = Z 0 −π rieiθ m Q k=1 √ x + reiθ_{− z} k dθ − Z π 0 rieiθ m Q k=1 √ x + reiθ_{− z} k dθ (11)

4. The equivalent path b∗_j :

Figure 47: b∗_j-cycle for 2N points

From Cauchy Theorem, we have Z bj 1 f (z)dz = Z b∗ j 1 f (z)dz (12)

where b∗_j is a path from zm to z2j in sheet-I and then from z2j to zm in sheet-II

(1) the path on the cut that is the path from z2s+2 to z2s+1, s = j, j + 1, ..., N − 2

on (+)edge in sheet-I and the path from z2s+1 to z2s+2, s = j, j + 1, · · · , N − 2

on (−)edge in sheet-II. (a) In theory

(i) z2s+2 +

→ z2s+1:

z − zk> 0 then arg(z − zk) = 0 then arg(

√

z − zk) = 0

then √z − zk= |z − zk|

1

2, k = 2s + 2, 2s + 3, · · · , m

z − zk< 0 then arg(z − zk) = −π then arg(

√ z − zk) = −π₂ then √z − zk= |z − zk| 1 2e− π 2i = −i|z − z_k| 1 2, k = 1, 2, · · · , 2s + 1 So we have f (z) = i2s+1 m Y k=1 |z − zk|− 1 2 (ii) z2s+2 −

L99 z2s+1 in (−) edge of sheet-II is as same as in (+)edge of

sheet-I, so consider z2s+2 +

→ z2s+1.

z − zk> 0 then arg(z − zk) = 0 ⇒ arg(

√

z − zk) = 0

then √z − zk= |z − zk|

1

2, k = 2s + 2, 2s + 1, ..., m

z − zk< 0 then arg(z − zk) = −π then arg(

√ z − zk) = −π₂ then √z − zk= |z − zk| 1 2e− π 2i = −i|z − z_k| 1 2, k = 1, 2, ..., 2s + 1

We found that same as above f (z) = i2s+1 m Y k=1 |z − zk|− 1 2

(b) In Mathematica: (i) z2s+2

+

→ z2s+1

z − zk> 0 then arg(z − zk) = 0 then arg(

√

z − zk) = 0

then √z − zk= |z − zk|

1

2 , k = 2s + 2, · · · , m

z − zk< 0 then arg(z − zk) = π then arg(

√ z − zk) = π₂ then √z − zk= |z − zk| 1 2e π 2i = i|z − z_k| 1 2 , k = 1, 2, · · · , 2s + 1 We have f (z) = (−i)2s+1 m Y k=1 |z − zk|− 1 2 (ii) z2s+2 − L99 z2s+1: z2s+2 + → z2s+1.

z − zk> 0 then arg(z − zk) = 0 then arg(

√

z − zk) = 0

then √z − zk = |z − zk|

1

2 , k = 2s + 2, 2s + 1, ..., m

z − zk< 0 then arg(z − zk) = π then arg(

√ z − zk) = π₂ then √z − zk = |z − zk| 1 2e π 2i = i|z − z_k| 1 2, k = 1, 2, ..., 2s + 1

So we obtain same value of f (z) as (i), but different with in theory f (z) = (−i)2s+1 m Y k=1 |z − zk|− 1 2

(c) Using Lemma 1 to modify the computation. (i) z ∈ z2s+2 + → z2s+1: arg(z − zk) = 0 then √ z − zk M ath. = √z − zk , k = 2s + 2, ..., m arg(z − zk) = −π then √ z − zk M ath. = −√z − zk , k = 1, 2, ..., 2s + 1 We obtain f (z)M ath.= (−1)2s+1f (z) = −f (z) (ii) z2s+2 +

→ z2s+1: in (−) edge of sheet-II is as same as in (+)edge of

sheet-I, z2s+2 + → z2s+1 arg(z − zk) = 0 then √ z − zk M ath. = √z − zk , k = 2s + 2, ..., m arg(z − zk) = −π then √ z − zk M ath. = −√z − zk , k = 1, 2, ..., 2s + 1 We have f (z)M ath.= (−1)2s+1f (z) = −f (z)

(2) In no cuts that is the path from z2s+1 to z2s, s = j, j + 1, ..., N − 2 in sheet-I

and the path from z2s to z2s+1, s = j, j + 1, ..., N − 2 in sheet-II.

(a) In theory: (i) z2s+1 → z2s:

z − zk> 0 then arg(z − zk) = 0 then arg(

√

z − zk) = 0

then √z − zk = |z − zk|

1

2, k = 2s + 1, . . . , m

z − zk< 0 then arg(z − zk) = −π then arg(

√ z − zk) = −π₂ then √z − zk = |z − zk| 1 2e− π 2i = −i|z − z_k| 1 2 , k = 1, 2, . . . , 2s Then we have f (z) = i2s m Y k=1 |z − zk|− 1 2 = (−1)s m Y k=1 |z − zk|− 1 2

(ii) The path z2s+1 L99 z2s is in sheet-II. We known that f (z)|(I) =

−f (z)|(II), so we first consider z2s+1 ← z2s

z − zk> 0 then arg(z − zk) = 0 then arg(

√

z − zk) = 0

then √z − zk = |z − zk|

1

2, k = 2s + 1, . . . , m

z − zk< 0 then arg(z − zk) = −π then arg(

√ z − zk) = −π₂ then √z − zk = |z − zk| 1 2e− π 2i = −i|z − z_k| 1 2 , k = 1, 2, . . . , 2s We have f (z)|z2s+1←z2s = i 2s Qm k=1 |z − zk|− 1 2 = (−1)s m Q k=1 |z − zk|− 1 2. So f (z)|z2s+1L99z2s = (−1) s+1 m Y k=1 |z − zk|− 1 2 (b) In Mathematica: (i) z2s+1 → z2s:

z − zk> 0 then arg(z − zk) = 0 then arg(

√

z − zk) = 0

then √z − zk = |z − zk|

1

2 , k = 2s + 1, . . . , m

z − zk< 0 then arg(z − zk) = π then arg(

√ z − zk) = π₂ then √1 z−zk = |z − zk| −1 2e− π 2i = −i|z − z_k|− 1 2, k = 1, 2, ..., 2s So f (z) = (−i)2s m Y k=1 |z − zk|− 1 2 = (−1)s m Y k=1 |z − zk|− 1 2

(ii) z2s+1 L99 z2s:

z − zk> 0 then arg(z − zk) = 0 then arg(

√

z − zk) = 0

then √z − zk = |z − zk|

1

2 , k = 2s + 1, ..., m

z − zk< 0 then arg(z − zk) = π then arg(

√ z − zk) = π₂ then √z − zk = |z − zk| 1 2e π 2i = i|z − z_k| 1 2 , k = 1, 2, ..., 2s

So we have same value of f (z) as (i), but different with in theory

f (z) = (−i)2s m Y k=1 |z − zk|− 1 2 = (−1)s m Y k=1 |z − zk|− 1 2

(c) Using Lemma 1 to modify the computation in Mathematica: (i) If z ∈ z2s+1 → z2s: arg(z − zk) = 0 then √ z − zk M ath. = √z − zk, k = 2s + 1, ..., m arg(z − zk) = −π then √ z − zk M ath. = −√z − zk, k = 1, 2, ..., 2s So f (z)M ath.= (−1)2sf (z) M ath.= f (z)

(ii) If z2s+1 L99 z2s , since f (z)|(II) = −f (z)|(I) consider z2s+1 ← z2s

arg(z − zk) = 0 ⇒ √ z − zk M ath. = √z − zk, k = 2s + 1, ..., m arg(z − zk) = −π ⇒ √ z − zk M ath. = −√z − zk, k = 1, 2, ..., 2s So if z ∈ z2s+1 L99 z2s f (z)M ath.= −(−1)2sf (z) M ath.= −f (z) Z b∗_j 1 f (z)dz = N −1 X s=j [(−1)s2 Z z2s z2s+1 m Y k=1 |z − zk|− 1 2dz] (13) M ath. = N −1 X s=j (2 Z z2s z2s+1 1 f (z)dz) (14)

3

The integrals of

_{f (z)}1

over a,b cycles for vertical cut

After knowing the integrals in horizontal cut, we will discuss the integrals for vertical cuts. In this case, we define that

z − zk =  reiθ_{, θ ∈ [−}3π 2 , π 2) iff z in sheet-I reiθ_{, θ ∈ [}π 2, 5π 2 ) iff z in sheet-II (15)

the cut in each sheet has two edges, label the starting edge with ”+” and the terming edge with ”−” and zk is the end point of the vertical cut.

Analysis the value of f(z) in sheet-I and sheet-II of theory. Example f (z) =√z. If z = ri ⊂ sheet-I z = |z|eiθ, θ ∈ [−3π 2 , π 2) then √ z = |z|12e iθ 2,θ 2 ∈ [− 3π 4 , π 4) (16) If z = ri ⊂ sheet-II z = |z|eiθ, θ ∈ [π 2, 5π 2 ) then √ z = |z|12e iθ 2,θ 2 ∈ [ π 4, 5π 4 ) (17) Figure 48: Example of f (z) =√z

If f (z) = s n Q k=1 (z − zk), then n Y k=1 (z − zk) = reiθ1, θ1 ∈ [− 3π 2 , π 2) in sheet-I n Y k=1 (z − zk) = reiθ2, θ2 ∈ [ π 2, 5π 2 ) in sheet-II From the idea of definition, reiθ1 _{= re}iθ2 _{and θ}

2 = θ1+ 2π. f (z)|(II)= √ reθ22 i = √ reθ1+2π2 i = √ reθ12ieπi = −f (z)|_(I) (18)

Discuss the difference between the value in theory and in Mathematica and find out how to modify the computation.

Figure 49: The value in sheet-I and Mathematica of √z

So we need to modify the computation in Mathematica s.t. the numerical result of Mathematica is identical to the numerical result of theory when θ ∈ [−3π₂ , −π].

Lemma 2. When z in sheet-I for vertical cut whose one of the end points is zk

√ z − zk M ath. =  − √ z − zk if arg(z − zk) ∈ [−3π₂ , −π], √ z − zk if arg(z − zk) ∈ (−π,π₂) Proof.

Let z in sheet-I and using polar form z − zk= reiθ. When θ ∈ (−π,π₂), the argument

θ into θ + 2π ∈ [π₂, π] where θ + 2π ∈ [π₂, π] and reθi_{= re}θi+2πi_{, but} In theory: √z − zk = √ reθ2i In Mathematica: √z − zk = √ reθ+2π2 i = − √ reθ2 So if θ ∈ [−3π₂ , −π] √ z − zk M ath. = −√z − zk

As same as horizontal cut. We first discuss the difference between the value in theory and the value in Mathematica. Compare their sign(f) is different or not? Using statement before about modify and get value, the result will be the same or not?

Example: The integrals of _{f (z)}1 over a,b cycles for vertical cut where f (z) = p(z − i)(z − 2i)(z − 3i)(z − 4i)(z − 5i)(z − 6i). Let f(z) =

6 Q k=1 √ z − zk where zk= ki, k = 1, 2, 3, 4, 5, 6

Figure 50: a and its equivalent path a∗

1. Compute R

a∗₁ 1

f (z)dz where a ∗

1 is equivalent path for a1 and a∗1 = the path along

vertical cut from i to 2i on (+)edge of sheet-I (called a∗₁₁) and then back from 2i to i on (−)edge of sheet-I (a∗₁₂)

(a) Analysis in theory:

Since z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − i) = −3₂π then arg√z − i = −3₄π

arg(z − ki) = −π₂ then arg√z − ki = −π₄, k = 2, . . . , 6

f (z) = ( 6 Y k=1 |z − ki|−12)e− 3 4π(e− 1 4π)5 = ( 6 Y k=1 |z − ki|−12)e− 8 4π = 6 Y k=1 |z − ki|−12 = R (Let 6 Q k=1 |z − ki|−1 2 = R)

(b) Analysis in mathematica (no matter in which sheet): Since z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki). arg(z − i) = π

2 then arg

√

z − i = π 4

arg(z − ki) = −π₂ then arg√z − ki = −π₄, k=2,3,4,5,6

f (z) = 6 Y k=1 |z − ki|−12e 1 4π(e− π 4)5 = 6 Y k=1 |z − ki|−12e− 4 4π = − 6 Y k=1 |z − ki|−12 = −R

Compare with (a) and (b), we find that when you want to get true value, the value which we get from Mathematica should multiply −1 that is sign(f (z)|(I))=-sign(f (z)|M ath.)

(c) Using the Lemma 2 to modify:

arg(z − i) = −3₂π than√z − iM ath.= −√z − i

arg(z − ki) = −π₂ than √z − kiM ath.= √z − ki, k=2, 3, 4, 5, 6

So f (z) M ath.= −f (z), the same result as above difference between theory and Mathematica, the difference is a minus.

(2) a∗₁₂ : Let z = ri, r : 1 → 2 and then dz = idr (a) Analysis in theory:

arg(z − i) = π₂ than arg(√z − i) = π₄

arg(z − ki) = −π₂ than arg(√z − ki) = −π₄, k=2,3,4,5,6

f (z) = 6 Y k=1 |z − ki|−12e π 4(e− π 4)5 = 6 Y k=1 |z − ki|−12e− 4 4π = − 6 Y k=1 |z − ki|−1₂

(b) Analysis in Mathematica (no matter in which sheet): Since z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − i) = π₂ than arg(√z − i) = π₄

arg(z − ki) = −π₂ than arg(√z − ki) = −π₄, k=2,3,4,5,6

f (z) = 6 Y k=1 |z − ki|−1_2eπ_4(e−π₄₎5 ₌ 6 Y k=1 |z − ki|−1_2e−4₄π = − 6 Y k=1 |z − ki|−12

Compare with (a) and (b) we find the value is same (c) Using Lemma 2 to modify:

arg(z − i) = π₂ than √z − iM ath.= √z − i arg(z − ki) = −π

2 than

√

z − kiM ath.= √z − ki, k = 2, 3, 4, 5, 6 Here we obtain f (z)M ath.= f (z), the same result as above.

Z a1 1 f (z)dz = Z a∗ 1 1 f (z)dz = −2 Z 2 1 i 6 Y k=1 |ri − ki|−12dr M ath. = 2 Z 2 1 i f (ri)dr = 0. + 0.871563i 2. Compute R_a∗ 2 1 f (z)dz where a ∗

2 is equivalent path for a2 and a∗2 = the path along

vertical cut from 4i to 3i on (+)edge of sheet-I (called a∗₂₁) and then back from 3i to 4i on (−)edge of sheet-I (a∗₂₂)

(1) a∗₂₁ : Let z = ri, r : 4 → 3, dz = idr

(a) Analysis in theory:

Since z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − ki) = −3₂π than arg(√z − ki) = −3₄π, k = 1, 2, 3 arg(z − ki) = −π₂ than arg(√z − ki) = −π₄, k = 4, 5, 6

f (z) = 6 Y k=1 |z − ki|−12(e− 3 4π)3(e− π 4)3 = 6 Y k=1 |z − ki|−12e− 12 4π = − 6 Y k=1 |z − ki|−12

(b) analysis in Mathematica (no matter in which sheet): Since z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − ki) = π₂ than arg(√z − i) = π₄, k=1,2,3 arg(z − ki) = −π₂ than arg(√z − ki) = −π₄, k=4,5,6

f (z) = 6 Y k=1 |z − ki|−1₂ (eπ4)3(e− π 4)3 = 6 Y k=1 |z − ki|−1₂

Compare with (a) and (b) we find that when you want to obtain true value, the value which we have from Mathematica should multiply −1, sign(f (z)|(I)) = −sign(f (z)|M athewatica)

(c) Using Lemma 2 to modify:

arg(z − i) = −3₂π than√z − iM ath.= −√z − i , k = 1, 2, 3 arg(z − ki) = −π₂ than √z − kiM ath.= √z − ki , k = 4, 5, 6

f (z) M ath.= −f (z)

same as the above result (2) a∗₂₂ : Let z = ri, r : 1 → 2, dz = idr

(a) Analysis in theory:

Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki). arg(z − ki) = π₂ than arg(√z − ki) = π₄, k=1,2,3

arg(z − ki) = −π₂ than arg(√z − ki) = −π₄, k=4,5,6 f (z) = 6 Y k=1 |z − ki|−12(e π 4)3(e− π 4) = 6 Y k=1 |z − ki|−12

(b) Analysis in Mathematica (no matter in which sheet):

Using polar form z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − ki) = π₂ than arg(√z − ki) = π₄, k=1,2,3 arg(z − ki) = −π₂ than arg(√z − ki) = −π₄, k=4,5,6

f (z) = 6 Y k=1 |z − ki|−12(e π 4)3(e− π 4)3 = 6 Y k=1 |z − ki|−12

Compare with (a) and (b) we find the value is same (c) Using Lemma 2 to modify:

arg(z − i) = π₂ then √z − iM ath.= √z − i, k=1,2,3 arg(z − ki) = −π₂ then √z − kiM ath.= √z − ki, k=4,5,6

f (z) M ath.= f (z)

same as the above result

Z a2 1 f (z)dz = Z a∗2 1 f (z)dz = 2 Z 4 3 i 6 Y k=1 |ri − ki|−12dr M ath. = 2 Z 4 3 i f (ri)dr = 0. + 1.74313i 3. Compute R_b∗ 2 1 f (z)dz where b ∗

2 is an equivalent path for b2 and b∗2 = the path along

vertical line from 5i to 4i on sheet-I (called b∗₂₁) and then back from 5i to 4i on sheet-II (b∗₂₂)

Figure 51: b and b∗

(a) Analysis in theory:

Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki). arg(z − ki) = −3

2π then arg(

√

z − ki) = −3

4π, k=1,2,3,4

arg(z − ki) = −π₂ then arg(√z − ki) = −π₄, k=5,6

f (z) = 6 Y k=1 |z − ki|−12(e− 3 4π)4(e− π 4)2 = 6 Y k=1 |z − ki|−12e− 14 4π = i 6 Y k=1 |z − ki|−12

(b) Analysis of Mathematica (no matter in which sheet):

Using polar form z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − ki) = π₂ then arg(√z − ki) = π₄, k=1,2,3,4 arg(z − ki) = −π₂ then arg(√z − ki) = −π₄, k=5,6

f (z) = 6 Y k=1 |z − ki|−12(e− π 4)2(e π 4)4 = 6 Y k=1 |z − ki|−12e 2 4π = i 6 Y k=1 |z − ki|−1₂

(c) Using Lemma 2 to modify:

arg(z − i) = π₂ then √z − iM ath.= √z − i, k=1, 2, 3, 4 arg(z − ki) = −π₂ then √z − kiM ath.= √z − ki, k=5, 6

f (z) M ath.= f (z) same as the above result

(2) b∗₂₂: We known that f (z)|(I) = −f (z)|(II) , so we can consider b∗∗22 is the path

along vertical line from 4i to 5i on sheet-I. Let z = ri, r : 4 → 5 then dz = idr

(a) Analysis in theory:

Using polar form z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − ki) = −3₂π then arg(√z − ki) = −3₄π, k=1,2,3,4 arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π, k=5,6

f (z) = 6 Y k=1 |z − ki|−12(e− 3 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e− 14 4π = i 6 Y k=1 |z − ki|−12 So f (z)|b∗ 22 = −f (z)|b∗∗22 = −i 6 Y k=1 |z − ki|−12

(b) Analysis in Mathematica (no matter in which sheet):

Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki). arg(z − ki) = 1₂π then arg(√z − ki) = 1₄π, k=1,2,3,4

arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π, k=5,6

f (z) = 6 Y k=1 |z − ki|−12(e 1 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e 2 4π = i 6 Y k=1 |z − ki|−12

Compare with (a) and (b) we find when you want get true value, the value which we obtain from Mathematica should multiply −1

(c) Using Lemma 2 to modify:

arg(z − i) = 1₂π then√z − iM ath.= √z − i, k=1,2,3,4 arg(z − ki) = −1₂π then √z − kiM ath.= √z − ki, k=5,6 f (z)|b∗∗₂₂ M ath. = f (z) then f (z)|b∗₂₂ = −f (z)|b∗∗₂₂ M ath. = −f (z) same as the above result

Z b2 1 f (z)dz = Z b∗₂ 1 f (z)dz = −2 Z 5 4 i 6 Y k=1 |z − ki|−12dr M ath. = −2 Z 5 4 i f (ri)dr = −1.48065 4. ComputeR b∗₁ 1 f (z)dz where b ∗

1is equivalent path for b1and b∗1 = b ∗ 2S b ∗ 11S b ∗ 13S b ∗ 14S b ∗ 12

where b∗₁₁= the path along vertical cut from 4i to 3i on (+)edge of sheet-I b∗₁₂ = the path along vertical cut from 3i to 4i on (−)edge of sheet-II, b∗₁₃ = the path along vertical line from 3i to 2i on sheet-I, b∗₁₄= the path along vertical line from 2i to 3i on sheet-II.

(1) b∗₁₁ ≡ a∗

21 : Done

(2) b∗₁₂ ≡ the path along vertical cut from 3i to 4i on (+)edge of sheet-I Let z = ri, r : 3 → 4, so dz = idr

(a) In theory:

Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki). arg(z − ki) = −3

2π then arg(

√

z − ki) = −3

4π, k=1,2

arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π, k=3,4,5,6

f (z) = 6 Y k=1 |z − ki|−12(e− 3 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e− 10 4π = −i 6 Y k=1 |z − ki|−12

(b) In Mathematica (no matter in which sheet):

Using polar form z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − ki) = 1₂π then arg(√z − ki) = 1₄π, k=1,2

z − i = −|z − ki|i arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π, k=3,4,5,6

f (z) = 6 Y k=1 |z − ki|−12(e− 1 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e− 2 4π = −i 6 Y k=1 |z − ki|−12

Compare with (a) and (b) we find that the value is same (c) Using Lemma 2 to modify:

arg(z − ki) = 1₂π then √z − kiM ath.= √z − ki, k=1,2 arg(z − ki) = −1₂π then √z − kiM ath.= √z − ki, k=3,4,5,6

f (z) M ath.= f (z)

same as the above result

(3) b∗₁₃: Let z = ri, r : 3 → 2, so dz = idr

(a) In theory:

Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki). arg(z − ki) = −3₂π then arg(√z − ki) = −3₄π, k=1,2

arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π, k=3,4,5,6

f (z) = 6 Y k=1 |z − ki|−12(e− 3 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e− 10 4π = −i 6 Y k=1 |z − ki|−12 (b) In Mathematica:

Using polar form z − ki = |z − ki|earg(z−ki)i_{, consider arg(z − ki).}

arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π, k=3,4,5,6 f (z) = 6 Y k=1 |z − ki|−12(e− 1 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e− 2 4π = −i 6 Y k=1 |z − ki|−12

Compare with (a) and (b) we find the value is same (c) Using Lemma 2 to modify:

arg(z − i) = 1₂π then√z − iM ath.= √z − i, k=1,2

arg(z − ki) = −1₂π then √z − kiM ath.= √z − ki, k=3,4,5,6

f (z) M ath.= f (z)

same as the above result

(4) b∗₁₄ : We known that f (z)|(I) = −f (z)|(II) , so we can consider b∗∗14= the path

along vertical line from 2i to 3i on sheet-I. Let z = ri, r : 2 → 3, so dz = idr

(a) Analysis in theory:

Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki). arg(z − ki) = 1₂π then arg(√z − ki) = 1₄π, k=1,2

arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π , k=3,4,5,6

f (z) = 6 Y k=1 |z − ki|−12(e 1 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e− 10 4π = −i 6 Y k=1 |z − ki|−12 So f (z)|b∗ 14 = −f (z)|b∗∗14 = i 6 Y k=1 |z − ki|−12

(b) In Mathematica (no matter in which sheet):

arg(z − ki) = 1₂π then arg(√z − ki) = 1₄π, k=1,2

arg(z − ki) = −1₂π then arg(√z − ki) = −1₄π, k=3,4,5,6

f (z) = 6 Y k=1 |z − ki|−12(e− 1 4π)4(e− 1 4π)2 = 6 Y k=1 |z − ki|−12e− 2 4π = −i 6 Y k=1 |z − ki|−1₂

Compare with (a) and (b) we find when you want get true value, the value which we got from Mathematica should multiply -1

(c) Using Lemma 2 to modify:

arg(z − i) = 1₂π then√z − iM ath.= √z − i, k=1,2

arg(z − ki) = −1₂π then √z − kiM ath.= √z − ki, k=3,4,5,6 f (z)|b∗∗ 14 M ath. = f (z) then f (z)|b∗ 14 = −f (z)|b∗∗14 M ath. = −f (z) same as the above result

By (1), (2), (3), (4) and Cauchy Integral Theorem Z b1 1 f (z)dz = Z b∗ 1 1 f (z)dz = −2i Z 5 4 6 Y k=1 |z − ki|−12dr − 2i Z 3 2 6 Y k=1 |z − ki|−12dr M ath. = −2 Z 5 4 i f (ri)dr − 2 Z 3 2 i f (ri)dr = −2.9613

We find that when compute the integral by Mathematica, the difference of integral form between theory and Mathematica can help us know how to modify. We also can use angle to decide how to modify. It will the same so we only use angle to decide how to modify below.

In general situation: Compute R _{f (z)}1 dz over a∗, b∗ for vertical cut where f (z) = s

m

Q

k=1

Figure 52: a-cycle and their equivalent path a∗

1. a-cycles: aj cycle is a cycle center at x with radius r enclosed [z2j, z2j−1] and doesn’t

intersect with other cuts. R_a

j 1 f (z)dz = R a∗ j 1

f (z)dz in sheet-I. The equivalent path

a∗_j =the path on a vertical cut from z2j to z2j−1 with (+)edge of sheet-I and then

from z2j−1 to z2j with (−)edge of sheet-I.

Using lemma to modify: (a) z ∈ z2j

+

→ z2j−1: Let z = ri, r : Im(z2j) → Im(z2j−1), so dz = idr

arg(z − zk) = −3π₂ then √ z − zk M ath. = −√z − zk, k = 1, 2, 3, ...2j − 1 arg(z − zk) = −π₂ then √ z − zk M ath. = √z − zk, k = 2j, 2j + 1, ..., m f (z) M ath.= (−1)2j−1f (z)M.= −f (z) (b) z ∈ z2j−1 −

→ z2j : Let z = ri, r = Im(z2j) → Im(z2j−1), so dz = idr

arg(z − zk) = −π₂ then √ z − zk M ath. = √z − zk,k=1,2,3,...2j-1 arg(z − zk) = −π₂ then √ z − zk M ath. = √z − zk, k = 2j, 2j + 1, ..., m f (z)M ath.= f (z)

By (a),(b) and Cauchy Theory: Z aj 1 f (z)dz = Z a∗ j 1 f (z)dz M ath. = 2 Z Im(z2j) Im(z2j−1) m Y k=1 1 √ z − zk idr (19) 2. b-cycles :

Figure 53: bj and b∗j of 2N − 1 points and 2N points

bj is a circle centered at x with radius r and enclosed the [z2N −1, z2j] and intersect at

the points on [z2j, z2j−1] and [z2N −1, z2N] or bj is a circle centered at x with radius r

and enclosed the [z2N −1, z2j] and intersect at the points on [z2j, z2j−1] and [z2N −1, ∞)

By Cauchy Theorem, we know that Z bj 1 f (z)dz = Z b∗ j 1 f (z)dz (20)

Equivalent path b∗_j is a path from zm to z2j in sheet-I and then from z2j to zm in