这是一份uwa西澳大学MATH2501的成功案例

From the point $p \in S^{1}$ is mapped onto the point:
$$
\phi_{1}(p)=\frac{2 a}{\tan \frac{\theta}{2}} \in \mathbb{R}^{1} .
$$
This is defined for $\theta \neq 0$, that is, $p \in M_{1}$ where $M_{1}$ is the chart excluding $\theta=0$. For the opposite stereographic projection (upwards from the south pole), we will get
$$
\phi_{2}(p)=\frac{2 a}{\cot \frac{\theta}{2}}=2 a \tan \frac{\theta}{2}
$$
which is defined for $\theta \neq \pi$, hence for $p \in M_{2}$.
Now consider $\phi_{2} \circ \phi_{1}^{-1}: \mathbb{R} \rightarrow \mathbb{R}$. We have:
$$
\begin{aligned}
\phi_{1}^{-1}(0,2 \pi) &=2 \cot ^{-1} \frac{x}{2 a} \
\phi_{2} \circ \phi_{1}^{-1} &=2 a \tan \left(\cot ^{-1} \frac{x}{2 a}\right) \
&=\frac{4 a^{2}}{x}
\end{aligned}
$$
This is the inversion map: $\mathbb{R}-{0} \rightarrow \mathbb{R}-{0}$ which is infinitely differentiable. The map $\phi_{1} \circ \phi_{2}^{-1}$ works similarly.

MATH2501 COURSE NOTES ：

There is an alternative way to look at this. For a given $a^{\prime} \in V^{*}$, we have a linear map
$$
\left\langle a^{\prime},\right\rangle: \quad V \rightarrow \mathbb{R}
$$
defined by
$$
\left\langle a^{\prime},\right\rangle: a \in V \rightarrow\left\langle a^{\prime}, a\right\rangle \in \mathbb{R}
$$
Thus the dual vector space to $V$ is the space of linear functionals on $V$.
Let us clarify what we mean by linear functionals. A functional is a map which takes an element of vector space to a number. In our case, clearly $\langle a,$, is a functional, and it has the additional property that:
$$
\left\langle a^{\prime},\right\rangle: \lambda a+\mu b \rightarrow \lambda\left\langle a^{\prime}, a\right\rangle+\mu\left\langle a^{\prime}, b\right\rangle
$$
which is just what we mean by linearity.
An important property of dual spaces is that if we have a map of vector spaces $V, W$ :
$$
f: V \rightarrow W
$$

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:
$$
\operatorname{Res}\left(f(z), z_{0}\right)=\lim {z \rightarrow z{0}}\left(\frac{1}{(n-1) !} \frac{\mathrm{d}^{n-1}}{\mathrm{~d} z^{n-1}}\left[\left(z-z_{0}\right)^{n} f(z)\right]\right)
$$