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

Marks: 8M

Year: Dec 2015

Let $f(z) = \frac{a}{z + 2} + \frac{bz + c}{z^2 + 1} \\ \therefore \frac{1}{(z^2 + 1)(z + 2)} = \frac{a(z^2 + 1) + (bz + c)(z + 2)}{(z^2 + 1)(z + 2)}$

Let z = -2

1 = a(5)

$\therefore a = \frac{1}{4}$

Let z = 0

$\therefore 1 = \frac{1}{5} + 2c \\ \therefore c = \frac{2}{5}$

Let z + 1

$1 = \frac{1}{5}(2) + (b + c)(3) \\ 1 = \frac{2}{5} + 3b + \frac{6}{3} \\ \therefore b = \frac{-1}{5} \\ \therefore f(z) = \frac{1}{5} \frac{1}{z + 2} - \frac{z - 2}{5(z^2 + 1)}$

Case (1) when $1 \lt |z| \lt 2$

i.e. $\frac{|z|}{2} \lt 1$ and $|z^2| \gt 1$

i.e. $\frac{1}{|z|^2} \lt 1$

Hence we write

$f(z) = \frac{1}{5} \times \frac{1}{2} \frac{1}{1 + \frac{z}{2}} - \frac{z - 2}{5} \cdot \frac{1}{z^2} \cdot \frac{1}{1 + \frac{1}{z^2}} \\ = \frac{1}{10} \bigg(1 + \frac{z}{2}\bigg)^{-1} - \bigg(\frac{z - 2}{5}\bigg) \frac{1}{z^2} \bigg(1 + \frac{1}{z^2}\bigg)^{-1} \\ = \frac{1}{10} \bigg[1 - \frac{z}{2} + \frac{z^2}{4}.....\bigg] - \bigg(\frac{z - 2}{5}\bigg) \times \bigg[\frac{1}{z^2} - \frac{1}{z^4} + \frac{1}{z^6}....\bigg]$

2nd case when $|z| \gt 2$

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