Prove that $\frac{1}{\sin^2 z } = \sum\limits_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $ I have the following problem: 

Find the constants $c_n$ so that 
$$ \frac{1}{\sin^2 z } = \sum_{n= -\infty} ^ {+\infty} \frac{c_n}{(z-\pi n)^2} $$
and the series converges uniformly on every bounded set after dropping finitely many terms. Justify all your claims.
Hint: Use Liouville's theorem to prove the equality. 

Let $c_n = 1 $ for all $n$. 
Let $\displaystyle f(z) := \frac{1}{\sin^2 z}$ and $\displaystyle g(z) := \sum_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $. 
Begin by showing that $h$ is an analytic function which converges uniformly on compact subsets of $\mathbb{C} \backslash \mathbb{Z}$. 
Suppose $K$ is a compact set which contains no integers. Define 
$$ \delta_n = \inf_{z \in K} |\pi-z/n| =\frac{1}{n} \inf_{z \in K} |\pi n-z|, $$
where the infimum exists because $K$ is compact. Also, compactness implies boundedness. Thus as $n \to \pm \infty$, we have $\delta_n \to \pi$. 
Therefore, for sufficiently large $n$, we have $\delta_n > 2$, and 
$$ \frac{1}{|z \pm n | ^2} \leq \frac{1}{\delta_n ^2 n^2} < \frac{1}{4n^2}. $$
By the Weierstrass M-test, $g$ converges absolutely uniformly on $K$. Since each term is analytic on $K$, we conclude that the series converges to an analytic function on $\mathbb{C} \backslash \mathbb{Z}$. 
Clearly the only poles of $g(z)$ are at $\pi n$ for each integer $n$, with corresponding principal part $\frac{1}{(z-\pi n)^2}$. 
For each integer $n$, we have $\sin^2 (\pi n) = \frac{d}{dz} \sin^2(\pi n) = 0$ and $\frac{d^2}{dz^z} \sin^2 (\pi n) \neq 0$, so $f(z) = 1/ \sin^2 (z)$ also has a pole of order two at each integer multiple of $\pi$, and no other poles.
Furthermore, the principal part of $f(z)$ is, using the Laurent formulas and contour integration, equal to $\displaystyle \frac{1}{(z-n\pi)^2}$. 
Thus $h(z) := f(z)-g(z)$ has removable singularities at the points $n\pi$. 
Note that both $f$ and $g$ are periodic with period $\pi$. That is, 
$$ f(z) = f(z +\pi) \quad \text{ and } \quad g(z) = g(z+\pi) \quad \text{ for all } z \in \mathbb{C}\backslash\mathbb{Z}. $$
Thus, since $h$ is bounded on the square $\{ z: |\operatorname{Re} z| < \pi, |\operatorname{Im} z |< \pi \}$ and periodic, we may conclude that $h$ is bounded on the set $\{ z: |\operatorname{Im} z |< \pi \}$. To show boundedness on the entire plane, we show that it holds on the vertical strip (with center removed) 
$$ S = \{ z: 0 \leq \operatorname{Re} z \leq \pi, | \operatorname{Im} z | \geq \pi \}. $$ 
For $z$ in $S$,
\begin{align*} 
\sum_{n= -\infty} ^ {+ \infty} |z-\pi n | ^{-2} 
&= \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(x-\pi n)^2 + y^2}  
&\leq \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi (n-1))^2 +y^2}  
\end{align*}
\begin{align*}
& = \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2}  
&< 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2}  
&\leq 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +\pi ^2}. 
\end{align*} 
Thus $g$ is bounded on the set. It is easy to show that $f$ is also bounded on $S$. Thus the difference $f-g$ is also bounded. By the periodicity of $h$, the function is bounded on the entire plane, and is constant by Liouville's theorem. 
I'm wondering if my reasoning so far is valid; and if so, how to show that the relevant constant is in fact zero. 
Thanks. 
 A: A discussion on the relevant constant:
Consider $z=it$.It is easy to figure out that $f(z)$ tends to zero when $t$ tends to infinity.So we just need to derive $\lim_{t\rightarrow\infty}g(it)=0$.
$\eqalign{
|g(it)|\leq\frac{1}{t^2}+\sum_{n=1}^{\infty}\frac{1}{t^2+\pi^2n^2}
}
$
The first term tends to zero when $t$ tends to infinity.
If $t\geq\pi^2N^2$ for some integer $N$ then $t^2+\pi^2n^2\geq\pi^2n^2t$ for all $n\leq N$.
Hence
$\eqalign{
|g(it)|\leq\frac{1}{t^2}+\frac{1}{t}\sum_{n=1}^{N}\frac{1}{\pi^2n^2}+\sum_{n=N+1}^{\infty}\frac{1}{\pi^2n^2}
}
$
Let $N$ tends to infinity and the result follows.
A: Equation $(8)$ from this answer says
$$
\begin{align}
\pi\cot(\pi z)
&=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\\
&=\frac1z+\sum_{k=1}^\infty\left(\frac{1}{z-k}+\frac{1}{z+k}\right)\tag{1}
\end{align}
$$
The convergence is uniform on compact subsets of $\mathbb{C}$, so we can differentiate to get
$$
-\pi^2\csc^2(\pi z)=-\frac{1}{z^2}-\sum_{k=1}^\infty\left(\frac{1}{(z-k)^2}+\frac{1}{(z+k)^2}\right)\tag{2}
$$
A wee bit o' manipulation, and a change of variables, gives us
$$
\csc^2(z)=\sum_{k=-\infty}^\infty\frac{1}{(z-k\pi)^2}\tag{3}
$$
