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

Marks: 4M

Year: Dec 2015

$|z| = 2$ is a circle with center at the origin and radius is $4z^3(z + 4) = 0$ given z = 0 and z = -4. Both the points lie inside the circle. Now by partial fraction. $\frac{1}{z^3(z + 4)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z^3} + \frac{D}{(z + 4)} \ \frac{1}{z^3(z + 4)} = \frac{Az^2 (z + 4) + Bz(z + 4) z (z + 4) + Dz^4}{z^3 (z + 4)} $ Let z = 0 we get, 1 = 4c $\therefore c = \frac{1}{4}$ Let z = -4 we get, 1 = $D(-4)^3 \hspace{0.5cm} \therefore D = \frac{-1}{64}$ Let z = 1 we get, $1 = 5A + 5B + \frac{5}{4} - \frac{1}{64} \ \frac{-15}{64} = 5A + 5B..............(1)$ Let z = -1 we get, $1 = 3A - 3B + \frac{3}{4} + \frac{1}{64} \ \frac{15}{64} = 3A - 3B............(2)$ Solving (1) and (2) we get, $\therefore A = \frac{1}{64} \hspace{1cm} B = \frac{-1}{16}$ $\therefore$ we get $\int \frac{1}{64z} + \frac{-1}{16z^2} + \frac{1}{4z^3} - \frac{1}{64(z + 4)}dz$ Hence by Cauchy's theorem, \begin{vmatrix} \hline 1st term f(z) = \frac{1}{64} & 2nd term f(z) = \frac{-1}{16} \\ \hline 3rd term f(z) = \frac{1}{4} & 4th term f(z) = \frac{-1}{64} \\ \hline \end{vmatrix} For $\therefore \int \frac{1}{64z}dz = \frac{2\pi i}{64} \ \therefore \int \frac{-1}{16z^2} = \frac{2\pi i}{1!} \frac{d}{dz} \frac{1}{16} = 0 \ \int \frac{1}{4z^3} = \frac{2\pi i}{1!} \frac{d}{dz} \frac{1}{4} = 0 \ \int \frac{-1 dz}{64z(z + 4)} = \frac{-2\pi i}{64}$ $\therefore$ Adding are the terms we get $\int \frac{1}{z^3 (z + 4)} = \frac{2\pi i}{64} - \frac{2\pi i}{64} = 0$