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

Marks: 6M

Year: May 2014

The circle |z| = 2 has center at the origin and radius 2.

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

$\therefore$ By Cauchy's Integral Theorem

$$f(3) = \oint_{c} \frac{3z^2 + 11z + 7}{z - 3}dz = 0$$

ii. The point z = 1 + i i.e. (1,1) lies inside the cirlce. Hence we take $\phi(z) = 3z^2 + 11z + 7$ which is analytic everywhere.

$\therefore f(a) = \oint_{c} \frac{\phi(z)}{z - a}dz = 2\pi i \phi(a) = 2\pi i(3a^2 + 11a + 7).................(1)$

Taking derivative of eq(1) we get

$f'(a) = 2\pi i (6a + 11)................(2)$

$$\therefore f'(1 + i) = 2\pi i(17 + 6i)$$

iii. The point z = 1 - i i.e. (1, -1) lies inside the circle. Hence we take $\phi(z) = 3z^2 + 11z + 7$ which is analytic everywhere.

Taking derivative of eq(2)

Now $f''(a) = 12\pi i.................(3)$

$$\therefore f"(1 - i) = 12\pi i$$

Hence,

$$f(3) = \oint_{c} \frac{3z^2 + 11z + 7}{z - 3}dz = 0 \\ f'(1 + i) = 2\pi i(17 + 6i) \\ f"(1 - i) = 12\pi i$$