I am a graduate student.I am studying complex analysis.I encountered the following problem in a lecture:
Find the residue of $f(z)=\frac{z-\sinh(z)}{z^2\sinh(z)} $ at $z=\pi i$.
Now,this problem is approached by the instructor as follows:
$\sinh z$ has a simple zero at $\pi i$ .From there the instructor concludes that $f(z)$ has a simple pole at $\pi i$.I understand that $1/f$ has a zero implies $f$ has a pole but $\sinh(z)$ has a simple zero should imply $\frac{1}{\sinh z}$ has a simple pole.But how does he conclude that $f(z)$ has a simple pole.I think the argument should be in this way.
$\frac{1}{\sinh z}$ has a simple pole implies that $\frac{1}{\sinh z}=\frac{g(z)}{z-\pi i}$ for some analytic function $g$ that does not vanish in a neighborhood of $\pi i$.Then $f(z)=(z-\pi i)^{-1} g(z).z^{-2}(z-\sinh z)=(z-\pi i)h(z)$ where $h$ is analytic in a neighborhood of $\pi i$,because $h(z)$ can have singularity only at $0$ due to $z^2$ in the denominator .But we can take a neighborhood of $\pi i$ not containing $0$.Am I correct?
I know that it is very easy question and I should be able to answer it on my own.Still,I want to make sure that I understand well.
Yes, that is correct. But it is simpler to say that, since $\sinh$ has a simple pole at $\pi i$ and $z^2$ is not $0$ at that point, then $z^2\sinh(z)$ also has a simple pole at $\pi i$. On the other hand, $\pi i$ is not a zero of $z-\sinh(z)$. Therefore, $f(z)$ has a simple pole at $\pi i$.