Engineering Mathematics Test 8

Concept:

If f(x, y) is harmonic then it must satisfy Laplace’s equation

\({{\nabla }^{2}}~f\left( x,~y \right)=0=\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}\)

Calculation:

Given function: f = 2x – x² + my²

So, for harmonic it should satisfy Laplace’s equation

\({{\nabla }^{2}}f=0=\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}\)

\(\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}=\frac{\partial \left( \frac{\partial f}{\partial x} \right)}{\partial x}=\frac{\partial }{\partial x}~\left( 2-2x \right)=-2\)

\(\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}=\frac{\partial \left( \frac{\partial f}{\partial y} \right)}{\partial y}=\frac{\partial \left( 2my \right)}{\partial y}=2m\)

\(\Rightarrow \frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}=-2+2m=0\)

(i) ln z

At, z = 0

The function is not analyse

(ii) e^1/z

Using the expansion formula for e^x

\({e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \)

\({e^{1/z}} = 1 + \left( {\frac{1}{z}} \right) + \frac{1}{{2!}}\left( {\frac{1}{{{z^2}}}} \right) + \frac{1}{{3!}}\left( {\frac{1}{{{z^3}}}} \right) + \ldots \)

At, z = 0, e^1/z is not analytic

(iii) \(\frac{1}{{1 - z}}\)

At, z = 1, the function is not analytic

(iv) cos z

expansion of cos z is:

\(\cos z = 1 - \frac{{{z^2}}}{{2!}} + \frac{{{z^4}}}{{4!}} - \frac{{{z^6}\;}}{{6!}} + \ldots \)

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

\(\mathop \oint \limits_C f\left( z \right)dz = 0\)

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

\(f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz\)

\({f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz\)

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

\(\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum\;of\;residues\;at\;the\;singualr\;points\;within\;C}}} \right]\)

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

\(Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]\)

2. If f(z) has a pole of order n at z = a, then

\(Res\;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}\)

Application:

\(\mathop \smallint \nolimits_C \frac{{\cos \pi {z^2}}}{{\left( {z - 1} \right)\left( {z - 2} \right)}}dz\)

Simple poles: z = 1, 2

Both the poles lie within the given region.

Residue at z = 1,

\(\mathop {\lim }\limits_{z \to 1} \frac{{\cos \pi {z^2}}}{{\left( {z - 1} \right)\left( {z - 2} \right)}} \cdot \left( {z - 1} \right) = \frac{{\cos \pi }}{{\left( { - 1} \right)}} = 1\)

Residue at z = 2

\(\mathop {\lim }\limits_{z \to 2} \frac{{\cos \pi {z^2}}}{{\left( {z - 1} \right)\left( {z - 2} \right)}}\left( {z - 1} \right) = \cos 4\pi = 1\)

Value of integral = 2πi [1 + 1] = 4πi

\(\mathop \smallint \nolimits_C f\left( z \right)dz\)

\( = \mathop \smallint \limits_0^\pi \frac{{\left( {{z^2} - 1} \right)}}{z}dz\)

\( = \mathop \smallint \limits_0^\pi \frac{{4{e^{2i\theta }} - 1}}{{2{e^{i\theta }}}}d\left( {2{e^{i\theta }}} \right)\)

\( = \left[ {2{e^{2i\theta }} - i} \right]_0^\pi \)

= -πi

Concept:

Let the real part u(x, y) of an analytic function f(z) = u + iv be given

To find the function v(x, y),

The total derivative of V is given by

\(dv = \frac{{\partial v}}{{\partial x}}dx + \frac{{\partial v}}{{\partial y}}dy\) 

By using CR equations, u_x = v_x and u_y = -v_x

\(dv = \frac{{ - \partial u}}{{\partial y}}dx + \frac{{\partial u}}{{\partial x}}dy\) 

The R.H.S of the above equation is in the form of Mdx + Ndy where

\(M = \frac{{ - \partial u}}{{\partial y}}\)  and \(N = \frac{{\partial u}}{{\partial x}}\) 

Since u is harmonic, it satisfies Laplace equation

\(\frac{{\partial M}}{{\partial y}} = \frac{{ - {\partial ^2}u}}{{\partial {y^2}}}\) and \(\frac{{\partial N}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) 

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0\) 

\( \Rightarrow \frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\) 

Hence the equation is on the exact equation.

Now, by integrating we get

\(v = \smallint - {u_y}dx\; + \smallint {u_x}\;dy + C\) 

Similarly when v(x, y) is given, then to find u(x, y)

\(u = \smallint {v_y}\;dx - \smallint {v_x}dy + C\) 

Calculation:

Given that, (x, y) = In (x² + y²)

\({u_x} = \frac{{2x}}{{\left( {{x^2} + {y^2}} \right)}},\;\;{u_y} = \frac{{2y}}{{\left( {{x^2} + {y^2}} \right)}}\) 

From CR equation,

\({u_x} = {u_y} = \frac{{2x}}{{\left( {{x^2} + {y^2}} \right)}}\) 

By integrating on both the sides,

\(v = \smallint \frac{{2x}}{{{x^2} + {y^2}}}dy\) 

\( \Rightarrow v = 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) + g\left( x \right)\) 

By differentiating w.r.t. ‘x’

\( \Rightarrow {v_x} = \frac{{ - 2y}}{{{x^2} + {y^2}}} + g'\left( x \right)\) 

\( \Rightarrow \; - {u_y} = \frac{{ - 2y}}{{{x^2} + {y^2}}} + g'\left( x \right)\) 

\( \Rightarrow \frac{{ - 2y}}{{{x^2} + {y^2}}} = \frac{{ - 2y}}{{{x^2} + {y^2}}} + g'\left( x \right)\) 

⇒ g’(x) = 0

⇒ g(x) = C

i) f(z) = Re(z) = Re(x + iy) = x

here u = x, v = 0

u_x = 1, v_y = 0

f(z) is not satisfying CR equations Hence f(z) is not analytic.

ii) f(z) = z̅ = x - fy

u = x, v = -y

u_x = 1, v_y = -1

u_x ≠ v_y

f(x) is not satisfying CR equations.

Hence f(z) is not analytic

iii) f(z) = Im (z) = Im (x + iy) = y

u = 0, v = y

u_x = 0, v_y = 1

f(x) is not satisfying CR equations.

Hence f(z) is not analytic

iv) f(z) = sin z = (x + iy)

= sin x cos iy + cos x sin iy

= sin x cos hy + i cos x sin hy

u = sin x cos hy, v = cos x sin hy

u_x = cos x cos hy

v_y = cos x cos hy

u_y = sin x sin hy

v_x = - sin x sin hy

u_x = v_y and u_y = - v_x

f(z) is satisfying CR equation.

\(f\left( z \right) = \frac{{z - 1}}{{z + 1}}\)

The given point is z = 1 ⇒ (z - 1) = 0

Let z – 1 = t ⇒ z = t + 1

\(f\left( z \right) = \frac{t}{{\left( {t + 2} \right)}} = \frac{t}{{2\left( {1 + \frac{t}{2}} \right)}}\)

\( = \frac{t}{2}{\left( {1 + \frac{t}{2}} \right)^{ - 1}}\)

\( = \frac{t}{2}\left( {1 - \frac{t}{2} + {{\left( {\frac{t}{2}} \right)}^2} - {{\left( {\frac{t}{2}} \right)}^3} + \ldots } \right)\)

\( = \frac{t}{2}\left( {1 - \frac{t}{2} + \frac{{{t^2}}}{4} - \frac{{{t^3}}}{8} + \ldots } \right)\)

\( = \frac{t}{2} - \frac{{{t^2}}}{4} + \frac{{{t^3}}}{8} - \frac{{{t^4}}}{{16}} + \ldots \)

\( = \frac{{\left( {z - 1} \right)}}{2} - \frac{{{{\left( {z - 1} \right)}^2}}}{4} + \frac{{{{\left( {z - 1} \right)}^3}}}{8} \ldots \)

Co-efficient of (z - 1)³ term is \(\frac{1}{8}\)

z = u + iv

\(2u + v = {e^x}\left( {\cos y-\sin y} \right)\)

\(2{u_x} + {v_x} = {e^x}\left( {\cos y - \sin y} \right)\)      ---- (1)  

\(2{u_y} + {v_y} = {e^x}\left( { - \sin y - \cos y} \right)\)      ----(2)

C.R. equations are

\({u_x} = {v_y},{u_y} = - {v_x}\)

Equation 1 becomes,

\(2{u_x} - {u_y} = {e^x}\left( {\cos y-\sin y} \right)\)      ----(3)

Equation (2) becomes

\(2{u_y} + {u_x} = {e^x}\;\left( { - \sin y - \cos y} \right)\)      ----(4)

From equations (3) and (4)

By doing operation 2 × (3) + (4)

\( \Rightarrow 4{u_x}-2{u_y} + 2{u_y} + {u_x} = {e^x}\left( {2\cos y-2\sin y-\sin y-\cos y} \right)\)

\( \Rightarrow {u_x} = \frac{1}{5}{e^x}\left( {\cos y - 3\sin y} \right)\)

From equation (3)

\( \Rightarrow {u_y} = \left[ {\frac{2}{5}{e^x}\left( {\cos y - 3\sin y} \right)} \right] - {e^x}\left( {\cos y - \sin y} \right)\)

\(f'\left( z \right) = {u_x} - i{u_y}\)

\( \Rightarrow f'\left( z \right) = \frac{1}{5}{e^x}\left( {\cos y - 3\sin y} \right) - i\left[ {\frac{2}{5}{e^x}\left( {\cos y - 3\sin y} \right) - {e^x}\left( {\cos y - \sin y} \right)} \right]\)

By Milne- Thomson’s method we express f^’(z) in terms of z on replacing x by z and y by 0.

\( \Rightarrow f'\left( z \right) = \frac{1}{5}{e^z}\left( 1 \right) - i\left[ {\frac{2}{5}{e^z}\left( 1 \right) - {e^z}\left( 1 \right)} \right]\)

\( \Rightarrow f'\left( z \right) = \frac{1}{5}{e^z} + i\frac{3}{5}{e^z} = \frac{{{e^z}}}{5}\left( {1 + 3i} \right)\)

By integrating w.r.t. z on both sides

\( \Rightarrow \smallint f'\left( z \right)dz = \smallint \frac{{{e^z}}}{5}\left( {1 + 3i} \right)dz\)

\({e^z} - 1 = \mathop \sum \limits_{n = 0}^\infty \frac{{{z^n}}}{{n!}} - 1\) 

\( = \mathop \sum \limits_{n = 1}^\infty \frac{{{z^n}}}{{n!}}\) 

\( = z\;\mathop \sum \limits_{n = 0}^\infty \frac{{{z^n}}}{{\left( {n + 1} \right)!}}\) 

e^x – 1 has a zero at ‘0’ of multiplicity one and hence f(z) has pole at 0 of order 1. So, the Laurent series f(z) is given by

\(f\left( z \right) = \mathop \sum \limits_{n = - 1}^\infty {a_n}{z^n}\)

\( = \frac{{a - 1}}{z} + {a_0} + {a_1}z + {a_2}{z^2} + {a_3}{z^3} + \ldots \) 

Since (e^z – 1) f(z) = 1

\( \Rightarrow \left( {\frac{{a - 1}}{z} + {a_0} + {a_1}z + {a_2}{z^2} + {a_3}{z^3}} \right)z\left( {1 + \frac{z}{2} + \frac{{{z^2}}}{6} + \frac{{{z^3}}}{4} + \frac{{{z^4}}}{{120}} + \ldots } \right) = 1\) 

\( \Rightarrow \left( {{a_{ - 1}} + {a_0}z + {a_1}{z^2} + {a_2}{z^3} + {a_3}{z^4} + \ldots } \right)\left( {1 + \frac{z}{2} + \frac{{{z^2}}}{6} + \frac{{{z^3}}}{{24}} + \frac{{{z^4}}}{{120}} + \ldots } \right) = 1\) 

By comparing both the sides,

a_-1 = 1

\({a_0} + \frac{{{a_{ - 1}}}}{2} = 0 \Rightarrow {a_0} = \frac{{ - 1}}{2}\) 

Similarly, \({a_1} = \frac{1}{{12}},\;{a_2} = 0,\;{a_3} = \frac{{ - 1}}{{720}}\)