Analytic continuation of $\sin(z)$

Question

Why $$\sin{ (z)} =\frac{e^{iz}-e^{-iz}}{2i}$$ the only analytic function, is equal to $\sin{(x)}$ for $z=x \in \mathbb{R}$?

Answer 1

Lemma Let $f,g$ be two entire functions which are equal on $\mathbb R$. Then, they are equal on $\mathbb C$.

Proof Since $f-g$ is analytic and the set of zeroes has an acumulation point, by the identity theorem for analytic functions $f-g=0$.

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Now apply this Lemma to $f(z)=\sin(z)$ and $g(z)=\frac{e^{iz}-e^{-iz}}{2i}$.

Alternatelly, you can argue that any function which is equal to $\sin(z)$ on $\mathbb R$ must be analytic on $\mathbb R$, and hence its Taylor series is the Taylor series of $\sin(z)$. But by Taylor Theorem, any such function must be equal to the Taylor series on $\mathbb C$.