Question

The residue of Res(3i) for the function R(z)=z^2-9/(z^2+9)^2

Answer 1

Answer:

0

Step-by-step explanation:

1) We have,

$R(z) = \frac{z^2-9}{(z^2+9)^2}\\ \\ = \frac{z^2-9}{(z+3i)^2(z-3i)^2}$

2)Clearly, there is 2nd order pole at z=3i.

So,

$g(z)=(z-3i)^2R(z)$

is analytic around z=3i.

3) Therefore,

$Res(R(z),z=3i) = \frac{1}{(2-1)!}*g'(3i)\\ \\ = 1*[\frac{-2*(z^2-9)}{(z+3i)^3}+\frac{2z}{(z+3i)^2}]_{z=3i} \\ \\ = \frac{-2*-18}{-216i}+\frac{2*3i*6i}{-36*6i} \\ \\=\frac{36-36}{-216i}=0$

Since, your question is  not clear so I just find the value of Res(3i).

Hope that works for you.