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

Marks: 6M

Year: Dec 2015

(1) Considering the contour $zf(z) \rightarrow 0$ as $|z| \rightarrow \infty$ we find the poles.

(2) $(z^6 + 1)$ given $z^6 = e^{(2n+ 1)\pi i}$

$\therefore z = e^{(2n + 1)\frac{i\pi}{6}}$ where n = 0, 1, 2, 3, 4, 5

$\therefore z = e^{\frac{i\pi}{6}}, e^{\frac{i\pi}{2}}, e^{\frac{5i\pi}{6}}, e^{\frac{7i\pi}{6}}, e^{\frac{3i\pi}{2}}, e^{\frac{11i\pi}{6}}$

Out of these 3 poles lie in the upper half plane.

Let $\alpha$ be one of these poles.

Reside (at z = $\alpha) \lim\limits_{z \rightarrow \alpha} \frac{(z - \alpha)z^2}{z^6 + 1}$

$\lim\limits_{z \rightarrow \alpha} \frac{(z - \alpha)2z + z^2}{6z^2}$, By L hospital rule

$= \frac{\alpha^2}{6\alpha^5} = \frac{\alpha^3}{6\alpha^6} = \frac{-\alpha^3}{5}$ as $\alpha^6 = -1$

Sum of the residue = $\frac{-1}{6} [a_{1}^3 + a_{2}^3 + a_{3}^3] \\ =\frac{-1}{6}\bigg[e^{\frac{i\pi}{2}} + e^{\frac{3i\pi}{2}} + e^{\frac{5i\pi}{2}}\bigg] \\ = \frac{-1}{6} \bigg[cos\frac{\pi}{2} + isin\frac{\pi}{2} + cos\frac{3\pi}{2} + isin\frac{3\pi}{2} + cos\frac{5\pi}{2} + isin\frac{5\pi}{2}\bigg] \\ = \frac{[i - i + i]}{-6} \\ = \frac{-i}{6}$

(3) $\int\limits_{-\infty}^{\infty} \frac{x^2}{x^6 + 1}dx = 2\pi i \bigg(\frac{-1}{6}\bigg) = \frac{\pi}{3}$