# Removable Singularities

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.
• $$\lim_{z\to z_0}(z-z_0)f(z) = 0$$.

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

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

## Exercises

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

• The principal part of $$f$$ at $$z=z_0$$ is gotten by expanding $$f(z) = \sum_{k\in {\mathbf{Z}}} c_k z^k$$ 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.

Write

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

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 $${\mathbf{Z}}$$, 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 $${\mathbf{Z}}\setminus\left\{{0}\right\}$$ 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 $${\mathbf{C}}\setminus{\mathbf{Z}}$$.

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