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

Marks: 6M

Year: May 2015

Given that $|z| = 2$

Thus the center of circle is at origin and radius is 2

Here $f(z) = 3z^2 + 7z + 1$

1) For $z_{0} = -3$ the point lies outside the region

Hence by Cauchy's theorem

f(-3) = 0

2) $z_{0}$ = i. Now i lies inside the circle

$\therefore f(a) = f(z_{0}) \times 2\pi i \\ = 2\pi i (3i^2 + 7i + 1) \\ = 2\pi i(7i - 2)$

3) $z_{0} = 1 - i$ lies inside the circle

$\therefore f'(1 - i) = \frac{2\pi i}{(1 - i)1} \times f'(2\mu \beta) \\ = 2\pi i \frac{d}{dz} (3z^2 + 7z + 1) \\ = \frac{2\pi i}{1} (6z + 7) \\ = 2\pi i[6(1 - i) + 7] \\ = 2\pi i[13 - 6i]$

Now putting z = 1 - i we get

$f"(1 - i) = \frac{2\pi i}{(1 - i)1} \times f"(z) \\ = 2\pi i \times \frac{d}{dz^2} (3z^2 + 7z + 1) \\ = 2\pi i \frac{d}{dz} (6z + 7) \\ = 2\pi i(6) \\ = 12\pi i$