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

Marks: 6M

Year: Dec 2014

Let $I = \int\limits_{0}^{2\pi} \frac{d\theta}{(2 + cos\theta)^2}........(1)$

put $z = e^{i\theta}$

then $dz = ie^{i\theta}d\theta \\ i.e. dz = izd\theta \\ \therefore d\theta = \frac{dz}{iz} \\ \therefore cos\theta = \frac{(e^{i\theta} + e^{-i\theta})}{2} \\ i.e. cos\theta = \frac{z^2 + 1}{2z}$

$\therefore$ C is |z| = 1

Substituting the above value of $cos\theta$ in eq(1) we get

$$I = \int\limits_{0}^{2\pi} \frac{1}{\bigg(2 + \frac{(z^2 + 1)^2}{2z}\bigg)^2} \cdot \frac{dz}{iz} \\ I = \frac{4}{i} \int\limits_{0}^{2\pi} \frac{z}{(z^2 + 4z + 1)^2}dz......(2)$$

The poles of the integrand are given by $z^2 + 4z + 1 = 0$

$\therefore$ The roots are $z = -2 \pm \sqrt{3}$

Let $\alpha = -2 + \sqrt{3}, \beta = -2 - \sqrt{3}$

Here pole $\alpha$, of order 2 lies inside C whereas $\beta$ lies outside the circle.

Hence by Cauchy Residues Theorem, we have

$\int f(z)dz = 2\pi i$ (sum of the residues).......over contour C .......(3)

$Res(z = \alpha) = \lim\limits_{z \rightarrow \alpha} \frac{d}{dz} \bigg[(z - \alpha)^2 \cdot \frac{z}{(z - \alpha)^2 (z - \beta)^2}\bigg] \\ = \lim\limits_{z \rightarrow \alpha} \frac{d}{dz} \bigg[\frac{z}{(z - \beta)^2}\bigg] \\ = \lim\limits_{z \rightarrow \alpha} \frac{[(z - \beta)^2 - z \cdot 2 (z - \beta)]}{(z - \beta)^4} \\ Res(z = \alpha) = \lim\limits_{z \rightarrow} \alpha \frac{-z - \beta}{(z - \beta)^3} \\ = - \frac{\alpha + \beta}{(\alpha - \beta)^3} \\ = - \frac{-4}{(2\sqrt{3})^3} \\ \therefore Res(z = \alpha) = \frac{1}{6\sqrt{3}}$

Hence $I = \frac{4}{i} \cdot 2\pi i \cdot \frac{1}{6\sqrt{3}}$............From eq(2) and eq(3)

$\therefore I = \frac{4\pi}{3\sqrt{3}}$

Hence $I = \int\limits_{0}^{2\pi} \frac{d\theta}{(2 + cos\theta)^2} = \frac{4\pi}{3\sqrt{3}}$