Removable Singularities

theorem (Riemann's removable singularity theorem):

If $z_0$ is an isolated singularity of $f(z)$ and ${\left\lvert {f(z)} \right\rvert}$ is bounded near $z_0$ , then $z_0$ is removable.

More generally, TFAE:

$f$ extends holomorphically over $z_0$ , i.e. there is a function $F$ such that ${ \left.{{F}} \right|_{{\Omega\setminus\left\{{z_0}\right\}}} } = f$
$f$ extends continuously over $z_0$ .
There exists some neighborhood of $z_0$ on which $f$ is bounded.
$\lim_{z\to z_0}(z-z_0)f(z) = 0$ .

remark:

Showing a singularity $z_0$ of $f$ is removable: it suffices to show

Expand $f(z) = \sum_{k\in {\mathbb{Z}}} c_k z^k$ and show $z_k=0$ for $k<0$ .
Show $\lim_{z\to z_0}f(z) \neq \infty$

Exercises

exercise (?):

Show that there is an entire function $h$ such that $\begin{align*} {\pi^2\over \sin^2(\pi z)} = \sum_{k\in {\mathbb{Z}}} {1\over (z-k)^2} + h(z) .\end{align*}$

#complex/exercise/completed

concept:

The principal part of $f$ at $z=z_0$ is gotten by expanding $f(z) = \sum_{k\in {\mathbb{Z}}} c_k z^k$ and taking $\sum_{k\leq 1} c_k z^k$ .
Common trick: to control a singularity, subtract off a function with the same principal part at that point.

solution:

Write

$f(z) = {\pi^2 \over \sin^2(\pi z)}$
$g(z) = \sum_{k\in {\mathbb{Z}}} {1\over (z-k)^2}$

Write the above equation as $f(z) = g(z) - h(z)$ and consider $h(z) \coloneqq f(z) - g(z)$ . Then $h$ is meromorphic with singularities precisely on the set ${\mathbb{Z}}$ , and are thus isolated. By the classification of isolated singularities, these can be removable, poles, or essential. If they are removable, then $h$ is entire.

Consider the singularity at $z_0 = 0$ . This is a pole of $f$ , and a computation shows it is order 2: $\begin{align*} f(z) &= \qty{\pi \over \sin(\pi z)}^2 \\ &= \qty{\pi \over \pi z - {1\over 3!}(\pi z)^3 + { \mathsf{O}} (z^5) }^2 \\ &= \qty{\pi \over \pi z (1 - {1\over 3!}(\pi z)^2 + { \mathsf{O}} (z^4))}^2 \\ &= {1\over z^2} \qty{1 \over 1 - {1\over 3!}(\pi z)^2 + { \mathsf{O}} (z^4)}^2 \\ &= {1\over z^2}\qty{1 + { \mathsf{O}} (z^2)}^2 \\ &= {1\over z^2} + { \mathsf{O}} (z^2) ,\end{align*}$ so $z=0$ is a zero of order 2 of $1/f$ . This expansion also shows that the principal part of $f$ at $z=0$ is ${1\over z^2}$ , which is precisely that of $g$ at $z=0$ , i.e. ${1\over (z-0)^2} = 1/z^2$ , Since $f-g$ subtracts off this part, $z=0$ becomes a removable singularity for $h$ since $\lim_{z\to 0} \qty{ f(z) - {1\over z^2}} = 1<\infty$ .

Now note that $f$ is periodic, and since the period is 1, a similar argument shows that the remaining singularities on ${\mathbb{Z}}\setminus\left\{{0}\right\}$ are all removable for $f-g$ . By Riemann’s removable singularity theorem, $f-g$ extends over these singularities, yielding an entire function $h$ that restricts to $f-g$ on ${\mathbb{C}}\setminus{\mathbb{Z}}$ .