$$f(z):= \frac{(z-1)^2(z+3)}{1 - \sin(\pi z/2)}$$
Classify all singularities of $f$. (p145 Frietag's).
I obtain that singularities are at $z= 1+4k$. Further if $z=1$, we have removable singularity, $z=-3$ we have simple pole and all other singularities are pole of order $2$. I made the substitution $w = z-r$, $r$ is a root, and taylor expand the cosine series.
Frietag's solution states that all singularities are simple other than at $z=1,-3$ which are removable.
Is he right or my proof wrong?
You are right about $-3$: it is a simple pole, since $f(z)=\frac{g(z)}{h(z)}$, with $g(z)=(z-1)^2(z+3)=16(z+3)+8(z+3)^2+(z+3)^3$ and $h(z)=1-\sin(\pi z/2)$. Furthermore, $h(-3)=h'(-3)=0$ and $h''(0)\neq0$.