这是一份uwa西澳大学MATH2501的成功案例

There are no other singularities in the finite complex plane. We examine the point at infinity.
$$
2 \sin \left(\frac{1}{z}\right)=\frac{1}{\zeta} \sin \zeta
$$
The point at infinity is a singularity. We take the limit $\zeta \rightarrow 0$ to demonstrate that it is a removable singularity.
$$
\lim {\zeta \rightarrow 0} \frac{\sin \zeta}{\zeta}=\lim {\zeta \rightarrow 0} \frac{\cos \zeta}{1}=1
$$
$$
\frac{\tan ^{-1}(z)}{z \sinh ^{2}(\pi z)}=\frac{\imath \log \left(\frac{z+z}{z-z}\right)}{2 z \sinh ^{2}(\pi z)}
$$

MATH2501 COURSE NOTES ：

Residues. Let $f(z)$ be single-valued an analytic in a deleted neighborhood of $z_{0}$. Then $f(z)$ has the Laurent series expansion
$$
f(z)=\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n},
$$
The residue of $f(z)$ at $z=z_{0}$ is the coefficient of the $\frac{1}{z-z_{0}}$ term:
$$
\operatorname{Res}\left(f(z), z_{0}\right)=a_{-1} .
$$
The residue at a branch point or non-isolated singularity is undefined as the Laurent series does not exist. If $f(z)$ has a pole of order $n$ at $z=z_{0}$ then we can use the Residue Formula:
$$
\operatorname{Res}\left(f(z), z_{0}\right)=\lim {z \rightarrow z{0}}\left(\frac{1}{(n-1) !} \frac{\mathrm{d}^{n-1}}{\mathrm{~d} z^{n-1}}\left[\left(z-z_{0}\right)^{n} f(z)\right]\right)
$$