Calculate residues at all isolated singularities of $f(z)=\frac{z^2+4}{(z+2)(z^2+1)^2}$.
So I found the isolated singularities to be $z=-2$ and $z=i$.
Then I found the residue at $z=-2$.
$\operatorname{res}(f,i)=\lim_{z\to -2}$ $(z+2)\left(\frac{z^2+4}{(z+2)(z^2+1)^2}\right)$
$=\lim_{z\to -2} \frac{z^2+4}{(z^2+1)^2}=\frac{8}{25}$
but I don't know how to do the other one.
I tried this, since it's a double pole:
$\operatorname{res}(f,i)=\lim_{z\to i} \left((z-i)^2\frac{z^2+4}{(z+2)(z^2+1)^2}\right)'$, but I got stuck because this didn't allow me to cancel the singularity. What do I do?
Best Answer
Let $g(z)=\dfrac{z^2+4}{(z+2)(z+i)^2}$. Then $g(i)=\dfrac{-6+3i}{20}$ and $g'(i)=-\dfrac{8+31i}{50}$. So, near $i$ you have$$g(z)=\frac{-6+3i}{20}-\dfrac{8+31i}{50}(z-i)+\cdots$$and therefore$$f(z)=\frac{g(z)}{(z-i)^2}=\frac{-6+3i}{20}(z-i)^{-2}-\dfrac{8+31i}{50}(z-i)^{-1}+\cdots$$So$$\operatorname{res}(i,f(z))=-\dfrac{8+31i}{50}$$and the computation of $\operatorname{res}(-i,f(z))$ is similar.