Is $f(z)=\frac{\sin(2z)}{e^z-1}$ holomorphic at $z=0$? The domain of $f$ is $\mathbb C$\ $\{0\}$.So it's not holomorphic at $0$?
2026-03-26 12:35:44.1774528544
Is $\frac{\sin(2z)}{e^z-1}$ holomorphic at $z=0$?
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It makes no sense to ask whether a function which is not defined at $0$ is holomorphic there or not. However, it is true that we can extend $f$ to an holomorphic function whoce domain contains $0$. That's so because$$\lim_{z\to0}\frac{\sin(2z)}{e^z-1}=2.$$So, $0$ is a removable singularity and you can therefore apply Riemann's theorem on removable singularities.