International Refereed Journal of Engineering and Science (IRJES) ISSN (Online) 2319-183X, (Print) 2319-1821 Volume 2, Issue 2(February 2013), PP.25-29 www.irjes.com

Schwarz-Christoffel Transformation On A Half Plane Abdualah Ibrahim Sultan Mathematics Department of Brawijaya University Malang, Indonesia. Mathematics Department of, Faculty of Science, Khoms, Al-Mergeb University , Khoms, Libya.

ABSTRACT :The Schwarz-Christoffel transformation has been studied by many reseachers due to a huge of its application in solving Laplace's equations and studying fluid flow phenomenas ( see [3],[4]). The proof of this theorem can find in (see for example [5]). Their method depend on the tangent of đ?&#x2018;&#x2021;đ?&#x2018;&#x2014; = 1 + đ?&#x2018;&#x2013;0 and the rotation of đ?&#x2018;&#x2021;đ?&#x2018;&#x2014; from 0 to đ?&#x153;&#x2039; on points (đ?&#x2018;Ľ, 0) . In this paper we would like to prove the ShwarzChristoffeltransformation directly by using the arguments identity of complex number, rotation of it from 0 to 2đ?&#x153;&#x2039; and conformal mapping. Keywords : Complex function, half plane, polygon, analytic function, transformation, Conformal mapping.

I.

INTRODUCTION

The Schwarz-Christoffel transformation known as a map from the upper half-plane to simply connected polygon with or without corners at infinity in the complex planes.The Schwarz-Christoffel transformation has been studied by many reseachers due to a huge of its application for example used in physical applications involving fluid movement, heat conduction and electrostatic pontential, as well asthe studying of Laplace's equations (see [3],[4]) and also from mathematical point of views ( see [1],[2],[6]). In this paper we study about the proof of the Schwarz-Christoffel theorem. The prove of this theorem can found in (see for example [5]). Their method depend on the tangent of đ?&#x2018;&#x2021;đ?&#x2018;&#x2014; = 1 + đ?&#x2018;&#x2013;0 and the rotation of đ?&#x2018;&#x2021;đ?&#x2018;&#x2014; from 0 to đ?&#x153;&#x2039; on points (đ?&#x2018;Ľ, 0). In this paper we would like to prove the Shwarz-Christoffel transformation directly by using the arguments identity of complex number, rotation of it from 0 to 2đ?&#x153;&#x2039; and conformal mapping.

II.

PRELIMINARIES

We need to review some important definitions and theorems which will be used later from complex analysis to understand the Schwarz-Christoffeltheorem.These definitions and theorems below can be found in the refences ( see [3], [4] ). Definition 2.1 A complex number is a number of the form đ?&#x2018;§ = đ?&#x2018;&#x17D; + đ?&#x2018;&#x2013;đ?&#x2018;? where the imaginary unit is defined as đ?&#x2018;&#x2013; = â&#x2C6;&#x2019;1andđ?&#x2018;&#x17D; is the real part of đ?&#x2018;§, đ?&#x2018;&#x17D; = đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;(đ?&#x2018;§), and đ?&#x2018;? is the imaginary part of đ?&#x2018;§ , đ?&#x2018;? = đ??źđ?&#x2018;&#x161;(đ?&#x2018;§). The set of all complex number is written by đ??ś. Definition 2.2. Arguments of complex number Let r and Î¸ be polar coordinates of the point đ?&#x2018;&#x17D;, đ?&#x2018;? that corresponds to a nonzero complex number đ?&#x2018;§ = đ?&#x2018;&#x17D; + đ?&#x2018;&#x2013;đ?&#x2018;?. Since đ?&#x2018;&#x17D; = cos đ?&#x153;&#x192;and đ?&#x2018;? = sin đ?&#x153;&#x192;, the number đ?&#x2018;§can be written in polar form as đ?&#x2018;§ = r (cos đ?&#x153;&#x192; + đ?&#x2018;&#x2013; sin đ?&#x153;&#x192;) = đ?&#x2018;§ đ?&#x2018;&#x2019; đ?&#x2018;&#x2013;đ?&#x153;&#x192; , where đ?&#x2018;§ is a positive real number called the modulus of đ?&#x2018;§, and đ?&#x153;&#x192; is real number called the argument of đ?&#x2018;§, and written by arg đ?&#x2018;§.The real numberđ?&#x153;&#x192; represents the angle and measured in radians. Theorem 2.1. Law of arguments of products . Let đ?&#x2018;?, đ?&#x2018;? â&#x2C6;&#x2C6; đ??ś. Then (1.1) arg đ?&#x2018;?đ?&#x2018;? = arg b + arg c. Theorem 2.2. Law of arguments of power Let đ?&#x2018;?, đ?&#x2018;? â&#x2C6;&#x2C6; đ??ś. Then (1.2) arg đ?&#x2018;? đ?&#x2018;? = đ?&#x2018;? arg đ?&#x2018;?. Definition 2.3.Complex Function. A complex function is a function đ?&#x2018;&#x201C; whose domain and range are subsets of the set đ??ś of complex numbers. Definition 2.4.Conformal mapping. Let đ?&#x2018;¤ = đ?&#x2018;&#x201C; (đ?&#x2018;§) be a complex mapping defined in a domain đ??ˇ and let đ?&#x2018;§0 be a point in đ??ˇ. Then we say that đ?&#x2018;¤ = đ?&#x2018;&#x201C; (đ?&#x2018;§) is conformal at đ?&#x2018;§0 if for every pair of smoothoriented curvesđ??ś1 and đ??ś2 in đ??ˇ intersecting at đ?&#x2018;§0 the angle between đ??ś1 and đ??ś2 atđ?&#x2018;§0 is equal tothe angle between the image curves đ??ś â&#x20AC;˛ 1 and đ??ś â&#x20AC;˛ 2 at đ?&#x2018;&#x201C; (đ?&#x2018;§0 ) in both magnitude and sense.

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Schwarz-Christoffel Transformation On A Half Plane

Figure 1.1 The angle betweenđ??ś1 and đ??ś2 (see [4]) In other words a mapping đ?&#x2018;¤ = đ?&#x2018;&#x201C; (đ?&#x2018;§) preserves both angle and shape but it cannot in general the size. Moreover a mapping that is conformal at every point in a domain D is called conformal in D, the conformal mapping relies on the properties of analytic function. The angle between any two intersecting arcs in the z-plane is equal to the angle between the images of the arcs in the w-plane under a linear mapping. Definition 2.5. Analytic Function.A function đ?&#x2018;&#x201C; (đ?&#x2018;§) is said to be analytic in a region đ?&#x2018;&#x2026; of the complex plane if đ?&#x2018;&#x201C; (đ?&#x2018;§) has a derivative at each point of đ?&#x2018;&#x2026;. In addition, analytic function is conformal at đ?&#x2018;§0 if and only if đ?&#x2018;&#x201C; â&#x20AC;˛ (đ?&#x2018;§) â&#x2030; 0. Definition 2.6.Polygon. A polygon is a plane figure that is bounded by closed path, composed of a finite sequence of straight-line segments (i.e. a closed polygonal chain or circuit). These segments are called its edges or sides, and the points where two edges meet are the polygonâ&#x20AC;&#x2122;s vertices (singular: vertex) or corners.

III.

MAIN THEOREM

Theorem (Schwarz-christoffel) Let đ?&#x2018;&#x192; be a polygon in the đ?&#x2018;¤-plane with vertices đ?&#x2018;¤1 , đ?&#x2018;¤2 , â&#x20AC;Ś , đ?&#x2018;¤đ?&#x2018;&#x203A; and exterior angles đ?&#x203A;źđ?&#x2018;&#x2DC; , where â&#x20AC;&#x201C; đ?&#x153;&#x2039; < đ?&#x203A;źđ?&#x2018;&#x2DC; < đ?&#x153;&#x2039; .There exists a one-to-one conformal mapping đ?&#x2018;¤ = đ?&#x2018;&#x201C; (đ?&#x2018;§) from the upper half plane, đ??źđ?&#x2018;&#x161; đ?&#x2018;§ > 0 onto đ??ş that satisfies the boundary conditions in equations đ?&#x2018;¤đ?&#x2018;&#x2DC; = đ?&#x2018;&#x201C; (đ?&#x2018;Ľđ?&#x2018;&#x2DC; ) for đ?&#x2018;&#x2DC; = 1,2 â&#x20AC;Ś . đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1 and đ?&#x2018;¤đ?&#x2018;&#x203A; = đ?&#x2018;&#x201C; â&#x2C6;&#x17E; , where đ?&#x2018;Ľ1 < đ?&#x2018;Ľ2 < đ?&#x2018;Ľ3 < â&#x2039;Ż < đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 < â&#x2C6;&#x17E;. Derivative đ?&#x2018;&#x201C; â&#x20AC;˛ (đ?&#x2018;§) is â&#x2C6;&#x2019; đ?&#x203A;ź1 â&#x2C6;&#x2019; đ?&#x203A;ź2 â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 (1.3) đ?&#x2018;&#x201C; â&#x20AC;˛ đ?&#x2018;§ = đ??´ đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 ) đ?&#x153;&#x2039; đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 đ?&#x153;&#x2039; â&#x20AC;Ś đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x153;&#x2039; and the function f can be expressed as an indefinite integral đ?&#x2018;&#x201C; đ?&#x2018;§ = đ??ľ + đ??´ â&#x2C6;Ť đ?&#x2018;&#x201C; â&#x20AC;˛ đ?&#x2018;§ = đ??´ đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 )

â&#x2C6;&#x2019; đ?&#x203A;ź1 đ?&#x153;&#x2039;

đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2

â&#x2C6;&#x2019; đ?&#x203A;ź2 đ?&#x153;&#x2039;

â&#x20AC;Ś đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A; â&#x2C6;&#x2019;1

â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x153;&#x2039;

(1.4)

đ?&#x2018;&#x2018;đ?&#x2018;§

where đ??´ and đ??ľ are suitably chosen constants. Two of the points {đ?&#x2018;Ľđ?&#x2018;&#x2DC; } may be chosen arbitrarily, and the constants đ??´ and đ??ľ determine the size and position of đ?&#x2018;&#x192;.

IV.

PROVE OF THE THEOREM

A transformation đ?&#x2018;¤ = đ?&#x2018;&#x201C; đ?&#x2018;§ constructed by mapping each point on the đ?&#x2018;Ľ axis in đ?&#x2018;§ -plane z on đ?&#x2018;¤ -plane, with đ?&#x2018;Ľ1 , đ?&#x2018;Ľ2 , â&#x20AC;Ś , đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 , đ?&#x2018;Ľđ?&#x2018;&#x203A; are the points on the đ?&#x2018;Ľ-axis and đ?&#x2018;Ľ1 < đ?&#x2018;Ľ2 < đ?&#x2018;Ľ3 < â&#x2039;Ż < đ?&#x2018;Ľđ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 < đ?&#x2018;Ľđ?&#x2018;&#x203A; where the points on the plane đ?&#x2018;§ is the domain of the transformation. These points are mapped in to points on the đ?&#x2018;&#x203A;-side, is đ?&#x2018;¤đ?&#x2018;&#x2014; = đ?&#x2018;&#x201C;(đ?&#x2018;Ľđ?&#x2018;&#x2014; ), đ?&#x2018;&#x2014; = 1, 2, . . . , đ?&#x2018;&#x203A; -1 and đ?&#x2018;¤đ?&#x2018;&#x203A; = đ?&#x2018;&#x201C; â&#x2C6;&#x17E; in đ?&#x2018;¤-plane. The function đ?&#x2018;&#x201C;chosen so that arg đ?&#x2018;&#x201C; â&#x20AC;˛ đ?&#x2018;§ different constant value (1.5) đ?&#x2018;&#x201C; â&#x20AC;˛ đ?&#x2018;§ = đ??´(đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 )â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; 1 (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 )â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; 2 â&#x20AC;Ś (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 )â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 where A is a complex constant and each đ?&#x2018;&#x2DC;đ?&#x2018;&#x2014; , with đ?&#x2018;&#x2014; = 1, 2, 3, đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1 are real constants. From equation (1.5) we taking the absolute value, then we have đ?&#x2018;&#x201C; â&#x20AC;˛ đ?&#x2018;§ = đ??´(đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 )â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; 1 (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 )â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; 2 â&#x20AC;Ś (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 )đ?&#x2018;&#x2DC; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; 1 â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; 2 = đ??´ (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 ) (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 ) â&#x20AC;Ś (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 )â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ??´ 1 1 = đ?&#x2018;&#x2DC;2 â&#x20AC;Ś đ?&#x2018;&#x2DC; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 , đ?&#x2018;&#x2DC;1 (đ?&#x2018;§â&#x2C6;&#x2019;đ?&#x2018;Ľ 1 )

(đ?&#x2018;§â&#x2C6;&#x2019;đ?&#x2018;Ľ 2 )

(đ?&#x2018;§â&#x2C6;&#x2019;đ?&#x2018;Ľ đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 )

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Schwarz-Christoffel Transformation On A Half Plane −𝑘

𝜋

3𝜋

by using (𝑧 − 𝑥𝑗 )−𝑘 𝑗 = 𝑧 − 𝑥𝑗 𝑗 exp −𝑖𝑘𝑗 𝜃𝑗 , (− < 𝜃𝑗 < 2 1, then 𝑓 ′ 𝑧 is analytic everywhere in the half plane, so 𝑓′ 𝑧 = <

𝐴 𝑒

−𝑖𝑘

𝑧−𝑥 1 𝐴 𝑧−𝑥 1

1 𝜃1 𝑘1

𝑒

−𝑖𝑘

−𝑖𝑘

1𝑗

𝑘2 𝜃𝑗

𝑘2

𝑧−𝑥 2 1

𝑘 1 𝑧−𝑥 2

By using 𝑒

2 𝜃2

=

𝑒 −𝑖𝑘 𝑛 −1

−𝑖𝑘

1 𝜃1

𝑒

−𝑖𝑘

2 𝜃2

(𝑧−𝑥 2 )𝑘 2

(𝑧−𝑥 1 )𝑘 1

𝑒 −𝑖𝑘 𝑛 −1

𝜃 𝑛 −1

(𝑧−𝑥 𝑛 −1 )𝑘 𝑛 −1

𝜃 𝑛 −1 𝑘 𝑛 −1

𝑧−𝑥 𝑛 −1 1

(1.6)

𝑘 𝑛 −1

𝑧−𝑥 𝑛 −1

=1

𝐴𝑒

)where𝜃𝑗 = arg 𝑧 − 𝑥𝑗 and j = 1, 2,..., n –

2

and 2

< 𝑧 − 𝑥𝑗 < 2 𝑍 and 1 1 2 < < 2𝑍 𝑍 𝑍 − 𝑥𝑗

𝑍

to equation (1.6) we have that 2𝐴 1 1 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1.6 < < … 𝑘 𝑘 1 2 𝑧 𝑧 𝑧 𝑘 𝑛 −1 2𝐴 = 𝑘 1 𝑘 2 +⋯+𝑘 𝑛 −1 . Since 𝑘1 + 𝑘2 + ⋯ + 𝑘𝑛−1 + 𝑥𝑛 = 2,−1< 𝑘𝑗 < 1 , (𝑗 = 1.2, … , 𝑛) then 𝑧

2𝐴 𝑧

𝑘 1 𝑘 2 +⋯+𝑘 𝑛 −1

2𝐴

=

𝑧

2−𝑘 𝑛

. It means, that there exist 𝑀 = 2𝐴 > 0 such that 𝑓′ 𝑧

𝑀

Using equation (1.7) and

𝑧

lim lim 𝐹 𝑧 = 𝑧→∞ 𝑧→∞ ≤

lim 𝑧→∞

𝑧0 𝑧

𝑧0

𝑓 ′ 𝑠 𝑑𝑠

≤ lim

𝑧 →∞ 𝑧0

𝑧

lim 𝑧→∞

𝑀

𝑧0

𝑧→∞

1 2−𝑘 𝑛

𝑑𝑠

2−𝑘 𝑛

𝑧 𝑧

1

0 𝑧

2−𝑘 𝑛

= 𝑀 lim ∫𝑧 𝑧 ∫𝑧 𝑧 0

(1.8) 𝑓 ′ 𝑠 𝑑𝑠

𝑓 ′ 𝑠 𝑑𝑠 𝑧

since 2 − 𝑘𝑛 > 1, then

(1.7)

.

2−𝑘 𝑛

𝑧

𝑑𝑠,

𝑑𝑠 → 0, that means the limit of the integral (1.7) exists as 𝑧 tend to infinity .

So there exists a number 𝑤𝑛 such that

𝑓 ∞ =: lim 𝐹(∞) = 𝑤𝑛 ,𝐼𝑚 𝑧 ≥ 0 . Next we will prove the equation 𝑧→∞

(1.3). After taking the argument to the equation (1.5) we have arg 𝑓 ′ 𝑧 = arg 𝐴 − 𝑘1 arg 𝑧 − 𝑥1 − 𝑘2 arg 𝑧 − 𝑥2 − … − 𝑘𝑛−1 arg⁡ (𝑧 − 𝑥𝑛−1 ) then by using identity (1.1),(1.2) yields arg 𝑓 ′ 𝑧 = arg 𝐴 + arg(𝑧 − 𝑥1 )−𝑘 1 (𝑧 − 𝑥2 )−𝑘 2 − ⋯ − arg(𝑧 − 𝑥𝑛−1 )−𝑘 𝑛 −1 arg 𝑓 ′ 𝑧 = arg 𝐴 (𝑧 − 𝑥1 )−𝑘 1 (𝑧 − 𝑥2 )−𝑘 2 … (𝑧 − 𝑥𝑛 −1 )−𝑘 𝑛 −1 𝑓 ′ 𝑧 = 𝐴 (𝑧 − 𝑥1 )−𝑘 1 (𝑧 − 𝑥2 )−𝑘 2 … (𝑧 − 𝑥𝑛−1 )−𝑘 𝑛 −1 −𝛼 1 Let −𝑘1 = then we have 𝑥

𝑓′ 𝑧 =

𝐴

(𝑧 − 𝑥1 )

𝑓′ 𝑧 = 𝑓 𝑧 +𝑐 = 𝑓 𝑧 =

− 𝛼1 𝑥

𝐴 (𝑧 − 𝑥1 )

− 𝛼1 𝑥

𝐴 (𝑧 − 𝑥1 )

𝐴 (𝑧 − 𝑥1 )

𝐴 (𝑧 − 𝑥1 )

(𝑧 − 𝑥2 )

− 𝛼1 𝑥

− 𝛼1 𝑥

If – 𝑐 = 𝐵 then 𝑓 𝑧 =𝐵+

− 𝛼2 𝑥

(𝑧 − 𝑥2 )

− 𝛼1 𝑥

− 𝛼2 𝑥

(𝑧 − 𝑥2 )

− 𝛼2 𝑥

− 𝛼2 𝑥

(𝑧 − 𝑥2 )

(𝑧 − 𝑥2 )

(𝑧 − 𝑥𝑛 −1 )

… (𝑧 − 𝑥𝑛 −1 )

− 𝛼 𝑛 −1 𝑥

… (𝑧 − 𝑥𝑛 −1 )

… (𝑧 − 𝑥𝑛 −1 )

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, and

− 𝛼 𝑛 −1 𝑥

− 𝛼 𝑛 −1 𝑥

… (𝑧 − 𝑥𝑛 −1 )

− 𝛼2 𝑥

− 𝛼 𝑛 −1 𝑥

−𝑐

− 𝛼 𝑛 −1 𝑥

𝑑𝑧

27 | Page

Schwarz-Christoffel Transformation On A Half Plane â&#x2C6;&#x2019; đ?&#x203A;ź1

â&#x2C6;&#x2019; đ?&#x203A;ź2

â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1

= đ??ľ + â&#x2C6;Ť đ??´ (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 ) đ?&#x2018;Ľ (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 ) đ?&#x2018;Ľ â&#x20AC;Ś (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 ) đ?&#x2018;Ľ đ?&#x2018;&#x2018;đ?&#x2018;§. Next, since the argument of a product of the complex numbers are equal to the number of arguments of each factor, from equation (1.5) we have (1.9) argđ?&#x2018;&#x201C; â&#x20AC;˛ đ?&#x2018;§ = argđ??´ â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;1 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;2 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 . The real numbers đ?&#x2018;Ľ1 , đ?&#x2018;Ľ2 , â&#x20AC;Ś , đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 , đ?&#x2018;Ľđ?&#x2018;&#x203A; take on the real axis z-plane. If đ?&#x2018;§ is a real number, then the number (1.10) đ?&#x2018; đ?&#x2018;&#x2014; = đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x2014; , is positive if đ?&#x2018;§ is greater than đ?&#x2018;Ľđ?&#x2018;&#x2014; and negative ifđ?&#x2018;§ is less than đ?&#x2018;Ľđ?&#x2018;&#x2014; and 0, if đ?&#x2018;§ > đ?&#x2018;Ľđ?&#x2018;&#x2014; (1.11) arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x2014; = đ?&#x153;&#x2039;, if đ?&#x2018;§ < đ?&#x2018;Ľđ?&#x2018;&#x2014; . If đ?&#x2018;§ = đ?&#x2018;Ľ and đ?&#x2018;Ľ < đ?&#x2018;Ľđ?&#x2018;&#x2014; , đ?&#x2018;&#x2014; = 1,2,3, â&#x20AC;Ś , đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1 then arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 = arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 = â&#x2039;Ż = arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 = đ?&#x153;&#x2039;.

(1.12)

If đ?&#x2018;§ = đ?&#x2018;Ľ and đ?&#x2018;Ľđ?&#x2018;&#x;â&#x2C6;&#x2019;1 < đ?&#x2018;Ľ < đ?&#x2018;Ľđ?&#x2018;&#x; , j = 1, 2, ..., r â&#x20AC;&#x201C; 1, r, r + 1, ..., n â&#x20AC;&#x201C;1 and let đ?&#x153;&#x2122; đ?&#x2018;&#x;â&#x2C6;&#x2019;1 = argđ?&#x2018;&#x201C; â&#x20AC;˛ (đ?&#x2018;§)

(1.13)

then according to equation (1.9) and (1.11) and (1.13) argđ?&#x2018;&#x201C; â&#x20AC;˛ đ?&#x2018;§ = argđ??´ â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;1 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ1 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;2 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľ2 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 argâ Ą (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 ) đ?&#x153;&#x2122; đ?&#x2018;&#x;â&#x2C6;&#x2019;1 = argđ??´ â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x; arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x; â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x;+1 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 argâ Ą (đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 ) = argđ??´ â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x; đ?&#x153;&#x2039; â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 đ?&#x153;&#x2039; â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+2 đ?&#x153;&#x2039; â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x153;&#x2039; = argđ??´ â&#x2C6;&#x2019; (đ?&#x2018;&#x2DC;đ?&#x2018;&#x; â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+2 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 )đ?&#x153;&#x2039; and đ?&#x153;&#x2122; đ?&#x2018;&#x; = argđ??´ â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x;+1 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+2 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x;+2 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 arg đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;Ľđ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 = argđ??´ â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 đ?&#x153;&#x2039; â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+2 đ?&#x153;&#x2039; â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+3 đ?&#x153;&#x2039; â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x153;&#x2039; = argđ??´ â&#x2C6;&#x2019; (đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+2 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+3 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 )đ?&#x153;&#x2039; Then đ?&#x153;&#x2122; đ?&#x2018;&#x; â&#x2C6;&#x2019; đ?&#x153;&#x2122; đ?&#x2018;&#x; â&#x2C6;&#x2019;1 = [argđ??´ â&#x2C6;&#x2019; (đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+2 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+3 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 )đ?&#x153;&#x2039;] â&#x2C6;&#x2019; [argđ??´ â&#x2C6;&#x2019; (đ?&#x2018;&#x2DC;đ?&#x2018;&#x; â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+1 â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x;+2 â&#x2C6;&#x2019; â&#x2039;Ż â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 )đ?&#x153;&#x2039;] đ?&#x153;&#x2122;

đ?&#x2018;&#x;

â&#x2C6;&#x2019;đ?&#x153;&#x2122;

đ?&#x2018;&#x; â&#x2C6;&#x2019;1 =đ?&#x2018;&#x2DC;đ?&#x2018;&#x; đ?&#x153;&#x2039;

(1.14)

(1.15) (1.16)

,with đ?&#x2018;&#x2014; = 1,2,3, â&#x2039;Ż , đ?&#x2018;&#x; â&#x2C6;&#x2019; 1, đ?&#x2018;&#x;, â&#x2039;Ż , đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1.

Such that, if đ?&#x2018;§ = đ?&#x2018;Ľ then đ?&#x2018;Ľđ?&#x2018;&#x2014; â&#x2C6;&#x2019;1 < đ?&#x2018;Ľ < đ?&#x2018;Ľđ?&#x2018;&#x2014; , đ?&#x2018;&#x2014; = 1,2, â&#x20AC;Ś , đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1 in the real axis đ?&#x2018;§-plane from the left point đ?&#x2018;Ľđ?&#x2018;&#x2014; to his right, then the vectorđ?&#x153;? in đ?&#x2018;¤-plane change with angle đ?&#x2018;&#x2DC;đ?&#x2018;&#x2014; đ?&#x153;&#x2039;, on the point of the image đ?&#x2018;Ľđ?&#x2018;&#x2014; ,as shown in Figure 1.2. angle đ?&#x2018;&#x2DC;đ?&#x2018;&#x2014; đ?&#x153;&#x2039; is outside the terms of the đ?&#x2018;§-plane at the pointđ?&#x2018;&#x160;đ?&#x2018;&#x2014; .

V

Y

W1 Wn

W

W2 k 2ď ° z

W4

ď ´ W3

x1

x2

t

x 3 x4

x n ď&#x20AC;­1

X

k 3ď ° U

Figure 1.2. Mapping đ?&#x2018;&#x201C; (đ?&#x2018;§) with đ?&#x2018;§ = đ?&#x2018;Ľ and đ?&#x2018;Ľđ?&#x2018;&#x2014; â&#x2C6;&#x2019;1 < đ?&#x2018;Ľ < đ?&#x2018;Ľđ?&#x2018;&#x2014; (see [3]) It is assumed that the sides of the đ?&#x2018;§-plane do not intersect each other and take opposite corners clockwise. Outside corners can be approximated by the angle betweenđ?&#x153;&#x2039; and - đ?&#x153;&#x2039;, so that â&#x2C6;&#x2019;1 < đ?&#x2018;&#x2DC;đ?&#x2018;&#x2014; < 1, If the terms of n-closed, the number of outer corner is 2đ?&#x153;&#x2039;.To prove it is seen to have sided closed (n + 1), see Figure 1.3

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Schwarz-Christoffel Transformation On A Half Plane V

k 2ď °

k1ď °

W2

W1 k n ď&#x20AC;Ť1ď °

Wn

k nď °

Wn ď&#x20AC;Ť1 U

Figure 1.3.Closed đ?&#x2018;&#x203A;-image has (đ?&#x2018;&#x203A; + 1) plane ( see [3]). It has known that đ?&#x2018;&#x2DC;đ?&#x2018;&#x2014; đ?&#x153;&#x2039; is the magnitude of the outer corner at the point đ?&#x2018;&#x160;đ?&#x2018;&#x2014; which is the image of a point đ?&#x2018;Ľđ?&#x2018;&#x2014; .If many terms have side n + 1, then đ?&#x2018;&#x2DC; 2 đ?&#x153;&#x2039; = đ?&#x153;&#x2122;2 â&#x2C6;&#x2019; đ?&#x153;&#x2122;1 đ?&#x2018;&#x2DC; 3 đ?&#x153;&#x2039; = đ?&#x153;&#x2122;3 â&#x2C6;&#x2019; đ?&#x153;&#x2122;2 đ?&#x2018;&#x2DC; 4 đ?&#x153;&#x2039; = đ?&#x153;&#x2122;4 â&#x2C6;&#x2019; đ?&#x153;&#x2122;3 â&#x2039;Ž đ?&#x2018;&#x2DC; đ?&#x2018;&#x203A; đ?&#x153;&#x2039; = đ?&#x153;&#x2122;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x153;&#x2122;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 While the angle đ?&#x2018;&#x2DC; 1 đ?&#x153;&#x2039; = đ?&#x153;&#x2122;1 â&#x2C6;&#x2019; đ?&#x153;&#x2122;đ?&#x2018;&#x203A;+1 and đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;+1 đ?&#x153;&#x2039; = đ?&#x153;&#x2122;đ?&#x2018;&#x203A;+1 + 2đ?&#x153;&#x2039; â&#x2C6;&#x2019; đ?&#x153;&#x2122;đ?&#x2018;&#x203A; . In terms of -n. đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A;+1 đ?&#x153;&#x2039; = 0 so that đ?&#x153;&#x2122;đ?&#x2018;&#x203A; = đ?&#x153;&#x2122;đ?&#x2018;&#x203A;+1 + 2đ?&#x153;&#x2039; . Thus, the đ?&#x2018;&#x2DC;1 đ?&#x153;&#x2039; + đ?&#x2018;&#x2DC;2 đ?&#x153;&#x2039; + đ?&#x2018;&#x2DC;3 đ?&#x153;&#x2039; + â&#x2039;Ż + đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A; đ?&#x153;&#x2039; = (đ?&#x153;&#x2122;1 â&#x2C6;&#x2019; đ?&#x153;&#x2122;đ?&#x2018;&#x203A;+1 ) + (đ?&#x153;&#x2122;2 â&#x2C6;&#x2019; đ?&#x153;&#x2122;1 ) + (đ?&#x153;&#x2122;3 â&#x2C6;&#x2019; đ?&#x153;&#x2122;2 ) + â&#x2039;Ż + (đ?&#x153;&#x2122;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x153;&#x2122;đ?&#x2018;&#x203A;+1 ) (đ?&#x2018;&#x2DC;1 + đ?&#x2018;&#x2DC;2 + đ?&#x2018;&#x2DC;3 + â&#x2039;Ż + đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A; )đ?&#x153;&#x2039; = (đ?&#x153;&#x2122;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x153;&#x2122;đ?&#x2018;&#x203A;+1 ) = 2đ?&#x153;&#x2039; (1.17) đ?&#x2018;&#x2DC;1 + đ?&#x2018;&#x2DC;2 + đ?&#x2018;&#x2DC;3 + â&#x2039;Ż + đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A; = 2 So it is clear that đ?&#x2018;&#x2DC;đ?&#x2018;&#x2014; with đ?&#x2018;&#x2014; = 1, 2, 3, . . . , đ?&#x2018;&#x203A;satisfy the condition đ?&#x2018;&#x2DC;1 + đ?&#x2018;&#x2DC;2 + đ?&#x2018;&#x2DC;3 + â&#x2039;Ż + đ?&#x2018;&#x2DC;đ?&#x2018;&#x203A; = 2 , â&#x2C6;&#x2019;1 < đ?&#x2018;&#x2DC;đ?&#x2018;&#x2014; < 1. End of proof.

V.

ACKNOWLEDGEMENTS

I would like to thankProf.Dr.Marjono and Dr. Bagus, for theencouragements. REFERENCES [1]. [2]. [3]. [4]. [5]. [6]. [7].

Mark J. Ablowitz Athanassios S.fokas. Complex variables : Introduction andapplications, Second edition.Cambridge texts in applied mathematics,2003. Howell, L.H. Computation of conformal maps by modified Schwarz-Christoffel transformations, 1990. James Ward Brown and Ruel V.Churchil. Complex Variables and Applictions, Eighth Edition, McGraw-Hill, 2009. Dennis G.Zill. Complex Analysis with Applications, Jones and Bartlett Publisher, 2003. Jhon Mathews and Russell Howell. Complex analysis for mathmatics and engineering sixth edition,Jones and Bartlett Publisher, 2012. Tobin A.Driscoll and Lloyd N. Thefethen. Schwarz-Christoffel Mapping, Cambridge Monographs on Applied and Computational Mathematics, 2002. Gonzalo Riera , Herna Ě n Carrasco , and Rube Ě n Preiss .The Schwarz-Christoffel conformal mapping for â&#x20AC;˛â&#x20AC;˛Polygonsâ&#x20AC;˛â&#x20AC;˛ with Infinitely Many Sides,InternationalJournal of Mathematics and Mathematical, Sciences Volume 2008, Hindawi Publishing Corporation, 2008.

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