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

Marks: 6M

Year: Dec 2015

Let z = $e^{i\theta}d\theta = \frac{dz}{iz}cos\theta = \frac{z^2 + 1}{2z}sin\theta = \frac{z^2 - 1}{2iz} \\ \therefore I = \int \frac{\bigg(\frac{z^2 - 1}{2iz}\bigg)^2}{a + b \bigg(\frac{z^2 + 1}{2z}\bigg)} \frac{dz}{iz} \\ = - \frac{1}{2i} \int \frac{(z^2 - 1)^2}{z^2(bz^2 + 2az + b)}dz$

Where is the circle |z| = 1

Now the poles of f(z) is given by z = 0 which is a pole of order 2 and

$z = \frac{-2a \pm \sqrt{4a^2 - 4b^2}}{2b}$

i.e.

$\alpha = \frac{-a + \sqrt{a^2 - b^2}}{b} \\ \beta = \frac{- a - \sqrt{a^2 - b^2}}{b}$

Which are simple poles since a > b > 0, $\alpha$ lies inside and $\beta$ lies outside.

The circle |z| = 1

Residue of f(z) (at z = 0) $\\ = z\varinjlim 0 \frac{d}{dz} \bigg[\frac{z^2 (z^2 - 1)^2}{z^2 (bz^2 + 2az + b)}\bigg] \\ = z\varinjlim 0 \frac{d}{dz} \bigg[\frac{(z^2 - 1)^2}{(bz^2 + 2az + b)}\bigg] \\ = z \varinjlim 0 \frac{(bz^2 + 2az + b)2 (z^2 - 1) 2z - (z^2 - 1)^2 (2bz + 2z)}{(bz^2 + 2az + b)^2} \\ = \frac{-2a}{b^2}$

Now Residue of b(z) at (z = $\alpha$ ) $\\ = z \varinjlim 0 \frac{(z - \alpha)(z^2 - 1)^2}{z^2 b (z - \alpha)(z - \beta)} \\ = z \varinjlim 0 \frac{(z^2 - 1)^2}{z^b (z - \beta)} = \frac{(\alpha^2 - 1)^2}{b\alpha^2 (\alpha - \beta)} \\ = \frac{\bigg[\alpha - \frac{1}{\alpha}\bigg]^2}{b (\alpha - \beta)} \\ = \frac{(\alpha - \beta)}{b (\alpha - \beta)} = \frac{1}{b} \frac{2 \sqrt{a^2} - b^2}{b}$

I $= 2\pi i \bigg(-\frac{1}{2i}\bigg) \bigg[\frac{-2a}{b^2} + \frac{2 \sqrt{a^2 - b^2}}{b}\bigg] \\ = \frac{2\pi}{b^2}[a - \sqrt{a^2 - b^2}]$