The question goes: An extension of the real function $\sin x$ into a complex analytic function is by defining $\sin z = z- z^3/3! + z^5/5!- \cdots$. Show that this is the only way6 to extend $\sin x$ to a complex analytic function on $\mathbb{C}$.
We are working on zeroes and singularities so it is probably related to this topic.
My trial: I know that $\sin z$ has order of zero $1$ from the given expansion $\sin z = z- z^3/3! + z^5/5!- \cdots$. I can also let $\sin z$ have another series expansion. But since the zeroes of $\sin z$ are $k\pi i$ which are all of multiplicity $1$, these two are the same function. I don't know if I am doing this correctly, and I don't know a sound way of getting the multiplicity of the zeroes of $\sin z$ either.
Can someone help me with this? Any help is appreciated. Thanks