= Solution
The map $f$ is bounded by one near the puncture, so the <Riemann removable singularity theorem> extends it holomorphically to a map $\widetilde f:\Delta\to\mathbb C$. The <maximum modulus principle> shows that its image still lies in $\Delta$: if $|\widetilde f(0)|=1$, the extension would be constant, contradicting its values in $\Delta\setminus\{0\}$ unless that constant also had modulus less than one.
If $f$ is biholomorphic, its inverse $g:\Delta\setminus\{0\}\to\Delta\setminus\{0\}$ also extends holomorphically. Were $a=\widetilde f(0)\ne0$, continuity of $\widetilde g\circ\widetilde f$ and the identity $g(f(z))=z$ on the punctured disk would give $g(a)=0$. This is impossible because $a\in\Delta\setminus\{0\}$ and $g$ takes values in the punctured disk. Hence
$$
\boxed{\widetilde f(0)=0.}
$$
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