Question: If $ \phi(\alpha)= \int \frac{ze^z}{z-\alpha} dz $ over the curve C where C is |z-2i| = 3

0

0

C = (0,2); r = 3

$\phi(\alpha) = 2 \pi i[\alpha e^\alpha]$ is $\alpha$ lies inside

= 0 if $\alpha$ lies outside

(i) z = 1 = (1,0)

Let A = (1,0), C = (0,2)

d(AC) = $ \sqrt{(1-0)^2 + (0-2)^2} = \sqrt{1+4} = \sqrt{5} \lt 3 $

Therefore, A = (1,0) lies outside.

$ \therefore \phi(1) = 2 \pi i[1e^1] = 2 \pi i e $

(ii) z = 2 = (2,0)

Let, B = (2,0); C = (0,2)

d(BD) = $ \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8} \lt 3 $

Therefore, B lies inside.

$ \phi'(\alpha) = 2 \pi i [\alpha e^\alpha + e^\alpha] = 2 \pi i [2e^2 + e^2] \\ = 2 \pi i (e^2)(2+1) = 2 \pi i(e^2)(3) = 6 \pi i e^2 $

(iii) z = 3 = (3,0)

D = (3,0); C = (0,2)

d(CD) = $ \sqrt{(3-0)^2 + (0-2)^2} = \sqrt{3^2 + (-2)^2} + \sqrt{9+4} = \sqrt{13} \gt 3 $

Therefore D lies outside.

$ \therefore \phi(3) = 0 $

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