Question: If f($\xi$) = $\int \frac{3z^2+2z+1}{z-\xi} dz $ over the curve C where C is the circle $x^2+y^2=4$.

0

0

Here, r = 2

$f(\xi) = 2 \pi i[3 \xi^2 + 2 \xi + 1]$ if $\xi$ lies inside

$f(\xi) = 0$ if $\xi$ lies outside

(i) z = 3 = (3,0) lies outside radius = 2. Therefore, f(3) = 0

(ii) z = (1-i) = (1,-1)

Let A = (1,-1) and c = (0,0) where c is the centre of circle $ x^2 + y^2 = 4 $

d(Ac) = $ \sqrt{(1-0)^2 + (-1-0)^2} = \sqrt{1+1} = \sqrt{2} \lt 2 $

Therefore, A lies inside the circle.

$ \therefore f'(\xi) = 2 \pi i (6 \xi + 2) \\ f'(1-i) = 2 \pi i (6 (1-i) + 2) \\ = 2 \pi i (6-6i + 2) \\ = 2 \pi i (8-6i) $

(iii) z =(1-i) lies inside

$ f''(\xi) = 2 \pi i(6) = 12 \pi i \\ f''(1-i) = 12 \pi i $

Please log in to add an answer.

Site

- Publications
- Advertise
- RSS
- Stats

Use of this site constitutes acceptance of our User
Agreement
and Privacy
Policy.