Can anyone give me a rigorus explantion to why simple roots are removable singularities in the function $f(z)=\text{sinh}\space z.$ I get the roots are $\pm n\pi i$, but why are $0, \pm \pi i$ and $\pm 2\pi i$ simple?
How about $\pm 3\pi i?$ Furthermore, it states in my answer solution that the function defined by:
$$h(z) = \frac{(z^2+\pi^2)(e^z-1)}{\text{sinh}\space z}$$ has the nearest pole to the origin of $\pm 3\pi i$, this also confuses me, why cant the poles infact be $\pm 2\pi i$ or $\pm \pi i$ or 0? Is it because they are removable? Then it goes back to my original question, why can these three roots be removable singularities in the first place?
I would appreciate the clarification.
If $f(z_0)=0$, then $f'(z_0) \ne 0$,
($f'(z) = \cosh z$)
Hence $z_0$ is a simple zero of $f$.
We have $g(z)=0$ for $z \in Z:=\{0, \pm i \pi , \pm 2 \pi i\}$ and each $z \in Z$ is a simple zero of $g$.
From 1. and 2, we get: each $z \in Z$ is a removable singularity of $h$.