engineeringmathematics-iv_unit-i

Functions

of complex variable

Course- B.Tech

Semester-IV Subject- ENGINEERING MATHEMATICS-IV

Unit- I

RAI UNIVERSITY, AHMEDABAD

Unit-I Functions of complex variable

Rai University | Ahmedabad

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Content Functions of a complex variable, Application of complex variable, Complex algebra, Analytic function, C-R equations, Theorem on C-R equation(without proof) ,C-R equation in polar form, Properties of Analytic Functions (orthogonal system), Laplace Equation, Harmonic Functions, Determination of Analytic function Whose real or Imaginary part is known : (1) When u is given, v can be determined (2) When v is given, u can be determined (3) By Milne-Thompson method and Finding Harmonic Conjugate functions, Conformal Mapping and its applications ,Define conformal Mapping, Some standard conformal transformations: (1) Translation (2) Rotation and Magnification (3) Inversion and Reflection



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1.1 Introduction— Functions of a complex variable provide us some powerful and widely useful tools in mathematical analysis as well as in theoretical physics. 1.2 Applications— 1. Some important physical quantities are complex variables (the wave-function , a.c. impedance Z()). 2. Evaluating definite integrals. 3. Obtaining asymptotic solutions of differentials equations. 4. Integral transforms. 5. Many Physical quantities that were originally real become complex as simple theory is made more general. The energy En En

0 + i (-1 the finite life time). 1.3 Complex Algebra—A complex number z(x, y) = x + iy . Where i = √−1. x is called the real part, labeled by Re z y is called the imaginary part, labeled by Im z Different forms of complex number— 1. Cartesian form— 푧(푥,푦) = 푥 + 푖푦 2. Polar form— z(r, ) = r(cos + i sin) 3. Euler form— z(r, ) = re Here, 푟 = 푥 + 푦 and = tan ( ) The choice of polar representation or Cartesian representation is a matter of convenience. Addition and subtraction of complex variables are easier in the Cartesian representation. Multiplication, division, powers, roots are easier to handle in polar form. 1.4 Derivative of the function of complex variable— The derivative of 푓(푧) is defined by—

푓 (푧) = 푙푖푚→

훿푓훿푧

= lim→

푓(푧 + 훿푧) − 푓(푧)훿푧

If 푧 = 푥 + 푖푦 and 푓(푧) = 푢 + 푖푣, let consider that—

yixz , viuf ,

yixviu

zf

Assuming the partial derivatives exist. Let us take limit by the two different approaches as in the figure. First, with y = 0, we let x0,

xvi

xu

zf

xz

00limlim

xvi

xu

For a second approach, we set x = 0 and then let y 0. This leads to –

yv

yui

yvi

yu

zf

yz

00limlim



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If we have the derivative, the above two results must be equal. Hence—

yv

yui

xvi

xu

Hence, the necessary and sufficient conditions for the derivative of the function 푓(푧) = 푢 + 푖푣 to exit for all values of z in a region R, are— (1) , , , are continuous functions of 푥 and 푦 in R;

(2) yv

xu

and xv

yu

.

1.5 Analytic function— A single valued function that possesses a unique derivative with respect to 푧 at all points on a region R is called an Analytic function of 푧 in that region. Necessary and sufficient condition to be a function Analytic— A function 푓(푧) = 푢 + 푖푣 is called an analytic function if— 1. 푓(푧) = 푢 + 푖푣 , such that 푢 and 푣 are real single valued functions of 푥 and 푦 and

, , , are continuous functions of 푥 and 푦 in R.

2. yv

xu

and

xv

yu

.

1.6 Cauchy-Riemann equations— If a function 푓(푧) = 푢 + 푖푣 is analytic in region R, then it must satisfies the relation

yv

xu

and

xv

yu

. These equations are called as Cauchy-

Riemann equation. Cauchy-Riemann equation in polar form—

v

rur and

rvu

r

1

1.7 Laplace’s equation— An equation in two variables 푥 and 푦 given by + = 0 is called as Laplace’s equation in two variables. 1.8 Harmonic functions— If 푓(푧) = 푢 + 푖푣 be an analytic function in some region R, then Cauchy-Riemann equations are satisfied. That implies—

yv

xu

... (1)

xv

yu

... (2)

Now, on differentiating (1) with respect to x and (2) with respect to y— = … (3)

= − … (4)

Assuming that = , adding equations (1) and (2)— 흏ퟐ풖흏풙ퟐ +

흏ퟐ풖흏풚ퟐ = ퟎ



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Similarly, on differentiating (1) with respect to 푦 and (2) with respect to 푥 and subtracting them—

흏ퟐ풗흏풙ퟐ +

흏ퟐ풗흏풚ퟐ = ퟎ

both 푢 and 푣 satisfy the Laplace’s equation in two variables, therefore 푓(푧) = 푢 + 푖푣 is called as Harmonic function. This theory is known as Potential theory. If 푓(푧) = 푢 + 푖푣 is an analytic function in which 푢(푥, 푦) is harmonic, then 푣(푥,푦)is called as Harmonic conjugate of 푢(푥, 푦). 1.9 Orthogonal system— Consider the two families of curves 푢(푥, 푦) = 푐 … (1) and 푣(푥, 푦) = 푐 … (2). Differentiating (1) with respect to 푥 –

= − = = 푚 [by using Cauchy s Riemann equations for analytic function]

Differentiating (2) with respect to 푥 –

= − = 푚

Here, 푚 푚 = −1 therefore every analytic function 푓(푧) = 푢(푥,푦) + 푖푣(푥,푦) represents two families of curves 푢(푥, 푦) = 푐 and 푣(푥, 푦) = 푐 form an orthogonal system. 1.10 Determination of Analytic function whose real or imaginary part is known— If the real or the imaginary part of any analytic function is given, other part can be determined by using the following methods— (a) Direct Method (b) Milne-Thomson’s Method

(c) Exact Differential equation method (d) Shortcut Method

1.10.1 Direct Method— Let 푓(푧) = 푢 + 푖푣 is an analytic function. If 푣 is given, then we can find 푢 in following steps— Step-I Find by using C-R equation =

Step-II Integrating with respect to 푥 to find 푢 with taking integrating constant 푓(푦)

Step-III Differentiate 푢 (from step-II) with respect to y. Evaluate containing 푓 (푦)

Step-IV Find by using C-R equation = −

Step-V by comparing the result of from step-III and step-IV, evaluate 푓 (푦) Step-VI Integrate 푓 (푦) and evaluate 푓(푦) Step-VII Substitute the value of 푓(푦) in step-II and evaluate 푢.



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Example— If 풇(풛) = 풖 + 풊풗 represents the analytic complex potential for an electric field and 풗 = 풙ퟐ − 풚ퟐ + 풙

풙ퟐ 풚ퟐ, determine the function 풖.

Solution— Given that – 푣 = 푥 − 푦 +

Hence— = 2푥 + ( )( ) ( )( )

= 2푥 + ( ) … (1)

Again, = −2푦 + ( ) … (2) So, 푢 and 푣 must satisfy the Cauchy’s-Riemann equations: = … (3) and = − … (4) Now, from equations (2) and (3)— = −2푦 + ( ) On integrating with respect to 푥— 푢 = −2푦 ∫ 푑푥 − 푦 ∫ ( ) 푑푥

푢 = −2푥푦 + + 푓(푦) Differentiating with respect to 푦— = −2푥 − ( ) + 푓 (푦) … (5) Again, from equations (1) and (4)— = −2푥 − ( ) … (6) On comparing (5) and (6)— 푓 (푦) = 0 , so 푓(푦) = 푘 Hence, 푢 = −2푥푦 + + 푘 ans 1.10.2 Milne-Thomson’s Method— Let a complex variable function is given by 푓(푧) = 푢(푥,푦) + 푖푣(푥,푦) … (1) Where— 푧 = 푥 + 푖푦 and 푥 = ̅ and 푦 = ̅ . Now, on writing 푓(푧) = 푢(푥, 푦) + 푖푣(푥, 푦) in terms of 푧 and 푧̅ –

푓(푧) = 푢 ̅ , ̅ + 푖푣 ̅ , ̅ Considering this as a formal identity in the two variables 푧 and 푧, and substituting 푧 = 푧̅ –

∴ 푓(푧) = 푢(푧, 0) + 푖푣(푧, 0) … (2) Equation (2) is same as the equation (1), if we replace 푥 by 푧 and 푦 by 0. Therefore, to express any function in terms of 푧, replace 푥 by 푧 and 푦 by 0. This is an elegant method of finding 푓(푧) , ‘when its real part or imaginary part is given’ and this method is known as Milne-Thomson’s Method.



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Example— If 풇(풛) = 풖 + 풊풗 represents the complex potential for an electric field and 풗 = 풙ퟐ − 풚ퟐ + 풙

풙ퟐ 풚ퟐ, determine the function 풖. (By using Milne-Thomson’s method)

Solution— Given that— 푓(푧) = 푢 + 푖푣 ∴ = + 푖 = + 푖 [by using cauchy − Riemann equation]

Hence, ∴ = + 푖 = −2푦 + ( ) + 푖 2푥 + ( ) by using Milne-Thomson’s method (replacing 푥 by z and 푦 by 0)—

휕푓휕푧

= 푖(2푧 −1푧

)

푓(푧) = 푖 푧 + + 푐

Hence, 푢 = Re 푖 푧 + + 푘 = Re 푖 푥 − 푦 + 푖(2푥푦) + − 푖 + 푐

= −2푥푦 + + 푐 ans

Example— Find the analytic function풇(풛) = 풖 + 풊풗, whose real part is 풙ퟑ − ퟑ풙풚ퟐ + ퟑ풙ퟐ −ퟑ풚ퟐ. Solution— Let 푓(푧) = 푢 + 푖푣 is an analytic function. Hence, from the question— 푢 = 푥 − 3푥푦 + 3푥 − 3푦 Now, = + 푖 = − 푖 [by using cauchy − Riemann equation]

Hence, ∴ = − 푖 = (3푥 − 3푦 + 6푥) + 푖(−6푥푦 − 6푦)

by using Milne-Thomson’s method (replacing 푥 by z and 푦 by 0)— = (3푧 + 6푧) + 푖(0)

푓(푧) = 푧 + 3푧 + 푖푐 Hence, 푣 = Im[푧 + 3푧 + 푖푐] = Im[푥 + 푖3푥 푦 − 3푥푦 − 푖푦 + 3푥 − 3푦 + 6푥푦 + 푖푐]

∴ 푣 = 3푥 푦 − 푦 + 푐 Therefore 푓(푧) = (푥 − 3푥푦 + 3푥 − 3푦 ) + 푖(3푥 푦 − 푦 + 푐) ans Example— Find the analytic function 풛 = 풖 + 풊풗, if 풖 − 풗 = (풙 − 풚)(풙ퟐ + ퟒ풙풚 + 풚ퟐ). Solution— Given that—

푢 − 푣 = (푥 − 푦)(푥 + 4푥푦 + 푦 ) − = 3푥 + 6푥푦 − 3푦 … (1)

− = 3푥 − 6푥푦 − 3푦 … (2)

On applying Cauchy-Riemann’s theorem in equation (2)— − − = 3푥 − 6푥푦 − 3푦 … (3)



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Now, from equations (1) and (3)— = 6푥푦 and = −3푥 + 3푦

Again, = + 푖 = 6푥푦 + 푖(−3푥 + 3푦 ) By using Milne-Thomson’s method— = 푖(−3푧 ) 푓(푧) = −푖푧 + 푐 = −푖(푥 + 푖3푥 푦 − 3푥푦 − 푖푦 ) + 푐 = (3푥 푦 − 푦 + 푐) + 푖(3푥푦 − 푥 ) ans

1.10.3 Exact Differential equation Method— Let 푓(푧) = 푢 + 푖푣 is an analytic function. Case-I If 풖(풙,풚) is given We know that— 푑푣 = 푑푥 + 푑푦

= − 푑푥 + 푑푦 [By C-R equations]

Let 푑푣 = 푀푑푥 + 푁푑푦, therefore 푀 = − and 푁 =

Again, = − and =

Therefore − = − + = 0 [∵ 푢 is a harmonic function]

∴ = [Satisfies the condition for

exact differential equation] Hence,

푣 = −휕푢휕푦

푑푥 + terms of 휕푢휕푥

not containg 푥 푑푦 + 푐

Case-II If 풗(풙,풚) is given We know that— 푑푢 = 푑푥 + 푑푦

= 푑푥 − 푑푦 [By C-R equations]

Let 푑푣 = 푀푑푥 + 푁푑푦, therefore 푀 = and 푁 = −

Again, = and = −

Therefore − = + = 0 [∵ 푣 is a harmonic function]

∴ = [Satisfies the condition for exact differential equation]



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Hence,

푢 =휕푣휕푦

푑푥 + terms of −휕푣휕푥

not containg 푥 푑푦 + 푐

Example— If 풖 = 풙ퟑ − ퟑ풙풚ퟐ + ퟑ풙 + ퟏ , determine 풗 for which 풇(풛) is an analytic function. Solution— Given that— 푢 = 푥 − 3푥푦 + 3푥 + 1 By using exact differential equation method, 푣 can be written as— 푣 = ∫ − 푑푥 + ∫ terms of not containg 푥 푑푦 + 푐

푣 = ∫ (6푥푦푑푥) + ∫(−3푦 + 3)푑푦 + 푐 푣 = 3푥 푦 − 푦 + 3푦 + 푐 Example— If 풖 = 풚ퟑ − ퟑ풙ퟐ풚 , determine 풗 for which 풇(풛) = 풖 + 풊풗 is an analytic function. Solution— Given that— 푢 = 푦 − 3푥 푦 By using exact differential equation method, 푣 can be written as— 푣 = ∫ − 푑푥 + ∫ terms of not containg 푥 푑푦 + 푐

푣 = ∫ (−3푦 + 3푥 )푑푥 + ∫(0)푑푦 + 푐 푣 = −3푥푦 + 푥 + 푐 1.10.4 Shortcut Method Case-I If 풖(풙,풚) is given If the real part of an analytic function 푓(푧) is given then—

푓(푧) = 2푢푧2

,푧2푖

− 푢(0,0) + 푖푐

Where, c is a real constant. Case-II If 풗(풙,풚) is given If the imaginary part of an analytic function 푓(푧) is given then—

푓(푧) = 2푖푣푧2

,푧2푖

− 푖푣(0,0) + 푐

Where, c is a real constant. Example— If 풇(풛) = 풖 + 풊풗 is an analytic function and 풖(풙,풚) = 퐬퐢퐧ퟐ풙

퐜퐨퐬퐡ퟐ풚 퐜퐨퐬ퟐ풙 , find 풇(풛).

Solution— Given that 푓(푧) is an analytic function and it’s real part is 푢(푥, 푦) = .

Hence, 푓(푧) = 2푢 , − 푢(0,0) + 푖푐

푓(푧) = 2 + 푖푐



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푓(푧) = 2 ( ) + 푖푐 [cosh(−푥) = cosh 푥]

푓(푧) = 2 ( ) + 푖푐 [cosh(푖푥) = cos 푥]

푓(푧) = 2 + 푖푐 푓(푧) = tan 푧 + 푖푐 Example— Determine the analytic function 풇(풛), whose imaginary part is ퟑ풙ퟐ풚 − 풚ퟑ. Solution— Given that 푣 = 3푥 푦 − 푦 , since the complex variable function 푓(푧) is analytic, hence it is given by—

푓(푧) = 2푖푣푧2

,푧2푖

− 푖푣(0,0) + 푐

푓(푧) = 2푖 3푧2

푧2푖

−푧2푖

+ 푐

푓(푧) = 2푖38푖푧 +

18푖푧 + 푐

푓(푧) = 2푖12푖푧 + 푐

푓(푧) = 푧 + 푐 Example— Determine the analytic function 풇(풛), whose imaginary part is 퐥퐨퐠(풙ퟐ + 풚ퟐ) +풙 − ퟐ풚. Solution— Given that 푣 = log(푥 + 푦 ) + 푥 − 2푦 푣 = + 1

푣 = − 2

Let 푓(푧) = 푢 + 푖푣, ∴ = 푢 + 푖푣 = 푣 + 푖푣 [By using C − R equations]

Hence, 푓 (푧) = −2 + 푖 + 1 푓(푧) = −2푧 + 2푖 log 푧 + 푖푧 + 푐 푓(푧) = 2푖 log 푧 + (푖 − 2)푧 + 푐 Example— Find the orthogonal trajectories of the family of curves given by the equation— 풙ퟑ풚 − 풙풚ퟑ = 풄ퟏ. Solution— given curve is— 푢 = 푥 푦 − 푥푦 = 푐 The orthogonal trajectories to the family of curves 푢 = 푐 will be 푣 = 푐 such that 푓(푧) = 푢 + 푖푣 becomes analytic. Hence, orthogonal trajectory 푣 = 푐 is given by—


푑푥 + 푡푒푟푚푠 표푓 휕푢휕푥

푛표푡 푐표푛푡푎푖푛푔 푥 푑푦 + 푐



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푣 = (– 푥 + 3푥푦 )푑푥 + (−푦 )푑푦 + 푐

푣 = −푥4

+3푥 푦

2−푦4

+ 푐

Example— Find the orthogonal trajectories of the family of curves given by the equation— 풆풙 퐜퐨퐬 풚 − 풙풚 = 풄.

Solution— given curve is— 푢 = 푒 cos 푦 − 푥푦 = 푐

The orthogonal trajectories to the family of curves 푢 = 푐 will be 푣 = 푐 such that 푓(푧) = 푢 + 푖푣 becomes analytic. Hence, orthogonal trajectory 푣 = 푐 is given by—


푑푥 + 푡푒푟푚푠 표푓 휕푢휕푥

푛표푡 푐표푛푡푎푖푛푔 푥 푑푦 + 푐

푣 = (푒 sin푦 − 푥)푑푥 + (−푦)푑푦 + 푐

푣 = 푒 sin푦 −푥2−푦2

+ 푐

Example— Show that the function 풖(풙,풚) = 풆풙퐜퐨퐬풚 is harmonic. Determine its harmonic conjugate 풗(풙,풚) and the analytic function 풇(풛) = 풖 + 풊풗. Solution— given that 푢(푥,푦) = 푒 cos푦 Now, 푢 = 푒 cos 푦, 푢 = 푒 cos푦 푢 = −푒 sin 푦, 푢 = −푒 cos푦 Here, 푢 + 푢 = 0, so 풖(풙,풚) is a harmonic function. Again, since 푓(푧) = 푢 + 푖푣 is an analytic function, so 푓(푧) is given by—

푓(푧) = 2푢푧2

,푧2푖

− 푢(0,0) + 푖푐

푓(푧) = 2푒 cos푧2푖

− 1 + 푖푐

푓(푧) = 2푒 cos−푖푧

2− 1 + 푖푐

푓(푧) = 2푒 cos푖푧2

− 1 + 푖푐

푓(푧) = 2푒12

푒 + 푒 − 1 + 푖푐

푓(푧) = (푒 + 1) − 1 + 푖푐 풇(풛) = 풆풛 + 풊풄 Now, 푓(푧) = 푒 + 푖푐 푓(푧) = 푒 푒 + 푖푐



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푓(푧) = 푒 (cos 푦 + 푖 sin푦) + 푖푐 푓(푧) = 푒 푐표푠 푦 + 푖 (푒 푠푖푛 푦 + 푐) Hence, 푣(푥,푦) = 푒 sin 푦 + 푐 1.11 Conformal Mapping— 1.11.1 Mapping— Mapping is a mathematical technique used to convert (or map) one mathematical problem and its solution into another. It involves the study of complex variables. Let a complex variable function 푧 = 푥 + 푖푦 define in 푍- plane have to convert (or map) in another complex variable function 푓(푧) = 푤 = 푢 + 푖푣 define in 푤- plane. This process is called as Mapping.

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1.11.2 Conformal Mapping— Conformal Mapping is a mathematical technique used to convert (or map) one mathematical problem and its solution into another preserving both angles and shape of infinitesimal small figures but not necessarily their size. The process of Mapping in which a complex variable function 푧 = 푥 + 푖푦 define in 푍- plane mapped to another complex variable function 푓(푧) = 푤 = 푢 + 푖푣 define in 푤- plane preserving the angles between the curves both in magnitude and sense is called as conformal mapping. The necessary condition for conformal mapping— if 푤 = 푓(푧) represents a conformal mapping of a domain D in the 푧 −plane into a domain D of the 푤 −plane then 푓(푧) is an analytic function in domain D. Application of Conformal mapping— A large number of problems arise in fluid mechanics, electrostatics, heat conduction and many other physical situations. There are different types of conformal mapping. Some standard conformal mapping are mentioned below— 1. Translation 2. Rotation 3. Magnification 4. Inversion 5. Reflection

Mapping



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1.11.2.1 Translation— A conformal mapping in which every point in 푧 −plane is translated in the direction of a given vector is known as Translation. Let a complex function 푧 = 푥 + 푖푦 translated in the direction of the vector 훼 = 푎 + 푖푏 , then that translation is given by— 푤 = 푧 + 훼. Example— Determine and sketch the image of |풁| = ퟏ under the transformation 풘 = 풛 + 풊. Solution— Let 푤 = 푢 + 푖푣 ∴ 푢 + 푖푣 = 푥 + 푖푦 + 푖 = 푥 + 푖(푦 + 1) Hence, 푥 = 푢 and 푦 = 푣 − 1 Again, from the question |푍| = 1 ∴ 푥 + 푦 = 1 푢 + (푣 − 1) = 1, which represent a circle in (푢, 푣) plane.

1.11.2.2 Rotation— A conformal mapping in which every point on 푧 −plane, let P(푥,푦) such that OP is rotated by an angle 훼 in anti-clockwise direction mapped into 푤 −plane given by 푤 = 푒 푧 is called as Rotation.

Example— Determine and sketch the image of rectangle mapped by the rotation of ퟒퟓ° in anticlockwise direction formed by 풙 = ퟎ,풚 = ퟎ, 풙 = ퟏ and 풚 = ퟐ. Solution— given rectangle is— 푥 = 0, 푦 = 0, 푥 = 1 and 푦 = 2 Let the required transformation is—

푤 = 푒 푧



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푤 = cos + 푖 sin (푥 + 푖푦)

푤 =√

+ 푖√

(푥 + 푖푦)

푤 =√

+ 푖√

푢 + 푖푣 =√

+ 푖√

∴ 푢 =√

and 푣 =√

Now, for 푥 = 0—

푢 =−푦√2

, 푣 =푦√2

⟹ 푢 + 푣 = 0

For 푦 = 0—

푢 =푥√2

, 푣 =푥√2

⟹ 푢 − 푣 = 0

For 푥 = 1—

푢 =1 − 푦√2

, 푣 =1 + 푦√2

⟹ 푢 + 푣 = √2

or 푦 = 2—

푢 =푥 − 2√2

, 푣 =푥 + 2√2

⟹ 푢 − 푣 = −2√2

1.11.2.3 Magnification (Scaling) — A conformal mapping in which every point on 푧 −plane mapped into 푤 −plane given by 푤 = 푐푧 (푐 > 0 is real) is called as magnification. In this process mapped shape is either stretched(푐 > 1) or contracted (0 < 푐 < 1) in the direction of Z. Example— Determine the transformation 풘 = ퟐ풛, where |풛| = ퟐ. Solution— given that |푧| = 2 ⟹ 푥 + 푦 = 4 , represents a circle on 푧 −plane having center at origin and radius equal to 2-units. Let the required transformation is—

푤 = 2푧 = 2(푥 + 푖푦) = 2푥 + 푖2푦 푢 + 푖푣 = 2푥 + 푖2푦

Hence, 푥 = and 푦 = ⟹ + = 4 ⟹ 푢 + 푣 = 16 represent a circle on 푤 −plane having center at origin and radius equal to 4-units.



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1.11.2.4 Inversion— A conformal mapping in which every point on 푧 −plane mapped into 푤 −plane given by 푤 = is called as inversion.

Example— Determine the transformation 풘 = ퟏ풛, where |풛| = ퟏ.

Solution— given that— |푧| = 1 ⟹ 푥 + 푦 = 1 … (1) Again, given that 푤 = = = ( )( )

=

푢 + 푖푣 = − 푖 = 푥 − 푖푦 [푥 + 푦 = 1]

Hence, 푢 = 푥 and 푣 = −푦 Now, from equation (1)—

푢 + 푣 = 1 1.11.2.5 Reflection — A conformal mapping in which every point on 푧 −plane mapped into 푤 −plane given by 푤 = 푧̅ . This represents the reflection about real axis. Example— Determine and sketch the transformation 풘 = 풛, where |풛 − (ퟐ + 풊)| = ퟒ. Solution— given that— |푧 − 2| = 4 |(푥 − 2) + 푖(푦 − 1)| = 4 (푥 − 2) + (푦 − 1) = 16 … (1) Again, from the question— 푤 = 푧̅ = 푥 − 푖푦 푢 + 푖푣 = 푥 − 푖푦 Hence, 푥 = 푢 and 푦 = −푣 Now, from equation (1)—

(푢 − 2) + (푣 + 1) = 16 Sketch—



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References— 1. Internet— 1.1 mathworld.wolfram.com 1.2 www.math.com 1.3 www.grc.nasa.gov 1.4 en.wikipedia.org/wiki/Conformal_map 1.5 http://www.webmath.com/ 2. Journals/Paper/notes— 2.1 Complex variables and applications-Xiaokui Yang-Department of Mathematics, Northwestern University 2.2 Conformal Mapping and its Applications-Suman Ganguli-Department of Physics, University of Tennessee, Knoxville, TN 37996- November 20, 2008 2.3 Conformal Mapping and its Applications-Ali H. M. Murid-Department of Mathematical Sciences-Faculty of Science-University Technology Malaysia- 81310-UTM Johor Bahru, Malaysia-September 29, 2012 3. Books— 3.1 Higher Engineering Mathematics-Dr. B.S. Grewal- 42nd edition-Khanna Publishers-page no 639-672 3.2 Higher Engineering Mathematics- B V RAMANA- Nineteenth reprint- McGraw Hill education (India) Private limited- page no 22.1-25.26 3.3 Complex analysis and numerical techniques-volume-IV-Dr. Shailesh S. Patel, Dr. Narendra B. Desai- Atul Prakashan 3.4 A Text Book of Engineering Mathematics- N.P. Bali, Dr. Manish Goyal- Laxmi Publications(P) Limited- page no 1033-1100

http://www.math.com/

http://www.grc.nasa.gov/

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