Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 8M

Year: Dec 2014

Consider $f(z) = \frac{z - 1}{z^2 - 2z - 3} = \frac{z - 1}{(z - 3)(z + 1)}$

Using partial fraction we get

$\frac{z - 1}{(z - 3)(z + 1)} = \frac{a}{z - 3} + \frac{b}{z + 1} \\ i.e. z - 1 = a(z + 1) + b(z - 3)$

put z = 3, we get $a = \frac{1}{2}$

put z = -1, we get $b = \frac{1}{2}$

$\frac{z - 1}{(z - 3)(z + 1)} = \frac{\frac{1}{2}}{z - 3} + \frac{\frac{1}{2}}{z + 1}$

Hence f(z) is not analytic at z = -1 and z = 3

$\therefore$ f(z) is analytic in

• |z| < 1
• 1 < |z| < 3
• |z| > 3

i. |z| < 1

$\because$ |z| < 1, |z| < 3

$\therefore f(z) = \frac{\frac{1}{2}}{z - 3} + \frac{\frac{1}{2}}{z + 1} \\ = \frac{1}{2} \cdot \frac{1}{-3} \cdot \frac{1}{ 1 - \frac{z}{3}} + \frac{1}{2} \frac{1}{1 + z} \\ = -\frac{1}{6} \bigg(1 - \frac{z}{3} \bigg)^{-1} + \frac{1}{2} (1 + z)^{-1} \\ = -\frac{1}{6} \bigg(1 + \frac{z}{3} + \bigg(\frac{z}{3}\bigg)^2 + \bigg(\frac{z}{3}\bigg)^3 + .....\bigg) + \frac{1}{2} (1 - z + z^2 - z^3 +.....) \\ f(z) = \frac{1}{3} - \frac{5}{9}z + \frac{13}{27} z^2 - ......$

Above is required Taylor's series

ii. 1 < |z| < 3

We get $\bigg| \frac{1}{z} \bigg| \lt 1$ and $\bigg|\frac{z}{3} \bigg| \lt 1$

$\therefore f(z) = \frac{\frac{1}{2}}{z - 3} + \frac{\frac{1}{2}}{z + 1} \\ = \frac{1}{2} \cdot \frac{1}{-3} \cdot \frac{1}{1 - \frac{z}{3}} + \frac{1}{2} \cdot \frac{1}{z} \cdot \frac{1}{1 + \frac{1}{z}} \\ = \frac{1}{-6} \cdot \frac{1}{1 - \frac{z}{3}} + \frac{1}{2z} \cdot \frac{1}{1 + \frac{1}{z}} \\ = -\frac{1}{6} \bigg(1 - \frac{z}{3} \bigg)^{-1} + \frac{1}{2z} \bigg(1 + \frac{!}{z} \bigg)^{-1} \\ = - \frac{1}{6} \bigg(1 + \frac{z}{3} + \bigg(\frac{z}{3}\bigg)^2 + \bigg(\frac{z}{3}\bigg)^3 + .......\bigg) + \frac{1}{2z} \bigg(1 - \bigg(\frac{1}{z} \bigg) + \bigg(\frac{1}{z} \bigg)^2 - \bigg(\frac{1}{z}\bigg)^3 + ...\bigg) \\ f(z) = -\frac{1}{6} \bigg(1 + \frac{z}{3} + \bigg(\frac{z}{3}\bigg)^2 + \bigg(\frac{z}{3}\bigg)^3 + ....\bigg) + \frac{1}{2} \bigg(\frac{1}{z} - \bigg(\frac{1}{z}\bigg)^2 + \bigg(\frac{1}{z}\bigg)^3 - \bigg(\frac{1}{z}\bigg)^4 + .....\bigg)$

This is required Laurent's series

iii. |z| > 3

$\because$ |z| > 3, clearly, |z| > 1

$\therefore \bigg|\frac{z}{3} \bigg| \gt 1$ and |z| > 1

i.e. $\bigg|\frac{3}{z}\bigg| \lt 1$ and $\bigg|\frac{1}{z}\bigg| \lt 1$

$\therefore f(z) = \frac{\frac{1}{2}}{z - 3} + \frac{\frac{1}{2}}{z + 1} \\ = \frac{1}{2z} \cdot \frac{1}{1 - \frac{3}{z}} + \frac{1}{2z} \cdot \frac{1}{1 + \frac{1}{z}} \\ = \frac{1}{2z} \bigg(1 - \frac{3}{z}\bigg)^{-1} + \frac{1}{2z} \bigg(1 + \frac{1}{z}\bigg)^{-1} \\ = \frac{1}{2z} \bigg(1 + \bigg(\frac{3}{z} \bigg) + \bigg(\frac{3}{z}\bigg)^2 + \bigg(\frac{3}{z}\bigg)^3 + ....\bigg) + \frac{1}{2z} \bigg(1 - \bigg(\frac{1}{z} \bigg) + \bigg(\frac{1}{z}\bigg)^2 - \bigg(\frac{1}{z}\bigg)^3 + ....\bigg) \\ = \frac{1}{z} + \frac{1}{z^2} + \frac{5}{z^3} + \frac{13}{z^4} + .........\\ f(z) = \frac{1}{z} + \frac{1}{z^2} + \frac{5}{z^3} + \frac{13}{z^4} + .........$

This is the required Laurent's series