Answer: Riemann's Removable Singularity via zf(z)→0

Key Idea / Intuition

The condition $z f(z) \to 0$ is exactly what forces the singularity to be "fake." The trick is to construct a function $g(z) = z^2 f(z)$ that is holomorphic on the entire disk including the origin — we can verify differentiability at $z=0$ directly from the limit condition. Once we know $g$ is holomorphic and vanishes to second order at $0$, we can write $g(z) = z^2 h(z)$ for some holomorphic $h$, and then $h$ is the desired extension of $f$.

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Formal Proof / Solution

Step 1: Define $g$ and show it is holomorphic on the full disk.

Set $g(z) = \begin{cases} z^2 f(z) & z \neq 0, \\ 0 & z = 0. \end{cases}$

For $z \neq 0$, $g$ is holomorphic (product of holomorphic functions). We need to check complex differentiability at $z = 0$.

Compute the difference quotient at the origin: $\frac{g(z) - g(0)}{z - 0} = \frac{z^2 f(z)}{z} = z f(z).$

By hypothesis, $z f(z) \to 0$ as $z \to 0$. Therefore: $g'(0) = \lim_{z \to 0} \frac{g(z) - g(0)}{z} = \lim_{z \to 0} z f(z) = 0.$

So $g$ is holomorphic on the entire disk $D(0, r)$, with $g(0) = 0$ and $g'(0) = 0$.

Step 2: Factor out $z^2$ from $g$.

Since $g$ is holomorphic on $D(0, r)$ with $g(0) = g'(0) = 0$, its Taylor series at $0$ has no constant or linear term: $g(z) = \sum_{n=2}^{\infty} a_n z^n = z^2 \sum_{n=0}^{\infty} a_{n+2} z^n.$

Define $h(z) = \sum_{n=0}^{\infty} a_{n+2} z^n,$ which converges on $D(0, r)$ and is holomorphic there.

Step 3: Conclude that $h$ extends $f$.

For $z \neq 0$: $h(z) = \frac{g(z)}{z^2} = \frac{z^2 f(z)}{z^2} = f(z).$

So $h$ is a holomorphic function on all of $D(0, r)$ that agrees with $f$ on the punctured disk. The singularity is removable. $\blacksquare$

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Remark: The classical statement of Riemann's theorem assumes only that $f$ is bounded near $0$. Boundedness implies $zf(z) \to 0$ (since $|zf(z)| \leq |z| \cdot M \to 0$), so the condition here is slightly weaker — it is in fact the sharp condition for removability.