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

Marks: 6M

Year: Dec 2014

The circle |z - 2i| = 3 has center at (0, 2) and radius 3

i. The point z = 1 lies inside the circle. Hence we take $f(z) = ze^z$ is analytic in and on C.

Hence by Cauchy's formula

$$\phi(1) = \oint_{c} \frac{ze^z}{z - 1}dz = 2\pi i f(z_{0})$$

here $z_{0} = 1, \therefore \phi(1) = 2\pi i(1 \cdot e^1) = 2\pi i e$

$$\therefore \phi(1) = 2\pi ie$$

ii. The point z = 2 lies inside the circle. Hence we take $f(z) = ze^z$ is analytic in and on C.

$\therefore \phi(\alpha) = \oint \frac{ze^z}{z - \alpha} dz = 2\pi if(\alpha) = 2\pi i [\alpha e^{\alpha}] \\ \therefore \phi'(\alpha) = 2\pi i[\alpha e^{\alpha} + e^{\alpha}] \\ i.e. \phi'(\alpha) = 2\pi ie^{\alpha} (\alpha + 1) \\ \therefore \phi'(2) = 2\pi ie^2(3) = 6\pi ie^2$

$$\therefore \phi'(2) = 2\pi ie^2(3) = 6\pi ie^2$$

iii. The point z = 3 lies outside the circle.

$\therefore$ By Cauchy's Integral Theorem

$$\phi(3) = \oint \frac{ze^z}{z - 3}dz = 0 \\ \therefore \phi(3) = 0$$

iv. The point z = 4 lies outside the circle.

$\therefore$ By Cauchy's Integral Theorem

$$\phi(4) = \oint \frac{ze^z}{z - 4}dz = 0 \\ \therefore \phi'(4) = 0$$

Hence

$$\phi(1) = 2\pi ie \\ \phi'(2) = 6\pi ie^2 \\ \phi(3) = 0 \\ \phi'(4) = 0$$