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