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?


2 Answers 2


By definition

$$\sin z=\frac1{2i}\left(e^{iz}-e^{-iz}\right)$$

so that

$$\sin z=0\iff e^{2iz}=1\iff z=k\pi\;,\;\;k\in\Bbb Z$$


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 $n$.

  • $\begingroup$ You're right. I did write "some integer $n$", which is the same as "any $n \in \mathbb{Z}$". $\endgroup$ Dec 17, 2021 at 15:13
  • $\begingroup$ Oh, your'e right, I got confused between integer and natural (I would understand it better in hebrew :) ). Anyway it is nice that you answered about something from 2014... $\endgroup$ Dec 17, 2021 at 15:21

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