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Here is a question from a practice exam:

Suppose $g(z)$ is a holomorphic function everywhere except the origin. Also suppose $$ |g'(z)|\leq \frac{1}{|z|^{3/2}} \quad \text{ for } 0<|z|\leq 1 $$ Prove that $z=0$ is a removable singularity.

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Since $|g'(z)z^2|<\sqrt{|z|}$, $0$ is a removable singularity of $g'(z)z^2$. And $g'(z)z^2\to 0$ as $z\to 0$, so $0$ is a zero of $g'(z)z^2$. Therefore $0$ is a removable singularity of $g'(z)z$. Let $g(z)=\dots+a_{-1}z^{-1}+a_0+a_1z+\dots$, since $0$ is a removable singularity of $zg'(z)$, we have $\dots+a_{-1}z^{-1}=0$, we are done.

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