Can someone verify my analysis for the following question.
$f(z)=\frac{z^\frac{1}{2}}{z^2 + 1}$,
find all isolated singular points, classify them and find residue of $f$ for each singular point.
$f(z)$ has 2 isolated singular points at $z=i$ and $z=-i$, simple poles at $z=i$ and $z=-i$. residue at $z=-i$ is $\frac{-i^\frac{-1}{2}}{2}$ and residue at $z=i$ is $\frac{i^\frac{-1}{2}}{2}$
It looks fine to me (even assuming you didn't have to choose branch for the square root function...)...yet it might be you're expected to write a simpler expression for those complex powers: $$z^2=i=e^{\pi i/2+2k\pi i}\implies z=e^{\pi i/4+k\pi i}=e^{\frac{\pi i}4\left(1+4k\right)}\implies z_k=\begin{cases}e^{\pi i/4}=\frac1{\sqrt2}(1+i)\\{}\\e^{5\pi i/4}=-\frac1{\sqrt2}(1+i)\end{cases}$$
Then you get, for example:
$$\frac{i^{-1/2}}2=\frac1{\sqrt2(1+i)}=\frac{1-i}{2\sqrt2}$$
and etc.