N相黎曼面上的路徑積分及微分方程上之應用

國 立 交 通 大 學

應 用 數 學 系

碩 士 論 文

N 相黎曼面上的路徑積分及微分方程上

之應用

Path Integrals on Riemann Surfaces of Genus N and

Its Applications on Differential Equations

研 究 生 ： 施 建 興

指 導 教 授 ： 李 榮 耀

N 相黎曼面上的路徑積分及微分方程上

之應用

Path Integrals on Riemann Surfaces of Genus N and Its

Applications on Differential Equations

研 究 生 : 施建興

Student

: Chien-Hsin

Shih

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 Reuqirements

for the Degree of Master

in

Applied Mathematics

July 2011

Hsinchu, Taiwan, Republic of China

N 相黎曼面上的路徑積分及微分方程上

之應用

研究生 : 施建興

指導教授 : 李榮耀

國立交通大學應用數學系研究所碩士班

摘要

假設

是一個

的多項式函數且

。 在

上是一個多值函數。在 ex

上我們利用適合的

建立 的

。則 是一個定義在

上的

單值函數。接著我們在 的代數結構上面做積分的運算。特別地，我們主要

針對兩種特別的路徑來積分，分別為

及

。運用

來計算這些積分。此外，我們將以上的方法應用在

微分方程上。

Path Integrals on Riemann Surfaces of Genus N and

Its Applications on Differential Equations

Author : Chien-Hsin Shih

Advisor : Jong-Eao Lee

Department of Applied Mathematics

National Chiao Tung University

Abstract

Let PN(u) be a polynomial of u and let f (u) =pPN(u). f is a 2-valued function defined

on the complex plane C. We construct the Riemann surface R by a proper cut-structure on the extended complex plane. Then f is a single-valued function on R. Then we do evaluations of path integrals on R with its algebraic structure for f . In particular, we evaluate integrals along two special paths, a − cycle and b − cycle, respectively. We apply the principle of deformation of paths to evaluate those integrals. Furthermore, we apply the above argument to differential equations.

誌 謝

終於卸下心中那重重的大石頭了，能完成此論文比什麼都好。當然，

首要感謝我的指導教授李榮耀教授，從撰寫論文到學位考試期間給了很大

的協助以及鼓勵，在最後口試前的緊要關頭，製作投影片上，給了對我非

常有用的意見，口試結束後也不忘再次向我提醒往後人生旅途路上需要改

進、需要注意的事項。還有也要感謝同學們的關心，在研究所求學期間能

夠交到你們這群好同學，也是未來一生中的好朋友，有你們的陪伴真的很

好。最後，一定要感謝我偉大的父親及母親，從小辛苦的栽培，雖然他們

所擁有學歷不高，但卻辛苦的賺錢，提供了對我而言很好的環境，從小時

候用心，細心的照顧，到長大後的關心，都讓我感到溫暖的愛。謹此獻給

世上最愛我、最關心我的父母親。

Contents

1 Introduction 1

1.1 Motivation . . . 1

1.2 Stereographic Projection . . . 2

1.3 Some Basic Definitions . . . 2

2 Riemann Surfaces and Cut Structures 4 2.1 The Riemann Surface for f (z) =√z . . . 4

2.2 The Riemann Surface for f (z) =p(z − r1)(z − r2) . . . 6

2.3 The Riemann Surface for f (z) =p(z − r1)(z − r2)(z − r3) . . . 7

2.4 Riemann Surfaces of Genus N . . . 9

2.5 To Draw Paths on Cut Planes and on Riemann Surfaces . . . 10

3 Integrals for Horizontal Cuts 13 3.1 Two Examples . . . 13

3.2 The Problem in Using Mathematica . . . 18

3.3 Evaluating Integrals Using Mathematica . . . 20

3.4 Generalization of Integrals Along Horizontal Cuts . . . 46

4 Integrals for Vertical Cuts 60 4.1 Cut Structures for Vertical Cuts . . . 60

4.2 The Problem in Using Mathematica . . . 60

4.3 Evaluating Integrals Using Mathematica . . . 62

4.4 Generalization of Integrals Along Vertical Cuts . . . 77

5 Integrals for Slant Cuts 94 5.1 Cut Structures for Slant Cuts . . . 94

5.2 The Problem in Using Mathematica . . . 95

5.3 Evaluating Integrals Using Mathematica . . . 97

5.4 Generalization of Integrals Along Slant Cuts . . . 107 6 An Application on Differential Equations 125 7 Conclusion 132

1

Introduction

1.1

Motivation

Let u be a twice differentiable function of t. Consider the following differential equation u00+ f (u) = 0, (1) where f is a polynomial of u. Multiply the equation by u0,

u00u0+ f (u)u0 = 0, (2) and integrate it using change of variables, we obtain

Z

u00(t)u0(t) dt + Z

f (u)u0(t) dt = E, where E is a constant. (3) =⇒ Z u0(t) du0(t) + Z f (u) du(t) = E, (4) =⇒ 1 2[u 0

(t)]2+ F (u(t)) = E, where F is an antiderivative of f . (5) =⇒ u0(t) = ±p2[E − F (u)] (6) We obtain the first order differential equation

du dt = u 0 (t) =p2[E − F (u)] , (7) or Z ₁ p2[E − F (u)] du = Z dt. (8) Since 2[E − F (u)] is a polynomial of u, it can be written as

2[E − F (u)] = (u − u1)(u − u2) . . . (u − un)

=

n

Y

k=1

(u − uk),

where u1, . . . , un are the complex roots of the equation 2[E − F (u)] = 0. Thus, equation

(8) can be written as Z ₁ pQn k=1(u − uk) du = Z dt (9) In order to solve for u, we need to evaluate the term

Z ₁ pQn

k=1(u − uk)

du.

In the denominator, the uk,s are possibly complex numbers, and they are also the branch

points or poles. Here, u : C → C, we will see later that the integrand is in fact a multiple-valued function. It is not so easy to evaluate the integral.

In this thesis, we will use the Riemann’s approach to evaluate the integrals of this kind. In addition, we will also discuss how to compute the integrals using the computer

1.2

Stereographic Projection

In this section, we give a short introduction to the concept of stereographic projection. The complex plane together with the point at infinity ∞ is called the extended complex plane or the extended z-plane. One can think of the complex plane as passing through the equator of a unit sphere centered at the point z = 0. To each point z in the plane there corresponds exactly one point P on the surface of the sphere. The point P is determined by the intersection of the line through the point z and the north pole N of the sphere with that surface. In like manner, to each point P on the surface of the sphere, other than the north pole N , there correspondings exactly one point z in the plane. By letting the point N of the sphere corresponds to the point ∞, we obtain a one to one corespondence between the points of the sphere and the points of the extended complex plane. The sphere is known as the Riemann sphere, and the correspondence is called a stereographic projection.

Figure 1

1.3

Some Basic Definitions

Definition 1. A function f of the complex variable z is analytic in an open set if it has a derivative at each point in that set.

Definition 2. A branch of a multiple-valued function f is any single-valued funtion F that is analytic in some domain at each point z of which the value F (z) is one of the values f (z).

Definition 3. A branch cut is a portion of a line or curve that is introduced in order to define a branch F of a multiple-valued function f . Any point that is common to all branch cuts of f is called a branch point.

Definition 4. A set of points z = (x, y) in the complex plane is said to be an arc if x = x(t), y = y(t), a ≤ t ≤ b,

It is convenient to describe the points of an arc C by means of the equation z = z(t), a ≤ t ≤ b,

where

z(t) = x(t) + iy(t).

Definition 5. An arc C is a simple arc if it does not cross itself; that is, C is simple if z(t1) 6= z(t2) when t1 6= t2. When the arc C is simple except for the fact that z(b) = z(a),

we say that C is a simple closed curve.

Definition 6. A contour, or piecewise smooth arc, is an arc consisting of a finite number of smooth arcs joined end to end. When only the initial and final values of z(t) are the same, a contour C is called a simple closed contour.

Definition 7. An analytic function w = w(z) is called an algebraic function if it satisfies a functional equation

a0(z)wn+ a1(z)wn−1+ · · · + an(z) = 0, a0(z) 6= 0, (10)

in which the ai(z) are polynomials in z with complex numbers as coefficients.

One simple example is the algebraic function, w = √z, defined by w2 − z = 0. It is not single-valued in the extended z-plane. In the next chapter, we will introduce a new surface on which to consider the algebraic function defined, and on which it is a single-valued function. This surface is called a Riemann surface.

2

Riemann Surfaces and Cut Structures

2.1

The Riemann Surface for f (z) =

√

z

We begin with the algebraic function f (z) =√z to explain how to construct the Riemann surface for f (z) such that f is a single-valued function on it.

Let z = rei(θ+2kπ)_{, r 6= 0, k ∈ Z. Then}

f (z) =√re12i(θ+2kπ) (11) =√re12iθeikπ (12) = ( √ re12iθ if k is even, −√re12iθ if k is odd. (13) Thus, f is a two-valued function in the extended z-plane. We use the following way to construct the Riemann surface for f (z).

If we cut the extended z-plane along the negative real axis (the branch cut is drawn using bold dashed line as in Figure 2) and restrict ourselves so as never to continue f (z) over this cut, we get two single-valued branches of f (z), namely,

f (z) =√re12iθ, −π ≤ θ < π, and

f (z) =√re12iθ, π ≤ θ < 3π.

To build the Riemann surface for f (z), we take two replicas of the z-plane cut along the negative real axis and call them sheet I and sheet II. The cut on each sheet has two edges. We label the edge of the third quadrant with a + and the edge of the second quadrant with a −. Then attach the + edge of the cut on sheet I to the − edge of the cut on sheet II, and attach the − edge of the cut on sheet I to the + edge of the cut on sheet II. Thus, whenever we cross the cut, we pass from one sheet to the other.

Figure 2

We imagine that the surface as two sheets lying over the extended z-plane, each cut along the negative axis. Using stereographic projection, we can consider the two sheets to be spheres. There is one cut on the surfaces of each sphere.

Figure 3

Figure 4

Now imagine that the spheres are made of rubber. By spreading the edges of the cuts, we can deform each sheet into a hemisphere. When each sheet is rotated so that the openings of the hemispheres face each other, the edges marked + and − face each other and the two hemispheres may be pasted together to give us a sphere. We call this surface the Riemann surface of genus 0 for f (z) =√z, denoted by R0 (Figure 6).

Figure 6

2.2

The Riemann Surface for f (z) =

p(z − r

)(z − r

)

In this section, we discuss how to construct the Riemann surface for the function f (z) = p(z − r1)(z − r2), r1 6= r2. We will find that this is essentially the same as the situation

for f (z) =√z.

The two point z = r1 and z = r2 are branch points of f (z) =p(z − r1)(z − r2). We

obtain two single-valued branches of f (z) by cutting the z-plane along the line segment joining r1 and r2. As in section 2.1, we have two replicas of the z-plane along this cut.

Joining them, we obtain a two-sheeted Riemann surface on which f (z) is single-valued.

Figure 7

If the surface were made of rubber, it could be deformed continuously into that of f (z) = √

z by moving r1 to ∞ and r2 to 0 and deforming the cut into the negative real axis.

Thus this new surface may also be mapped topologically into a sphere.

Figure 9

Figure 10

Also, this surface is the Riemann surface of genus 0 for f (z) =p(z − r1)(z − r2).

2.3

The Riemann Surface for f (z) =

p(z − r

)(z − r

)(z − r

)

Now, we look another example whose Riemann surface is defferent from the ones in the earlier examples.

Let f (z) =p(z − r1)(z − r2)(z − r3) be the algebraic function defined by w2 = (z −

r1)(z − r2)(z − r3), where r1, r2, r3 are distinct. For each i, let z = ri + rei(θ+2kπ), r 6= 0,

k ∈ Z be in the cut plane. We have √ z − ri = q (ri+ rei(θ+2kπ)) − ri (14) = √ rei(θ+2kπ) ₍₁₅₎ = ( √ re12iθ if k is even, −√re12iθ if k is odd. (16) Thus, we go from one point to the other by continuing f (z) over any closed path winding once around one of the roots r1, r2, r3,

√

z − ri changes sign when the argument θ =

arg(z − ri) changes by 2π.

We cut the z-palne from r1 to ∞ and from r2 to r3. Then we take two copies of the

Figure 11

Figure 12

Stretch each cut into a circular hole and rotate the spheres until the holes face each other, as in Figure 13.

Figure 13

We may join them together so that each + edges is attached to the − edge of the corre-sponding cut on the other sphere, as in Figure 14.

Figure 14

Thus, The two-sheeted Riemann surface can be mapped topologically onto a torus. This surface is called the Riemann surface of genus 1 for f (z), denoted by R1.

2.4

Riemann Surfaces of Genus N

We now generalize the results from section 2.1 to section 2.3. Let f (z) =pP (z) =p(z − r1)(z − r2) · · · (z − rn),

where r1, r2, . . . , rn are the roots of the polynomial P (z) of order n.

If the number of roots is even, say n = 2N + 2, we can separate the branch points into pairs, (r1, r2), (r3, r4), . . . , (r2N +1, r2N +2). This gives us n₂ = N + 1 cuts in the cut plane

drawn in Figure 15.

Figure 15 There are N holes in the Riemann surface drawn in Figure 16.

If the number of roots is odd, say n = 2N +1, there must be a cut from ∞ to r1. The

re-maining branch points r2, . . . , r2N +1 can be separated into pairs, (r2, r3), . . . , (r2N, r2N +1).

This will give us n+1₂ = N + 1 cuts in the cut plane dranw in Figure 17.

Figure 17 There are also N holes in the Riemann surface drawn in Figure 18.

Figure 18 The surface in which there are N holes is called the Riemann surface of genus N , denoted by RN, as in Figure 16 and Figure 18.

2.5

To Draw Paths on Cut Planes and on Riemann Surfaces

In this section, we explain how to draw the paths on cut planes and on Riemann surfaces. In the cut planes, we use solid lines to draw a path on sheet I and use dash lines to draw a path on sheet II.

In the Riemann surfaces, we use dash lines to draw a path on the back of the surfaces and use solid lines to draw a path on the front of the surfaces. Let f (z) =√z.

19

Let (I, +) denote the + edge of sheet I, (I, −) denote the − edge of sheet I, (II, +) denote the + edge of sheet II, and (II, −) denote the − edge of sheet II. The path from point A to point B denotes that start from A in sheet II, cross through the cut from (II, −) to (I, +), to B. γ1 denotes a path in sheet I, from a point in (I, +) to a point in (I, −). γ2

denotes a path in sheet II, from a point in (II, −) to a point in (II, +). The corresponding paths in Riemann surface is drawn in following figure.

Figure 20

a − cycle is a closed path that encloses a finite cut (the endpoint of cut is a finite number). b − cycle is a closed path that starts from + edge of a cut (it maybe finite cut or infinte cut) without encloed by any a − cycle, to + edge of another cut encloed by a a − cycle. Then the path crosses through − edge of this cut and goes into sheet II, and finally arrives to the − edge of the starting cut.

Let f (z) =pz(z − 1)(z − 2)(z − 3)(z − 4). The a − cycles and b − cycles in cut plane and their corresponding paths in Riemann surface are drawn in the following figures.

Figure 22 The numbers of a − cycles and b − cycles must be the same. In next few chapters, we aim to evaluate the integrals along a − cycles and b − cycles.

3

Integrals for Horizontal Cuts

3.1

Two Examples

We give a few of examples to explain how to evaluate path integrals. We will use the principle of deformation of paths (Theorem 1). It tells us that if a simple closed con-tour(piecewise smooth arc) C1 is continously deformed into another simple closed coutour

C2, always passing through points at which a function f is analytic, then the value of the

integral of f over C1 never changes.

The circle

z = z0+ Reiθ, −π ≤ θ < π,

is a circle centered at the point z0 and with radius R. It is a simple closed curve, oriented

in the counterclockwise direction.

Cauchy-Goursat Theorem. If a function f is analytic at all points interior to and on a simple closed contour C, then

Z

C

f (z) dz = 0.

Theorem 1. Let C1 and C2 denote positively oriented simple closed contours, where C2 is

interior to C1. If a function f is analytic in the closed region consisting of those contours

and all points between them, then Z C1 f (z) dz = Z C2 f (z) dz. 1

C

C

Example 1. Let f (z) =√z and let γ be the positively oriented (counterclockwise oriented) circular path z = eiθ, −π ≤ θ < π. This is a path looked like a circle centered at the point 0 with radius 1. Evaluate the integral R_γf (z) dz.

Solution.

Figure 24 z ∈ γ =⇒ z = eiθ, −π ≤ θ < π =⇒√z = e12iθ, dz = ieiθdθ. Then, Z γ f (z) dz = Z π −π

f (eiθ)ieiθ dθ = Z π −π √ eiθ_ieiθ _dθ = −4 3i . (2) Deformation of path Figure 25

Let C = γ ∪ −γ∗∗∪ −γ∗_{. Since C is a simple closed contour,} Z C f (z) dz = 0 =⇒ Z γ f (z) dz + Z −γ∗∗ f (z) dz + Z −γ∗ f (z) dz = 0 =⇒ Z γ f (z) dz +  − Z γ∗∗ f (z) dz  +  − Z γ∗ f (z) dz  = 0 =⇒ Z γ f (z) dz = Z γ∗ f (z) dz + Z γ∗∗ f (z) dz. (17) In (1), we have evaluated R

γf (z)dz. Now we evaluate the value of the right hand side of

equation (17).

Since the points along the path γ∗ is in the + edge of the cut plane, the points on γ∗ has the angle −π.

z ∈ γ∗ =⇒ z = rei(−π), r : 1 → 0 =⇒√z =√re12i(−π) = −i √ r, dz = −dr. Then, Z γ∗ f (z) dz = Z 0 1 (−i√r)(−dr) = i Z 0 1 √ r dr = −2 3i .

Similarly, the points along the path γ∗∗ is in the − edge of the cut plane. So the points on γ∗∗ has the angle π.

z ∈ γ∗∗ =⇒ z = reiπ, r : 0 → 1 =⇒√z =√re12iπ = i √ r, dz = −dr. Then, Z γ∗∗ f (z) dz = Z 1 0 i√r(−dr) = −i Z 1 0 √ r dr = −2 3i .

Finally, we obtain the result Z γ∗ f (z) dz + Z γ∗∗ f (z) dz = (−2 3i) + (− 2 3i) = −4 3i = Z γ f (z) dz. From (1) and (2), we have verified equation (17).

Example 2. Let f (z) =p(z − 1)(z − 2) and let γ be the positively oriented circular path z = 3₂ + eiθ, −π ≤ θ < π. Evaluate the integral R_γf (z) dz.

Solution.

Step 1. Draw the cut plane

Figure 26

If a point goes from point A , crossing through the left part of 1 on real axis ({x ∈ R|x < 1}), to point B, it crosses two branch cuts. Then √z − 1 changes sign one time and √

z − 2 also changes sign one time. So f (z) totally changes sign two times. If a point goes from point C, crossing through the line segment between 1 and 2 ({x ∈ R|1 < x < 2}), to point D, it crosses only one branch cut. Then √z − 1 does not change sign but √z − 2 changes sign one time. So f (z) totally changes sign only one times. Thus, there is only one cut between 1 and 2 (Figure 25).

Step 2. Evaluate the integrals (1) Integral along the circle

Figure 27 z ∈ γ =⇒ z = 3 2+ e iθ_{, −π ≤ θ < π} =⇒ dz = ieiθdθ. Then, Z γ f (z) dz = Z π −π r (3 2 + e iθ_{) − 1} r (3 2 + e iθ_{) − 2 ie}iθ _dθ = −0.785395i . (2) Deformation of path Figure 28

Since the points along the path γ∗ is in the + edge of the cut plane, the points on γ∗ has the angle −π. z ∈ γ∗ =⇒ z = 2 + rei(−π) = 2 − r, r : 1 → 0 =⇒√z − 2 =√re12i(−π)= −i √ r, dz = −dr. Then, Z γ∗ f (z) dz = Z 0 1 p 1 + rei(−π)_(−i√_r)(−dr) = Z 0 1 √ 1 − r(−i√r)(−dr) = i Z 0 1 √ 1 − r√r dr = −0.392699i .

Similarly, the points along the path γ∗∗ is in the − edge of the cut plane. So the points on γ∗∗ has the angle π.

z ∈ γ∗∗=⇒ z = 2 + reiπ= 2 − r, r : 0 → 1 =⇒√z − 2 =√re12iπ = i √ r, dz = −dr. Then, Z γ∗∗ f (z) dz = Z 1 0 √ 1 − r(i√r)(−dr) = −i Z 1 0 √ 1 − r√r dr = −0.392699i . We obtain Z γ∗ f (z) dz + Z γ∗∗ f (z) dz = (−0.392699i) + (−0.392699i) = −0.785398i .

Again, we verify equation (17). Note that, Z γ f (z) dz = Z γ∗ f (z) dz + Z γ∗∗ f (z) dz = i Z 0 1 √ 1 − r√r dr +  −i Z 1 0 √ 1 − r√r dr  = i Z 0 1 √ 1 − r√r dr +  i Z 0 1 √ 1 − r√r dr  = 2i Z 0 1 √ 1 − r√r dr = 2 Z γ∗ f (z) dz . That is, Z γ f (z) dz = 2 Z 1−→2+ f (z) dz (18)

3.2

The Problem in Using Mathematica

Before we use Mathematica to compute the integrals, we need to know what phenomena will happens. Let z = reiθ_{. We use the notation, arg z to denote the argument of the}

complex number z. So, arg z = θ. Let (I) denote sheet I and let (II) denote sheet II. Let w = f (z) =√z. In theoretical aspect, z ∈ (I) =⇒ −π ≤ arg z < π =⇒ −π 2 ≤ 1 2arg z < π 2.

f maps the points on sheet I into the right-half plane {z ∈ C| − π2 ≤ arg z < π 2}.

Figure 29 And, z ∈ (II) =⇒ π ≤ arg z < 3π =⇒ π 2 ≤ 1 2arg z < 3π 2 . f maps the points on sheet II into the left-half plane {z ∈ C|π2 ≤ arg z <

3π 2 }.

Figure 30

If you compute √z using Mathematica, you can discover that the range of f (z) = √z are as same as the range described above except the points along the + edge of sheet I ,that is, {z ∈ C|arg z = −π}. For example, suppose that z = −2 ∈ (I, +), where (I, +) denotes the + edge of sheet I.

z ∈ (I, +) =⇒ arg z = −π =⇒ arg√z = −π 2 =⇒ −2 = 2ei(−π) =⇒ f (z) =√−2 = 2ei(−π)
12 =√2ei(−π2) = − √ 2i.

But in Mathematica, √−2 = √2i. This value needs to time −1 to obtain the correct value. Therefore,

z ∈ (I, +) =⇒√_{z = (−1) · math} √z
,

where math (√z) means the value of√z computed by Mathematica. We use the notation math (·) to denote the value of “ · ” computed by Mathematica.

2 -i 2 i 2 -theoretical Mathematica

Thus, if you want to compute an integral along the + edge in sheet I, e.g., R₋₁ +

−→0

√ zdz (integration along the line segment {z ∈ (I, +)| − 1 < z < 0} from −1 to 0), you must multiply the result value computed in Mathematica by −1 to obtain the correct value. That is, Z −1−→0+ √ z dz = (−1) · math Z 0 −1 √ z dz  = −0.666667i.

Suppose that f (z) = √z. Let θ2 = θ1 + 2π and let z1 = reiθ1, −π ≤ θ1 < π and

z2 = reiθ2, π ≤ θ2 < 3π. Then z1 and z2 are the same points in the complex plane C, but

in the cut plane, z1 ∈ (I) and z2 ∈ (II) (Figure 32).

Figure 32 f (z2) = √ z2 = √ re12iθ2 = √ re12i(θ1+2π) = √ re12iθ1+iπ =√re12iθ1eiπ = √ re12iθ1 · (−1) = −√z₁ = −f (z₁) (19) This tells us that √z|II = −

√

z|I, the value of

√

z in sheet II is the value of√z in sheet I multiplied by −1. Thus, if g(z) =ph(z) where h(z) = (z − z1)(z − z2) · · · (z − zk) =

Qk

j=1(z − zj), we can assume that h(z) = Reiθ for some positive real number R and θ.

Let h(z)|I = Reiθ1, −π ≤ θ1 < π and h(z)|II = Reiθ2, π ≤ θ2 < 3π, where θ2 = θ1+ 2π.

g(z)|II = p h(z)|II = √ Reiθ2 = √ Rei(θ1+2π)= √ Reiθ1eiπ = √ Reiθ1 · (−1) = (−1) ·ph(z)| I = (−1) · g(z)|I (20)

3.3

Evaluating Integrals Using Mathematica

In Example 1 and Example 2, we use the analytic method to evaluate the integral. In this section, we will explain how to modify the value computed in Mathematica.

Example 3. Evaluate the integral in Example 1 Using Mathematica . Solution. (1)Along −1−→ 0 (z ∈ γ+ ∗_{)(Figure 25)} arg z = −π =⇒√_{z = (−1) · math} √z
Z γ∗ f (z) dz = Z −1−→0+ f (z) dz = Z 0 −1(−1) · math √ z
dz = (−1) · math Z 0 −1 √ z dz  . (2)Along −1←− 0 (z ∈ γ− ∗∗_{)(Figure 25)} arg z = π =⇒√_{z = math} √z
Z γ∗∗ f (z) dz = Z −1←−0− f (z) dz = Z −1 0 math √ z
dz = math Z −1 0 √ z dz  . Then, Z γ f (z) dz = Z γ∗ f (z) dz + Z γ∗∗ f (z) dz = (−1) · math Z 0 −1 √ z dz  + math Z −1 0 √ z dz  = (−1) · math Z 0 −1 √ z dz  + (−1) · math Z 0 −1 √ z dz  = (−2) · math Z 0 −1 √ z dz  (21) = −1.33333i .

This agrees with the valueR_γf (z)dz = −4₃i = −1.33333i in Example 1. Example 4. Evaluate the integral in Example 2 Using Mathematica . Solution.

(1)Along 1−→ 2+

arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1

Thus, f (z) =√z − 1√_{z − 2 = (−1) · math} √z − 1√z − 2
. Then, Z γ∗f (z) dz = (−1) · math Z 2 1 √ z − 1√z − 2  . (2)Along 1←− 2− arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 2) = π =⇒ √_{z − 2 = math} √z − 2
. Thus, f (z) =√z − 1√_{z − 2 = math} √z − 1√z − 2
. Then, Z γ∗∗f (z) dz = math Z 1 2 √ z − 1√z − 2  = (−1) · math Z 2 1 √ z − 1√z − 2  = Z γ∗ f (z) dz. So, Z γ f (z) dz = Z γ∗ f (z) dz + Z γ∗∗ f (z) dz = 2 Z γ∗ f (z) dz = (−2) · math Z 2 1 √ z − 1√z − 2  = −0.785398i .

This value also agrees with the answer in Example 2.

In the next example, we evaluate an integral along a positively oriented simple closed curve in which there are two branch cuts.

Example 5. Suppose that f (z) = p(z − 1)(z − 2)(z − 3)(z − 4) and γ is a positively oriented simple closed curve that encloses all cuts (Figure 33). Evaluate the integral R

γf (z)dz.

Figure 33 According to the deformation of path, we have

Z γ f (z) dz = Z 1−→4+ f (z) dz + Z 1←−4− f (z) dz (22) = Z 1−→2+ f (z) dz + Z 2−→3+ f (z) dz + Z 3−→4+ f (z) dz  + Z 3←−4− f (z) dz + Z 2←−3− f (z) dz + Z 1←−2− f (z) dz  (23) = Z 1−→2+ f (z) dz + Z 3−→4+ f (z) dz  + Z 3←−4− f (z) dz + Z 1←−2− f (z) dz  . (24)

Because of the two paths 2 −→ 3 and 2+ ←− 3 are not along any branch cut, the two− integralsR

2−→3+ f (z)dz and

R

2←−3− f (z)dz in equation (23) are canceled by each other. Thus,

we only investigate the four integrals in equation (24). Theoretical Evaluation

(1)Along 1−→ 2+

z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒ z − 1 = (1 − r)ei0 =⇒√z − 1 =√1 − re12i0= √ 1 − r z − 2 < 0 =⇒ arg(z − 2) = −π =⇒ z − 2 = rei(−π) =⇒√z − 2 =√re12i(−π) = −i √ r

z − 3 < 0 =⇒ arg(z − 3) = −π =⇒ z − 3 = (1 + r)ei(−π)

=⇒√z − 3 =√1 + re12i(−π)= −i √

1 + r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒ z − 4 = (2 + r)ei(−π)

=⇒√z − 4 =√2 + re12i(−π)= −i √ 2 + r Z 1−→2+ f (z) dz = Z 0 1 √

1 − r(−i√r)(−i√1 + r)(−i√2 + r)(−dr) = i3 Z 0 1 √ 1 − r√r√1 + r√2 + r dr = −i Z 0 1 √ 1 − r√r√1 + r√2 + r dr = 0.76002i .

From the procedure above, we find that we can simplify the representation of√z − k, k = 1, 2, 3, 4. We only substitute z = 2 − r directly into√z − k for each k and remember the following rules :

z ∈ (I, +) =⇒ (

z − k > 0 =⇒√z − k =p(2 − r) − k

z − k < 0 =⇒√z − k =p−(k − z) = (−i)√k − z = −ipk − (2 − r) The minus sign, “−”, is necessary because z − k ∈ (I, +) and arg(z − k) = −π. It is the cause of the factor (−i) appearing.

z ∈ (I, −) =⇒ (

z − k > 0 =⇒√z − k =p(2 − r) − k

z − k < 0 =⇒√z − k =p−(k − z) = i√k − z = ipk − (2 − r) It is important that we must make the number inside square roots to be positive. Thus, we can also write

z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 =p(2 − r) − 1 =√1 − r z − 2 < 0 =⇒ arg(z − 2) = −π =⇒√z − 2 =p(2 − r) − 2 =√−r = −i√r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 =p(2 − r) − 3 = p−(1 + r) = −i√1 + r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 =p(2 − r) − 4 = p−(2 + r) = −i√2 + r

(2)Along 3−→ 4+ z = 4 + rei(−π)= 4 − r, r : 1 −→ 0 =⇒ dz = −dr z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 = p(4 − r) − 1 =√3 − r z − 2 > 0 =⇒ arg(z − 2) = 0 =⇒√z − 2 = p(4 − r) − 2 =√2 − r z − 3 > 0 =⇒ arg(z − 3) = 0 =⇒√z − 3 = p(4 − r) − 3 =√1 − r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = p(4 − r) − 4 = −i√r Z 3−→4+ f (z) dz = Z 0 1 √ 3 − r√2 − r√1 − r(−i√r)(−dr) = i Z 0 1 √ 3 − r√2 − r√1 − r√r dr = −0.76002i . (3)Along 3←− 4− z = 4 + rei(−π)= 4 − r, r : 0 −→ 1 =⇒ dz = −dr z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 =p(4 − r) − 1 =√3 − r z − 2 < 0 =⇒ arg(z − 2) = 0 =⇒√z − 2 =p(4 − r) − 2 =√2 − r z − 3 < 0 =⇒ arg(z − 3) = 0 =⇒√z − 3 =p(4 − r) − 3 =√1 − r z − 4 < 0 =⇒ arg(z − 4) = π =⇒√z − 4 =p(4 − r) − 4 =√−r = i√r Z 3←−4− f (z) dz = Z 1 0 √ 3 − r√2 − r√1 − r(i√r)(−dr) = −i Z 1 0 √ 3 − r√2 − r√1 − r√r dr = −i  − Z 0 1 √ 3 − r√2 − r√1 − r√r dr  = i Z 0 1 √ 3 − r√2 − r√1 − r√r dr = Z 3−→4+ f (z) dz = −0.76002i .

(4)Along 1←− 2− z = 2 + rei(−π)= 2 − r, r : 0 −→ 1 =⇒ dz = −dr z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 = p(2 − r) − 1 =√1 − r z − 2 < 0 =⇒ arg(z − 2) = π =⇒√z − 2 = p(2 − r) − 2 =√−r = i√r z − 3 < 0 =⇒ arg(z − 3) = π =⇒√z − 3 = p(2 − r) − 3 = p−(1 + r) = i√1 + r z − 4 < 0 =⇒ arg(z − 4) = π =⇒√z − 4 = p(2 − r) − 4 = p−(2 + r) = i√2 + r Z 1←−2− f (z) dz = Z 1 0 √

1 − r(i√r)(i√1 + r)(i√2 + r)(−dr) = −i3 Z 1 0 √ 1 − r√r√1 + r√2 + r dr = i Z 1 0 √ 1 − r√r√1 + r√2 + r dr = −i Z 0 1 √ 1 − r√r√1 + r√2 + r dr = Z 1−→2+ f (z) dz = 0.76002i . Z γ f (z) dz = Z 1−→2+ f (z) dz + Z 3−→4+ f (z) dz + Z 3←−4− f (z) dz + Z 1←−2− f (z) dz = 2 Z 1−→2+ f (z) dz + 2 Z 3−→4+ f (z) dz = 2 Z 1−→2+ f (z) dz + Z 3−→4+ f (z) dz  = 2 (0.76002i) + (−0.76002i)
= 0 . Using Mathematica (1)Along 1−→ 2+ arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 2) = −π =⇒ √_{z − 2 = (−1) · math} √z − 2
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
.

Z 1−→2+ f (z) dz = (−1)3_{· math} Z 2 1 √ z − 1√z − 2√z − 3√z − 4 dz  = (−1) · math Z 2 1 √ z − 1√z − 2√z − 3√z − 4 dz  . (2)Along 3−→ 4+ arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 2) = 0 =⇒ √_{z − 2 = math} √z − 2
arg(z − 3) = 0 =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
. Z 3−→4+ f (z) dz = (−1) · math Z 4 3 √ z − 1√z − 2√z − 3√z − 4 dz  . (3)Along 3←− 4− arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 2) = 0 =⇒ √_{z − 2 = math} √z − 2
arg(z − 3) = 0 =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = π =⇒ √_{z − 4 = math} √z − 4
. Z 3←−4− f (z) dz = math Z 3 4 √ z − 1√z − 2√z − 3√z − 4 dz  = (−1) · math Z 4 3 √ z − 1√z − 2√z − 3√z − 4 dz  = Z 3−→4+ f (z) dz. (4)Along 1←− 2− arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 2) = π =⇒ √_{z − 2 = math} √z − 2
arg(z − 3) = π =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = π =⇒ √_{z − 4 = math} √z − 4
. Z 1←−2− f (z) dz = math Z 1 2 √ z − 1√z − 2√z − 3√z − 4 dz  = (−1) · math Z 2 1 √ z − 1√z − 2√z − 3√z − 4 dz  = Z 1−→2+ f (z) dz.

Therefore, Z γ f (z) dz = 2 Z 1−→2+ f (z) dz + 2 Z 3−→4+ f (z) dz = (−2) · math Z 2 1 √ z − 1√z − 2√z − 3√z − 4 dz  + (−2) · math Z 4 3 √ z − 1√z − 2√z − 3√z − 4 dz  = (−2) · math Z 2 1 √ z − 1√z − 2√z − 3√z − 4 dz + Z 4 3 √ z − 1√z − 2√z − 3√z − 4 dz  = −3.15797 × 10−15+ 1.77636 × 10−15i . (25) To compare the values in equation (24) and equation (25), the two values in fact are the same. Note that −3.15797 × 10−15 and 1.77636 × 10−15 are very small numbers, so we can say them to be 0.

In the next example, we evaluate an integral along one b − cycle.

Example 6. Let f (z) = p(z − 1)(z − 2)(z − 3) and let γ be the oriented positively cir-cular path

z = ( ₃

2 + e

iθ _{if − π ≤ θ < 0, (in sheet I)} 3

2 + e

iθ _{if 2π ≤ θ < 3π. (in sheet II)}

Solution.

Figure 34 1. Integral along the circle

Since f (z)|II = −f (z)|I, we have Z γ f (z) dz = Z 0 −π f (z) dz + Z 3π 2π f (z) dz = Z 0 −π f (z) dz + (−1) Z π 0 f (z) dz = Z 0 −π f (z) dz − Z π 0 f (z) dz.

z = 3 2 + e iθ =⇒ dz = ieiθdθ Z γ f (z) dz = Z 0 −π r (3 2+ e iθ_{) − 1} r (3 2+ e iθ_{) − 2} r (3 2 + e iθ_{) − 3 ie}iθ _dz − Z π 0 r (3 2+ e iθ_{) − 1} r (3 2+ e iθ_{) − 2} r (3 2 + e iθ_{) − 3 ie}iθ _dz = −0.958512 . 2. Deformation of path Theoretical Evaluation (1) Along 1−→2 z = 2 + rei(−π)= 2 − r, r : 1 −→ 0 =⇒ dz = −dr z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 =p(2 − r) − 1 =√1 − r z − 2 < 0 =⇒ arg(z − 2) = −π =⇒√z − 2 =p(2 − r) − 2 =√−r = −i√r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 =p(2 − r) − 3 = p−(1 + r) = −i√1 + r Z 1−→2 f (z) dz = Z 0 1 √ 1 − r(−i√r)(−i√1 + r)(−dr) = −i2 Z 0 1 √ 1 − r√r√1 + r dr = Z 0 1 √ 1 − r√r√1 + r dr. (2) Along 1L992 Since f (z)|II = −f (z)|I, we have R 1L992f (z) dz = − R 1←−2f (z) dz. Therefore, we first

evaluate the value of the integral in sheet I then we multiply the value by −1. z = 2 + rei(−π)= 2 − r, r : 0 −→ 1 =⇒ dz = −dr z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 = p(2 − r) − 1 =√1 − r z − 2 < 0 =⇒ arg(z − 2) = π =⇒√z − 2 = p(2 − r) − 2 =√−r = i√r z − 3 < 0 =⇒ arg(z − 3) = π =⇒√z − 3 = p(2 − r) − 3 = p−(1 + r) = i√1 + r

Z 1←−2 f (z) dz = Z 1 0 √ 1 − r(i√r)(i√1 + r)(−dr) = −i2 Z 1 0 √ 1 − r√r√1 + r dr = Z 1 0 √ 1 − r√r√1 + r dr. Thus, Z 1L992 f (z) dz = − Z 1←−2 f (z) dz = − Z 1 0 √ 1 − r√r√1 + r dr = Z 0 1 √ 1 − r√r√1 + r dr = Z 1−→2 f (z) dz From (1) and (2), we obtain

Z γ f (z) dz = Z 1−→2 f (z) dz + Z 1L992 f (z) dz (26) = Z 1−→2 f (z) dz + Z 1−→2 f (z) dz (27) = 2 Z 1−→2 f (z) dz (28) = 2 Z 0 1 √ 1 − r√r√1 + r dr (29) = −0.958512 . (30) Using Mathematica

According to equation (26) to equation (28), we have Z γ f (z) dz = 2 Z 1−→2 f (z) dz. (31) Thus we only evaluateR_1−→2f (z) dz.

Along 1−→2 : arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 2) = −π =⇒ √_{z − 2 = (−1) · math} √z − 2
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
. Z 1−→2 f (z) dz = (−1)2_{· math} Z 2 1 √ z − 1√z − 2√z − 3 dz  = math Z 2 1 √ z − 1√z − 2√z − 3 dz  .

So, Z γ f (z) dz = 2 Z 1−→2 f (z) dz = 2 · math Z 2 1 √ z − 1√z − 2√z − 3 dz  = −0.958512 .

We give the final example to evaluate integrals along all a-cycles and b-cycles. Example 7. Suppose that

f (z) =p(z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6)

=p(z − (−5))(z − (−3))(z − (−1))(z − 1)(z − 3)(z − 4)(z − 6) .

Let a1, a2, a3 be three a − cycles and let b1, b2, b3 be three b − cycles drawing in Figure

35. Evaluate the six integrals R_a

kf (z)dz and R

bkf (z)dz, k = 1, 2, 3 using the method of deformation of path.

Figure 35 Solution.

Suppose that there is an a − cycle encloses a cut which is between z1 to z2. From

equation (18) in example 2, we have the result Z a−cycle f (z) dz = 2 Z z1 + −→z2 f (z) dz (32) From equation (21) in example 3, we also have

Z a−cycle f (z) dz = (−2) · math Z z2 z1 f (z) dz  (33) 1. To evaluate R_a 1f (z)dz Theoretical Evaluation Along −3−→ −1 :+ z = −1 + rei(−π)= −1 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√4 − r z + 3 > 0 =⇒ arg(z + 3) = 0 =⇒√z + 3 =√2 − r z + 1 < 0 =⇒ arg(z + 1) = −π =⇒√z + 1 =√−r = −i√r z − 1 < 0 =⇒ arg(z − 1) = −π =⇒√z − 1 = p−(2 + r) = −i√2 + r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 = p−(4 + r) = −i√4 + r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = p−(5 + r) = −i√5 + r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = p−(7 + r) = −i√7 + r Z −3−→−1+ f (z) dz = −(−i)5 Z 0 2 √ 4 − r√2 − r√r√2 + r√4 + r√5 + r√7 + r dr = i Z 0 2 √ 4 − r√2 − r√r√2 + r√4 + r√5 + r√7 + r dr. Z a1 f (z)dz = 2 Z −3−→−1+ f (z) dz = 2i Z 0 2 √ 4 − r√2 − r√r√2 + r√4 + r√5 + r√7 + r dr = −144.283i . Using Mathematica Along −3−→ −1 :+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = −π =⇒ √_{z + 1 = (−1) · math} √z + 1
arg(z − 1) = −π =⇒ √_{z − 1 = (−1) · math} √z − 1
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
arg(z − 6) = −π =⇒ √_{z − 6 = (−1) · math} √z − 6
.

Z −3−→1+ f (z) dz = (−1)5_{· math} Z −1 −3 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = (−1) · math Z −1 −3 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −72.1417i . Z a1 f (z)dz = 2 Z −3−→−1+ f (z) dz = (−2) · math Z −1 −3 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −144.283i . 2. To evaluate R_a 2f (z)dz Theoretical Evaluation Along 1−→ 3 :+ z = 3 + rei(−π)= 3 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√8 − r z + 3 > 0 =⇒ arg(z + 3) = 0 =⇒√z + 3 =√6 − r z + 1 > 0 =⇒ arg(z + 1) = 0 =⇒√z + 1 =√4 − r z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒ √z − 1 = √2 − r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 = √−r = −i√r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = p−(1 + r) = −i√1 + r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = p−(3 + r) = −i√3 + r Z 1−→3+ f (z) dz = −(−i)3 Z 0 2 √ 8 − r√6 − r√4 − r√2 − r√r√1 + r√3 + r dr = −i Z 0 2 √ 8 − r√6 − r√4 − r√2 − r√r√1 + r√3 + r dr. Z a2 f (z)dz = 2 Z 1−→3+ f (z) dz = −2i Z 0 2 √ 8 − r√6 − r√4 − r√2 − r√r√1 + r√3 + r dr = 88.2841i . Using Mathematica

Along 1−→ 3 :+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = 0 =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
arg(z − 6) = −π =⇒ √_{z − 6 = (−1) · math} √z − 6
. Z 1−→3+ f (z) dz = (−1)3_{· math} Z 3 1 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = (−1) · math Z 3 1 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = 44.142i . Z a2 f (z)dz = 2 Z 1−→3+ f (z) dz = (−2) · math Z 3 1 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = 88.2841i . 3. To evaluate R a3f (z)dz Theoretical Evaluation Along 4−→ 6 :+ z = 6 + rei(−π)= 6 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√11 − r z + 3 > 0 =⇒ arg(z + 3) = 0 =⇒√z + 3 =√9 − r z + 1 > 0 =⇒ arg(z + 1) = 0 =⇒√z + 1 =√7 − r z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 =√5 − r z − 3 > 0 =⇒ arg(z − 3) = 0 =⇒√z − 3 =√3 − r z − 4 > 0 =⇒ arg(z − 4) = 0 =⇒√z − 4 =√2 − r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 =√−r = −i√r Z 4−→6+ f (z) dz = −(−i) Z 0 2 √ 11 − r√9 − r√7 − r√5 − r√3 − r√2 − r√r dr = i Z 0 2 √ 11 − r√9 − r√7 − r√5 − r√3 − r√2 − r√r dr.

Z a3 f (z)dz = 2 Z 4−→6+ f (z) dz = 2i Z 0 2 √ 11 − r√9 − r√7 − r√5 − r√3 − r√2 − r√r dr = −198.138i . Using Mathematica Along 4−→ 6 :+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = 0 =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 3) = 0 =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = 0 =⇒ √_{z − 4 = math} √z − 4
arg(z − 6) = −π =⇒ √_{z − 6 = (−1) · math} √z − 6
. Z 4−→6+ f (z) dz = (−1) · math Z 6 4 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −99.0688i . Z a3 f (z)dz = 2 Z 4−→6+ f (z) dz = (−2) · math Z 6 4 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −198.138i .

From example 6 and equation (31), we have Z b−cycle f (z) dz = 2 Z z1 + −→z2 f (z) dz, (34) and Z b−cycle f (z) dz = 2 · math Z z2 z1 f (z) dz  . (35) 4. To evaluate R b1f (z)dz

Theoretical Evaluation Along −5−→ −3 :+ z = −3 + rei(−π)= −3 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√2 − r z + 3 < 0 =⇒ arg(z + 3) = −π =⇒√z + 3 =√−r = −i√r z + 1 < 0 =⇒ arg(z + 1) = −π =⇒√z + 1 =p−(2 + r) = −i√2 + r z − 1 < 0 =⇒ arg(z − 1) = −π =⇒√z − 1 = p−(4 + r) = −i√4 + r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 = p−(6 + r) = −i√6 + r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = p−(7 + r) = −i√7 + r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = p−(9 + r) = −i√9 + r Z −5−→−3+ f (z) dz = −(−i)6 Z 0 2 √ 2 − r√r√2 + r√4 + r√6 + r√7 + r√9 + r dr = Z 0 2 √ 2 − r√r√2 + r√4 + r√6 + r√7 + r√9 + r dr Z b1 f (z)dz = 2 Z −5−→−3+ f (z) dz = 2 Z 0 2 √ 2 − r√r√2 + r√4 + r√6 + r√7 + r√9 + r dr = −291.688 . Using Mathematica Along −5−→ −3 :+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = −π =⇒ √_{z + 3 = (−1) · math} √z + 3
arg(z + 1) = −π =⇒ √_{z + 1 = (−1) · math} √z + 1
arg(z − 1) = −π =⇒ √_{z − 1 = (−1) · math} √z − 1
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
arg(z − 6) = −π =⇒ √_{z − 6 = (−1) · math} √z − 6
. Z −5−→−3+ f (z) dz = (−1)6 _{· math} Z −3 −5 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = math Z −3 −5 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −145.844 .

Z b1 f (z)dz = 2 Z −5−→−3+ f (z) dz = 2 · math Z −3 −5 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −291.688 . 5. To evaluate R_b 2f (z)dz Figure 38 Z b2 f (z) dz = Z −5−→1+ f (z) dz + Z −5L991− f (z) dz = Z −5−→−3+ f (z) dz + Z −3−→−1+ f (z) dz + Z −1−→1+ f (z) dz + Z −1L991− f (z) dz + Z −3L99−1− f (z) dz + Z −5L99−3− f (z) dz Theoretical Evaluation (1)Along −5−→ −3+ z = −3 + rei(−π)= −3 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√2 − r z + 3 < 0 =⇒ arg(z + 3) = −π =⇒√z + 3 =√−r = −i√r z + 1 < 0 =⇒ arg(z + 1) = −π =⇒√z + 1 =p−(2 + r) = −i√2 + r z − 1 < 0 =⇒ arg(z − 1) = −π =⇒√z − 1 = p−(4 + r) = −i√4 + r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 = p−(6 + r) = −i√6 + r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = p−(7 + r) = −i√7 + r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = p−(9 + r) = −i√9 + r Let u1(r) = √ 2 − r√r√2 + r√4 + r√6 + r√7 + r√9 + r. Then Z f (z) dz = −(−i)6 Z 0 u1(r) dr = Z 0 u1(r) dr

(2)Along −5_{L99 −3}−

The same procedure in example 6, we first evaluate −5←− −3.− Z −5←−−3− f (z) dz = Z 2 0 i6u1(r) (−dr) = −i6 Z 2 0 u1(r) dr = Z 2 0 u1(r) dr. Then Z −5L99−3− f (z) dz = − Z −5←−−3− f (z) dz = − Z 2 0 u1(r) dr = Z 0 2 u1(r) dr = Z −5−→−3+ f (z) dz. From (1) and (2), we have

Z −5−→−3+ f (z) dz + Z −5L99−3− f (z) dz = 2 Z −5−→−3+ f (z) dz (36) = 2 Z 0 2 u1(r) dr. (37) (3)Along −3−→ −1+ z = −1 + rei(−π)= −1 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√4 − r z + 3 > 0 =⇒ arg(z + 3) = 0 =⇒√z + 3 =√2 − r z + 1 < 0 =⇒ arg(z + 1) = −π =⇒√z + 1 = −i√r z − 1 < 0 =⇒ arg(z − 1) = −π =⇒√z − 1 = −i√2 + r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 = −i√4 + r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = −i√5 + r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = −i√7 + r Let u2(r) = √ 4 − r√2 − r√r√2 + r√4 + r√5 + r√7 + r. Then Z −3−→−1+ f (z) dz = −(−i)5 Z 0 2 u2(r) dr = i Z 0 2 u2(r) dr. (4)Along −3_{L99 −1}− Z −3←−−1− f (z) dz = −i5 Z 2 0 u2(r) dr = −i Z 2 0 u2(r) dr. Then Z −3L99−1− f (z) dz = − Z −3←−−1− f (z) dz = i Z 2 0 u2(r) dr = −i Z 0 2 u2(r) dr = − Z −3−→−1+ f (z) dz.

From (3) and (4), we have Z −3−→−1+ f (z) dz + Z −3L99−1− f (z) dz (38) = Z −3−→−1+ f (z) dz +  − Z −3−→−1+ f (z) dz  (39) = 0. (40) (5)Along −1−→ 1+ z = 1 + rei(−π)= 1 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√6 − r z + 3 > 0 =⇒ arg(z + 3) = 0 =⇒√z + 3 =√4 − r z + 1 > 0 =⇒ arg(z + 1) = 0 =⇒√z + 1 =√2 − r z − 1 < 0 =⇒ arg(z − 1) = −π =⇒√z − 1 = −i√r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 = −i√2 + r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = −i√3 + r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = −i√5 + r Let u3(r) = √ 6 − r√4 − r√2 − r√r√2 + r√3 + r√5 + r. Then Z −1−→1+ f (z) dz = −(−i)4 Z 0 2 u3(r) dr = − Z 0 2 u3(r) dr. (6)Along −1_{L99 1}− Z −1←−1− f (z) dz = −i4 Z 2 0 u3(r) dr = − Z 2 0 u3(r) dr. Then Z −1L991− f (z) dz = − Z −1←−1− f (z) dz = Z 2 0 u3(r) dr = − Z 0 2 u3(r) dr = Z −1−→1+ f (z) dz. From (5) and (6), we have

Z −1−→1+ f (z) dz + Z −1L991− f (z) dz = 2 Z −1−→1+ f (z) dz (41) = −2 Z 0 2 u3(r) dr. (42)

According to equation (36), (37), (38), (39), (40), (41), and (42), Z b2 f (z) dz = 2 Z −5−→−3+ f (z) dz + 2 Z −1−→1+ f (z) dz = 2 Z 0 2 u1(r) dr +  −2 Z 0 2 u3(r) dr  = −291.688 + 101.116 = −190.572 .

Using Mathematica (1)Along −5−→ −3+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = −π =⇒ √_{z + 3 = (−1) · math} √z + 3
arg(z + 1) = −π =⇒ √_{z + 1 = (−1) · math} √z + 1
arg(z − 1) = −π =⇒ √_{z − 1 = (−1) · math} √z − 1
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
arg(z − 6) = −π =⇒ √_{z − 6 = (−1) · math} √z − 6
. Z −5−→−3+ f (z) dz = (−1)6_{· math} Z −3 −5 f (z) dz  = math Z −3 −5 f (z) dz  (2)Along −5_{L99 −3}− arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = π =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = π =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = π =⇒ √_{z − 1 = math} √z − 1
arg(z − 3) = π =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = π =⇒ √_{z − 4 = math} √z − 4
arg(z − 6) = π =⇒ √_{z − 6 = math} √z − 6
. Z −5←−−3− f (z) dz = math Z −5 −3 f (z) dz  = −math Z −3 −5 f (z) dz  = − Z −5−→−3+ f (z) dz Z −5L99−3− f (z) dz = − Z −5←−−3− f (z) dz = Z −5−→−3+ f (z) dz From (1) and (2), Z −5−→−3+ f (z) dz + Z −5L99−3− f (z) dz = Z −5−→−3+ f (z) dz + Z −5−→−3+ f (z) dz = 2 Z −5−→−3+ f (z) dz = 2 · math Z −3 −5 f (z) dz  = 2 · math Z −3 −5 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −291.688 .

(3)Along −3−→ −1+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = −π =⇒ √_{z + 1 = (−1) · math} √z + 1
arg(z − 1) = −π =⇒ √_{z − 1 = (−1) · math} √z − 1
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
arg(z − 6) = −π =⇒ √_{z − 6 = (−1) · math} √z − 6
. Z −3−→−1+ f (z) dz = (−1)5_{· math} Z −1 −3 f (z) dz  = −math Z −1 −3 f (z) dz  (4)Along −3_{L99 −1}− arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = π =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = π =⇒ √_{z − 1 = math} √z − 1
arg(z − 3) = π =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = π =⇒ √_{z − 4 = math} √z − 4
arg(z − 6) = π =⇒ √_{z − 6 = math} √z − 6
. Z −3←−−1− f (z) dz = math Z −3 −1 f (z) dz  = −math Z −1 −3 f (z) dz  = Z −3−→−1+ f (z) dz Z −3L99−1− f (z) dz = − Z −3←−−1− f (z) dz = − Z −3−→−1+ f (z) dz From (3) and (4), Z −3−→−1+ f (z) dz + Z −3L99−1− f (z) dz = Z −3−→−1+ f (z) dz − Z −3−→−1+ f (z) dz = 0 . (5)Along −1−→ 1+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = 0 =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = −π =⇒ √_{z − 1 = (−1) · math} √z − 1
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
√ √

Z −1−→1+ f (z) dz = (−1)4_{· math} Z 1 −1 f (z) dz  = math Z 1 −1 f (z) dz  (6)Along −1_{L99 1}− arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = 0 =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = π =⇒ √_{z − 1 = math} √z − 1
arg(z − 3) = π =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = π =⇒ √_{z − 4 = math} √z − 4
arg(z − 6) = π =⇒ √_{z − 6 = math} √z − 6
. Z −1←−1− f (z) dz = math Z −1 1 f (z) dz  = −math Z −1 −1 f (z) dz  = − Z −1−→1+ f (z) dz Z −1L991− f (z) dz = − Z −1←−1− f (z) dz = Z −1−→1+ f (z) dz From (5) and (6), Z −1−→1+ f (z) dz + Z −1L991− f (z) dz = Z −1−→1+ f (z) dz + Z −1−→1+ f (z) dz = 2 Z −1−→1+ f (z) dz = 2 · math Z 1 −1 f (z) dz  = 2 · math Z 1 −1 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −101.116 . Thus, Z b2 f (z) dz = Z −5−→−3+ f (z) dz + Z −5L99−3− f (z) dz  + Z −1−→1+ f (z) dz + Z −1L991− f (z) dz  = −291.688 + 101.116 = −190.572 . 6. To evaluate R_b 3f (z)dz

Figure 39 Z b3 f (z) dz = Z −5−→4+ f (z) dz + Z −5L994− f (z) dz = Z −5−→−3+ f (z) dz + Z −3−→−1+ f (z) dz + Z −1−→1+ f (z) dz + Z 1−→3+ f (z) dz + Z 3−→4+ f (z) dz + Z 3L994− f (z) dz + Z 1L993− f (z) dz + Z −1L991− f (z) dz + Z −3L99−1− f (z) dz + Z −5L99−3− f (z) dz We only need to evaluate the four integrals,R

1−→3+ f (z)dz, R 1L993− f (z)dz, R 3−→4+ f (z)dz, and R

3L994− f (z) dz. The other six integrals are as same as the six integrals evaluated in 5.

Theoretical Evaluation (1)Along 1−→ 3+ z = 3 + rei(−π)= 3 − r, r : 2 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√8 − r z + 3 > 0 =⇒ arg(z + 3) = 0 =⇒√z + 3 =√6 − r z + 1 > 0 =⇒ arg(z + 1) = 0 =⇒√z + 1 =√4 − r z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 = √2 − r z − 3 < 0 =⇒ arg(z − 3) = −π =⇒√z − 3 = −i√r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = −i√1 + r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = −i√3 + r Let u4(r) = √ 8 − r√6 − r√4 − r√2 − r√r√1 + r√3 + r. Then Z 1−→3+ f (z) dz = −(−i)3 Z 0 2 u4(r) dr = −i Z 0 2 u4(r) dr. (2)Along 1_{L99 3}− Z 1←−3− f (z) dz = −i3 Z 2 0 u4(r) dr = i Z 2 0 u4(r) dr.

Then Z 1L993− f (z) dz = − Z 1←−3− f (z) dz = −i Z 2 0 u4(r) dr = i Z 0 2 u4(r) dr = − Z 1−→3+ f (z) dz. From (1) and (2), we have

Z 1−→3+ f (z) dz + Z 1L993− f (z) dz = Z 1−→3+ f (z) dz +  − Z 1−→3+ f (z) dz  = 0. (43) (3)Along 3−→ 4+ z = 4 + rei(−π)= 4 − r, r : 1 −→ 0 =⇒ dz = −dr z + 5 > 0 =⇒ arg(z + 5) = 0 =⇒√z + 5 =√9 − r z + 3 > 0 =⇒ arg(z + 3) = 0 =⇒√z + 3 =√7 − r z + 1 > 0 =⇒ arg(z + 1) = 0 =⇒√z + 1 =√5 − r z − 1 > 0 =⇒ arg(z − 1) = 0 =⇒√z − 1 = √3 − r z − 3 > 0 =⇒ arg(z − 3) = 0 =⇒√z − 3 = √1 − r z − 4 < 0 =⇒ arg(z − 4) = −π =⇒√z − 4 = −i√r z − 6 < 0 =⇒ arg(z − 6) = −π =⇒√z − 6 = −i√2 + r Let u5(r) = √ 9 − r√7 − r√5 − r√3 − r√1 − r√r√2 + r. Then Z 3−→4+ f (z) dz = −(−i)2 Z 0 1 u5(r) dr = Z 0 1 u5(r) dr. (4)Along 3_{L99 4}− Z 3←−4− f (z) dz = −i2 Z 1 0 u5(r) dr = Z 1 0 u5(r) dr. Then Z 3L994− f (z) dz = − Z 3←−4− f (z) dz = − Z 1 0 u5(r) dr = Z 0 1 u5(r) dr = Z 3−→4+ f (z) dz. From (3) and (4), we have

Z 3−→4+ f (z) dz + Z 3L994− f (z) dz = 2 Z 3−→4+ f (z) dz = 2 Z 0 1 u5(r) dr. (44)

So, we have Z b3 f (z) dz = 2 Z −5−→−3+ f (z) dz + 2 Z −1−→1+ f (z) dz + 2 Z 3−→4+ f (z) dz = 2 Z 0 2 u1(r) dr +  −2 Z 0 2 u3(r) dr  + 2 Z 0 1 u5(r) dr = −291.688 + 101.116 + (−30.8213) = −221.393 . Using Mathematica (1)Along 1−→ 3+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = 0 =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 3) = −π =⇒ √_{z − 3 = (−1) · math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
arg(z − 6) = −π =⇒ √_{z − 6 = (−1) · math} √z − 6
. Z 1−→3+ f (z) dz = (−1)3_{· math} Z 3 1 f (z) dz  = −math Z 3 1 f (z) dz  (2)Along 1_{L99 3}− Z 1←−3− f (z) dz = math Z 1 3 f (z) dz  = −math Z 3 1 f (z) dz  = Z 1−→3+ f (z) dz Z 1L993− f (z) dz = − Z 1←−3− f (z) dz = − Z 1−→3+ f (z) dz From (1) and (2), Z 1−→3+ f (z) dz + Z 1L993− f (z) dz = Z 1−→3+ f (z) dz − Z 1−→3+ f (z) dz = 0 . (3)Along 3−→ 4+ arg(z + 5) = 0 =⇒ √_{z + 5 = math} √z + 5
arg(z + 3) = 0 =⇒ √_{z + 3 = math} √z + 3
arg(z + 1) = 0 =⇒ √_{z + 1 = math} √z + 1
arg(z − 1) = 0 =⇒ √_{z − 1 = math} √z − 1
arg(z − 3) = 0 =⇒ √_{z − 3 = math} √z − 3
arg(z − 4) = −π =⇒ √_{z − 4 = (−1) · math} √z − 4
√ √

Z 3−→4+ f (z) dz = (−1)2_{· math} Z 4 3 f (z) dz  = math Z 4 3 f (z) dz  (4)Along 3_{L99 4}− Z 3←−4− f (z) dz = math Z 3 4 f (z) dz  = −math Z 4 3 f (z) dz  = − Z 3−→4+ f (z) dz Z 3L994− f (z) dz = − Z 3←−4− f (z) dz = Z 3−→4+ f (z) dz From (3) and (4), Z 3−→4+ f (z) dz + Z 3L994− f (z) dz = Z 3−→4+ f (z) dz + Z 3−→4+ f (z) dz = 2 Z 3−→4+ f (z) dz = 2 · math Z 4 3 f (z) dz  = 2 · math Z 4 3 p (z + 5)(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6) dz  = −30.8213 . Thus, Z b3 f (z) dz = Z −5−→−3+ f (z) dz + Z −5L99−3− f (z) dz  + Z −1−→1+ f (z) dz + Z −1L991− f (z) dz  = + Z 3−→4+ f (z) dz + Z 3L994− f (z) dz  = −291.688 + 101.116 + (−30.8213) = −221.393 .

3.4

Generalization of Integrals Along Horizontal Cuts

We evaluate the integrals on the Riemann surface of genus N. If the Riemann surface is of genus N, then there are 2N + 1 or 2N + 2 branch points.

Figure 40 1 ≤ k ≤ N in Figure 40. Let f (z) =p(z − z1)(z − z2) · · · (z − z2N +1) = 2N +1 Y j=1 pz − zj. (1) To evaluateR_a kf (z)dz Figure 41 Theoretical Evaluation Along z2k + −→ z2k+1, let d = |z2k+1− z2k|. z = z2k+1+ rei(−π)= z2k+1− r, r : d −→ 0 =⇒ dz = −dr For j = 1, 2, · · · , 2k, arg(z − zj) = 0 =⇒pz − zj = q (z2k+1− r) − zj For j = 2k + 1, 2k + 2, · · · , 2N + 1, arg(z − zj) = −π =⇒pz − zj = −i q zj − (z2k+1− r) Let u(r) = 2k Y j=1 q (z2k+1− r) − zj ! _{2N +1} Y j=2k+1 q zj − (z2k+1− r) ! .

Then Z z2k + −→z2k+1 f (z) dz = −(−i)(2N +1)−(2k+1)+1 Z 0 d u(r) dr = −(−i)2N −2k+1 Z 0 d u(r) dr = i2N −2k+1 Z 0 d u(r) dr = i2N −2k· i Z 0 d u(r) dr = (i2)N −k· i Z 0 d u(r) dr = (−1)N −k· i Z 0 d u(r) dr Thus, Z ak f (z) dz = 2 Z z2k + −→z2k+1 f (z) dz (45) = (−1)N −k· 2i Z 0 d u(r) dr (46) Note that the value of the integral is a pure imaginary number.

Using Mathematica For j = 1, 2, · · · , 2k, arg(z − zj) = 0 =⇒pz − zj = math pz − zj
For j = 2k + 1, 2k + 2, · · · , 2N + 1, arg(z − zj) = −π =⇒pz − zj = (−1) · math pz − zj
Then, Z z2k + −→z2k+1 f (z) dz = (−1)(2N +1)−(2k+1)+1 _{· math} Z z2k+1 z2k f (z) dz  = (−1) · math Z z2k+1 z2k f (z) dz  Thus, Z ak f (z) dz = 2 Z z2k + −→z2k+1 f (z) dz (47) = (−2) · math Z z2k+1 z2k f (z) dz  (48) (2) To evaluateR_b kf (z)dz

Figure 42 Before our compuation, we first discuss the integrals of the two kinds of path drawn in Figure 43 and Figure 44.

Class 1. Along a path that there is a cut on it

Figure 43 Since f (z)|II = −f (z)|I, Z zm−1 − L99zm f (z) dz = − Z zm−1 − ←−zm f (z) dz = − Z zm−1 + −→zm f (z) dz So, we have Z zm−1−→z+ m f (z) dz + Z zm−1 − L99zm f (z) dz = 0. (49)

Class 2. Along a path that there is no cut on it

Figure 44 Since f (z)|II = −f (z)|I, Z zm−1 − L99zm f (z) dz = − Z zm−1 − ←−zm f (z) dz = −  − Z zm−1 + −→zm f (z) dz  = Z + f (z) dz

So, we have Z zm−1−→zm f (z) dz + Z zm−1L99zm f (z) dz = 2 Z zm−1−→zm f (z) dz. (50)

Thus, we only need to evaluate the integralsR

zj +

−→zj+1

f (z) dz for j = 1, 2, · · · , 2k − 1 and add them together. That is,

Z bk f (z) dz = 2 Z z1−→z2 f (z) dz + Z z3−→z4 f (z) dz + · · · + Z z2k−1−→z2k f (z) dz ! = 2 k X m=1 Z z2m−1−→z2m f (z) dz (51) Figure 45 Theoretical Evaluation Along z2m−1−→z2m, let d = |z2m− z2m−1|. z = z2m+ rei(−π)= z2m− r, r : d −→ 0 =⇒ dz = −dr For j = 1, 2, · · · , 2m − 1, arg(z − zj) = 0 =⇒pz − zj = q (z2m− r) − zj For j = 2m, 2m + 1, · · · , 2N + 1, arg(z − zj) = −π =⇒pz − zj = −i q zj − (z2m− r) Let u(r) = 2m−1 Y j=1 q (z2m− r) − zj ! _{2N +1} Y j=2m q zj − (z2m− r) ! . Then Z z2m−1−→z2m f (z) dz = −(−i)(2N +1)−2m+1 Z 0 d u(r) dr = −(−i)2N −2m+2 Z 0 d u(r) dr = (−1)N −m Z 0 d u(r) dr

Thus, Z bk f (z) dz = 2 k X m=1 Z z2m−1−→z2m f (z) dz (52) = 2 k/2 X m=1 (−1)N −m Z 0 d u(r) dr (53)

Note that the value of the integral is a real number. Using Mathematica For j = 1, 2, · · · , 2m − 1, arg(z − zj) = 0 =⇒pz − zj = math pz − zj
For j = 2m, 2m + 1, · · · , 2N + 1, arg(z − zj) = −π =⇒pz − zj = (−1) · math pz − zj
Then, Z z2m−1 + −→z2m f (z) dz = (−1)(2N +1)−2m+1_{· math} Z z2m z2m−1 f (z) dz  = math Z z2m z2m−1 f (z) dz  Thus, Z bk f (z) dz = 2 k X m=1 Z z2m−1−→z2m f (z) dz (54) = 2 k X m=1 math Z z2m z2m−1 f (z) dz  (55)

Case 2. The number of branch points is even (2N + 2 branch points)

1 ≤ k ≤ N in Figure 46. Let f (z) =p(z − z1)(z − z2) · · · (z − z2N +2) = 2N +2 Y j=1 pz − zj. (1) To evaluateR akf (z)dz Figure 47 Theoretical Evaluation Along z2k+1 + −→ z2k+2, let d = |z2k+2− z2k+1|. z = z2k+2+ rei(−π)= z2k+2− r, r : d −→ 0 =⇒ dz = −dr For j = 1, 2, · · · , 2k + 1, arg(z − zj) = 0 =⇒pz − zj = q (z2k+2− r) − zj For j = 2k + 2, 2k + 3, · · · , 2N + 2, arg(z − zj) = −π =⇒pz − zj = −i q zj − (z2k+2− r) Let u(r) = 2k+1 Y j=1 q (z2k+2− r) − zj ! _{2N +2} Y j=2k+2 q zj− (z2k+2− r) ! . Then Z z2k+1 + −→z2k+2 f (z) dz = −(−i)(2N +2)−(2k+2)+1 Z 0 d u(r) dr = −(−i)2N −2k+1 Z 0 d u(r) dr = (−1)N −k· i Z 0 d u(r) dr Thus, Z ak f (z) dz = 2 Z z2k+1 + −→z2k+2 f (z) dz (56) = (−1)N −k· 2i Z 0 d u(r) dr (57)

Note that the value of the integral is a pure imaginary number. Using Mathematica For j = 1, 2, · · · , 2k + 1, arg(z − zj) = 0 =⇒pz − zj = math pz − zj
For j = 2k + 2, 2k + 3, · · · , 2N + 2, arg(z − zj) = −π =⇒pz − zj = (−1) · math pz − zj
Then, Z z2k+1 + −→z2k+2 f (z) dz = (−1)(2N +2)−(2k+2)+1_{· math} Z z2k+2 z2k+1 f (z) dz ! = (−1) · math Z z2k+2 z2k+1 f (z) dz ! Thus, Z ak f (z) dz = 2 Z z2k+1 + −→z2k+2 f (z) dz (58) = (−2) · math Z z2k+2 z2k+1 f (z) dz ! (59) (2) To evaluateR_b kf (z)dz Figure 48 Using similar arguments in class 1 and class 2 of Case 1, we also can see that

Z bk f (z) dz = 2 Z z2−→z3 f (z) dz + Z z4−→z5 f (z) dz + · · · + Z z2k−→z2k+1 f (z) dz ! = 2 k X m=1 Z z2m−→z2m+1 f (z) dz (60)

Figure 49 Theoretical Evaluation Along z2m−→z2m+1, let d = |z2m+1− z2m|. z = z2m+1+ rei(−π)= z2m+1− r, r : d −→ 0 =⇒ dz = −dr For j = 1, 2, · · · , 2m, arg(z − zj) = 0 =⇒pz − zj = q (z2m− r) − zj For j = 2m + 1, 2m + 2, · · · , 2N + 2, arg(z − zj) = −π =⇒pz − zj = −i q zj − (z2m− r) Let u(r) = 2m Y j=1 q (z2m− r) − zj ! _{2N +2} Y j=2m+1 q zj− (z2m− r) ! . Then Z z2m−→z2m+1 f (z) dz = −(−i)(2N +2)−(2m+1)+1 Z 0 d u(r) dr = −(−i)2N −2m+2 Z 0 d u(r) dr = (−1)N −m Z 0 d u(r) dr Thus, Z bk f (z) dz = 2 k X m=1 Z z2m−→z2m+1 f (z) dz (61) = 2 k X m=1 (−1)N −m Z 0 d u(r) dr (62) Note that the value of the integral is a real number.

Using Mathematica For j = 1, 2, · · · , 2m, arg(z − zj) = 0 =⇒pz − zj = math pz − zj
For j = 2m + 1, 2m + 2, · · · , 2N + 2, arg(z − zj) = −π =⇒pz − zj = (−1) · math pz − zj

Then, Z z2m−→z2m+1 f (z) dz = (−1)(2N +2)−(2m+1)+1_{· math} Z z2m+1 z2m f (z) dz  = math Z z2m+1 z2m f (z) dz  Thus, Z bk f (z) dz = 2 k X m=1 Z z2m−→z2m+1 f (z) dz (63) = 2 k X m=1 math Z z2m+1 z2m f (z) dz  (64) Example 8. Let f (z) = 2N +2 Y j=1 pz − zj.

Suppose that Im(z2j−1) = Im(z2j), j = 1, 2, · · · , N + 1. The cuts are drawn in Figure 50.

Let Re(zj) = xj, j = 1, 2, · · · , 2N + 2 and Im(z2j−1) = Im(z2j) = yj, j = 1, 2, · · · , N + 1.

Evaluate R_a

kf (z)dz and R

Figure 51 Figure 52 Solution. (1) To evaluateR_a kf (z)dz (Figure 51) Theoretical Evaluation Along z2k+1 + −→ z2k+2, let d = |z2k+2− z2k+1|. z = z2k+2+ rei(−π)= z2k+2− r, r : d −→ 0 =⇒ dz = −dr For j = 2k + 2, arg(z − z2k+2) = −π =⇒ √ z − z2k+2= −i √ r For other j, arg(z − zj) 6= −π =⇒pz − zj = q (z2k+2− r) − zj Let u(r) = 2k+1 Y j=1 q (z2k+2− r) − zj ! √ r
2N +2 Y j=2k+3 q zj − (z2k+2− r) ! . Then Z z2k+1 + −→z2k+2 f (z) dz = −(−i) Z 0 d u(r) dr = i Z 0 d u(r) dr Thus, Z ak f (z) dz = 2i Z 0 d u(r) dr .

Using Mathematica Along z2k+1 + −→ z2k+2. For j = 2k + 2, arg(z − z2k+2) = −π =⇒ √ z − z2k+2= (−1) · math √ z − z2k+2
For other j, arg(z − zj) 6= −π =⇒pz − zj = math pz − zj
Then, Z z2k+1 + −→z2k+2 f (z) dz = (−1) · math Z z2k+2 z2k+1 f (z) dz ! Thus, Z ak f (z) dz = 2 Z z2k+1 + −→z2k+2 f (z) dz = (−2) · math Z z2k+2 z2k+1 f (z) dz ! (2) To evaluateR_b kf (z)dz (Figure 52) For j = 2, 3, · · · , k, Z z2j−1 + 99Kz2j f (z) dz + Z z2j−1 − ←−z2j f (z) dz = − Z z2j−1 + −→z2j f (z) dz + Z z2j−1 + −→z2j f (z) dz = 0 . So, Z bk f (z) dz = Z z1 + 99Kz2 f (z) dz + Z z1 − L99z2 f (z) dz  + k X j=1 Z z2j−1−→z2j+2 f (z) dz + Z z2j−1L99z2j+2 f (z) dz ! + Z z2k+1 + −→z2k+2 f (z) dz + Z z2k+1 − ←−z2k+2 f (z) dz ! = (−1) · 2 Z z1 + −→z2 f (z) dz + 2 k X j=1 Z z2j−1−→z2j+2 f (z) dz = 2i Z 0 d 2k+1 Y q (z2k+2− r) − zj ! √ r
2N +2 Y q zj − (z2k+2− r) ! dr .

Theoretical Evaluation Along z1 + −→ z2, let d = |z2− z1|. z = z2+ rei(−π)= z2− r, r : d −→ 0 =⇒ dz = −dr For j = 2, arg(z − z2) = −π =⇒ √ z − z2 = −i √ r For other j, arg(z − zj) 6= −π =⇒pz − zj = q (z2 − r) − zj Let u(r) =p(z2− r) − z1  √ r
2N +2 Y j=3 q zj − (z2− r) ! . Then Z z1−→z+ 2 f (z) dz = −(−i) Z 0 d u(r) dr = i Z 0 d u(r) dr Along z2j−1 + −→ z2j+2, let θ = − arctan|z2j−1− z2j+1| |z2j+1− z2j+2| , d = |z2j−1− z2j+2|. z = z(r) = z2j−1+ reiθ, r : d −→ 0 =⇒ dz = eiθdr Then, Z z2j−1−→z2j+2 f (z) dz = Z d 0 2N +2 Y j=1 q |z(r) − zj| eiθdr . Thus, Z bk f (z) dz = (−1) · 2 Z z1 + −→z2 f (z) dz + 2 k X j=1 Z z2j−1−→z2j+2 f (z) dz + Z ak f (z) dz = (−1) · 2i Z 0 d u(r) dr + 2 k X j=1 Z d 0 2N +2 Y j=1 q |z(r) − zj| eiθdr + Z ak f (z) dz . Using Mathematica Along z1 + −→ z2, let d = |z2− z1|. For j = 2, arg(z − z2) = −π =⇒ √ z − z2 = (−1) · math √ z − z2
For other j, arg(z − zj) 6= −π =⇒pz − zj = math pz − zj

Then, Z z1 + −→z2 f (z) dz = (−1) · math Z z2 z1 f (z) dz  Along z2j−1 + −→ z2j+2, let θ = − arctan|z2j−1− z2j+1| |z2j+1− z2j+2| , d = |z2j−1− z2j+2|. z = z2j−1+ reiθ, r : d −→ 0 =⇒ dz = eiθdr Then, Z z2j−1−→z2j+2 f (z) dz = math Z d 0 f (z2j−1+ reiθ)eiθdr  . Thus, Z bk f (z) dz = (−1) · 2 Z z1 + −→z2 f (z) dz + 2 k X j=1 Z z2j−1−→z2j+2 f (z) dz + Z ak f (z) dz = (−1) · 2 · (−1) · math Z z2 z1 f (z) dz  + 2 k X j=1 math Z d 0 f (z2j−1+ reiθ)eiθdr  + (−2) · math Z z2k+2 z2k+1 f (z) dz ! = 2 · math Z z2 z1 f (z) dz  + 2 k X j=1 math Z d 0 f (z2j−1 + reiθ)eiθdr  + (−2) · math Z z2k+2 z2k+1 f (z) dz ! .

4

Integrals for Vertical Cuts

4.1

Cut Structures for Vertical Cuts

We first define the branches for vertical cuts. Let f (z) = √z and let z = reiθ_{, where}

θ = arg z. We define two single-valued branches of f as f (z) =√re12iθ, −3π 2 ≤ θ < π 2, and f (z) =√re12iθ, π 2 ≤ θ < 5π 2 . And we define sheet I and sheet II as

sheet I = {z ∈ C| −3π 2 ≤ arg z < π 2}, and sheet II = {z ∈ C|π 2 ≤ arg z < 5π 2 }.

To Label the second quadrant with a + and label the first quadrant with a −.

Figure 53

Then we can use the same method used in section 2.1 to construct the Riemann surface for f. (see Figure 6)

4.2

The Problem in Using Mathematica

We use (I) to denote sheet I and (II) to denote sheet II. We can see z ∈ (I) =⇒ −3π 2 ≤ arg z < π 2 =⇒ − 3π 4 ≤ 1 2arg z < π 4. f maps the points on sheet I into the region {z ∈ C| −3π4 ≤ arg z <

π 4}.