Riemann's theorem on removable singularities

Statement

Let $c \in \mathbb{C}$, let $r > 0$, and let $f$ be holomorphic on the punctured disc $0 < |z - c| < r$. Assume that $f$ is bounded near $c$: there are $M \ge 0$ and $\delta \in (0, r)$ with $|f(z)| \le M$ for $0 < |z - c| < \delta$. Then $c$ is a removable singularity of $f$. Moreover, the limit $\lim_{z \to c} f(z)$ exists, and the extension of $f$ by this value at $c$ is holomorphic on $|z - c| < r$.

Remarks

The following are equivalent: $c$ is removable; $f$ has a continuous extension to $c$; $f$ is bounded near $c$; $\lim_{z \to c} (z - c) f(z) = 0$. Proof idea: $h(z) = (z - c)^2 f(z)$, with $h(c) = 0$, is complex differentiable at $c$ with $h'(c) = 0$. So $h(z) = \sum_{n \ge 2} a_n (z - c)^n$, and $f(z) = \sum_{n \ge 2} a_n (z - c)^{n-2}$. For example, $\frac{\sin z}{z}$ is bounded near $0$, so $0$ is removable.

Explanations

L1 human

If a holomorphic function stays bounded near a single bad point, then the point is not really bad. You can give the function a value there, and the function becomes holomorphic at that point.