× Close
Join the Ques10 Community
Ques10 is a community of thousands of students, teachers, and academic experts, just like you.
Join them; it only takes a minute
Sign up
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

Find the values of:

(i) f(3)

(ii) f’(1-i)

(iii) f’’(1-i)

Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 103 views
ADD COMMENTlink
modified 10 weeks ago  • written 5 months ago by gravatar for Manan Bothra Manan Bothra0
0

Here, r = 2

enter image description here

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

ADD COMMENTlink
written 10 weeks ago by gravatar for Manan Bothra Manan Bothra0
Please log in to add an answer.


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