# the zeros of $\sin(z)$, where $z$ is a complex number

How do I find the zeros of $\sin(z)$, where $z$ is a complex number?

I know that along the real line we have zeros along $k\pi$, where $k$ is an integer. But what about the rest of the plane? The taylor series: $$\sum_{n=0}^{\infty}(-1)^n \dfrac{z^{2n+1}}{(2n+1)!},$$ doesn't really tell me that much.

How do I find the other zeros?

$$\sin z=\frac1{2i}\left(e^{iz}-e^{-iz}\right)$$
$$\sin z=0\iff e^{2iz}=1\iff z=k\pi\;,\;\;k\in\Bbb Z$$
Use the formula $$\sin z = \frac{e^{iz} - e^{-iz}}{2i}.$$ It shows that $\sin z = 0$ iff $e^{iz} = e^{-iz}$ iff $e^{2iz} = 1$ iff $2iz = 2\pi i n$ for some integer $n$ iff $z = n\pi$ for some integer $n$.
• You're right. I did write "some integer $n$", which is the same as "any $n \in \mathbb{Z}$". Dec 17, 2021 at 15:13