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.

Solution 1:

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.

Solution 2:

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 $$ \csc^2(z)=\sum_{k=-\infty}^\infty\frac{1}{(z-k\pi)^2}\tag{3} $$