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

Marks: 8M

Year: May 2015

Let $f(z) = \frac{2z - 3}{z^2 - 4z + 3} = \frac{a}{z - 1} + \frac{b}{z - 3}$

By partial fraction

2z - 3 = a(z - 3) + b(z - 1)

When z = 1 -1 = a(-2)

$\therefore a = \frac{1}{2}$

When z = 3 $b = \frac{3}{2}$

$f(z) = \frac{1}{2} \bigg(\frac{1}{z - 1}\bigg) + \frac{3}{2} \frac{1}{(z - 3)} \\ = \frac{1}{2} \bigg[\frac{1}{(z - 4) + 3}\bigg] + \frac{3}{2[(z - 4) + 1]}$

When (z - 4) < 1 we write

$f(z) = \frac{1}{2[(z - 4) + 3]} + \frac{3}{2[(z - 4) + 1]}$ as

$= \frac{1}{2.3 \bigg[1 + \bigg(2 - \frac{4}{3}\bigg)\bigg]} + \frac{3}{2[1 + (2 - 4)]} \\ = \frac{1}{6} \bigg[1 + \bigg(\frac{2 - 4}{3}\bigg)\bigg]^{-1} + \frac{3}{2}[1 + (2 - 4)]^{-1} \\ = \frac{1}{3} \bigg[1 - \bigg(\frac{2 - 4}{3}\bigg) + \bigg(\frac{2 - 4}{3}\bigg)^2 + .....\bigg] + \frac{3}{2} [1 - (2 - 4) + (2 - 4)^2 + .....]$

When (2 - 4) > 3 we write

$f(z) = \frac{1}{2[(z - 4) + 3]} + \frac{3}{2[(z - 4) + 1]}$ as

$= \frac{1}{2(z - 4) \bigg[1 + \frac{3}{z - 4}\bigg]} + \frac{3}{2(z - 4) \bigg[1 + \frac{1}{z - 4}\bigg]} \\ = \frac{1}{2(z - 4)} \bigg[1 + \frac{3}{z - 4}\bigg]^{-1} + \frac{3}{2(z - 4)} \bigg[1 + \frac{1}{z - 4}\bigg]^{-1} \\ = \frac{1}{2(z - 4)} \bigg[1 - \bigg(\frac{3}{z - 4}\bigg) + \bigg(\frac{3}{z - 4}\bigg)^2 + .....\bigg] + \frac{3}{2(z - 4)} \bigg[1 - \bigg(\frac{1}{z - 4}\bigg) + \bigg(\frac{1}{z - 4}\bigg)^2 + ......\bigg]$