Engineering Mathematics Test 9

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

Concept:

If u(x, y) is harmonic function, then it must satisfy Laplace’s equation u_xx + u_yy = 0

Calculation:

Given:

u(x, y) = ebx cos 5y

∴ u_x = be^bx cos 5y

∴ u_xx = b²e^bx cos 5y

∴ u_y = -5e^bx sin 5y

∴ u_yy = -25e^bx cos 5y

∵ for harmonic function, uxx + uyy = 0

∴ b² e^bx cos 5y – 25e^bx cos 5y = 0

∴ e^bx cos 5y (b² - 25) = 0

∴ b² – 25 = 0

∴ b = ± 5

\(f(z) = \frac{{\phi(z)}}{{\psi(z)}}\)

\(\cot hz = \frac{{\cos hz}}{{\sin hz}}\)

sin hz = 0

z = nπi

Residue

\(\begin{array}{*{20}{c}} {Res\;f\left( z \right)}\\ {z = {z_0}} \end{array} = \frac{{\phi \left( z \right)}}{{\psi '\left( z \right)}}\)

\(\begin{array}{*{20}{c}} {Res\;f\left( z \right)}\\ {z = {\rm{n\pi i}}} \end{array} = \frac{{\cos hz}}{{\cos hz}}\)

f(z) = z̅ = x - iy

u = x, v = -y

u_x = 1, v_y = -1

u_y = 0, v_x = 0

u_x ≠ v_y, as f(z) is not satisfying CR equations, f(x) is not analytic.

g(z) = |z|²

\(\left| z \right| = \sqrt {{x^2} + {y^2}}\)

g(z) = x² + y²

u_x = 2x, v_x = 0

u_y = 2y, v_y = 0

at z = 0, u_x = v_y and u_y = -v_x

g(x) is satisfying CR equations at z = 0 but it is not satisfying CR equations at neighborhood of z = 0.

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.

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:

\(f\left( z \right) = \frac{{z - 8}}{{z\left( {z - 4} \right)}}\)

Poles: z = 0, 4

Zeros: z = 8

Residue at f(z) at z = 0 is,

\(= \mathop {{\rm{lt}}}\limits_{z \to 0} zf\left( z \right)\)

\(= \mathop {{\rm{lt}}}\limits_{z \to 0} \frac{{z - 8}}{{z - 4}} = 2\)

Residue at f(z) at z = 4 is,

\(= \mathop {{\rm{lt}}}\limits_{z \to 4} \left( {z - 4} \right)f\left( z \right)\)

\(= \mathop {{\rm{lt}}}\limits_{z \to 4} \frac{{z - 8}}{z} = - 1\)

Value of the integral = 2πi [sum of residues]

= 2πi [2 - 1] = 2πi

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

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

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

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

z^-3 and z² are not continuous everywhere and hence f(z) is also not continuous everywhere.

At z = 0, we have

\(\mathop {\lim }\limits_{z \to 0} \left| {f\left( z \right)} \right| = \mathop {\lim }\limits_{z \to 0} \left| {\frac{{{z^{ - 3}}}}{{{z^2}}}} \right| = \mathop {\lim }\limits_{z \to 0} \left| 1/z^5 \right| \) = ∞ 

So f is not continuous at 0.

The complex derivative if exists, given by

\(\mathop {\lim }\limits_{z \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{z - 0}} = \mathop {\lim }\limits_{z \to 0} \frac{{{z^{ - 3}}}}{{{z^2}}}\)

Let z = x + yi, if y = 0 and x → 0, then

\(\mathop {\lim }\limits_{z \to 0} \frac{{{z^{ - 3}}}}{{{z^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^{-3}}}}{{{x^2}}} = +∞\)

On the other hand, if x = 0 and y → 0, then

\(\mathop {\lim }\limits_{iy \to 0} \frac{{{z^{ - 3}}}}{{{z^2}}} = \mathop {\lim }\limits_{iy \to 0} \frac{{{{\left( { yi} \right)}^{-3}}}}{{{{\left( {yi} \right)}^2}}} = - ∞\)

Hence, f’(0) does not exist.

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

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 \(\frac{1}{8}\)