22
$\begingroup$

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?

$\endgroup$

2 Answers 2

26
$\begingroup$

By definition

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

so that

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

$\endgroup$
13
$\begingroup$

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 $n$.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.