Complex number $|z|$=1

Question

Are there only finitely many complex numbers $z$ with $|z|=1$?

I think there are infinitely many such numbers.

$\cos\theta+i\sin\theta$ for $\theta\in[0,2\pi)$ – yoyo Nov 26 '20 at 02:45 — yoyo, Nov 26 '20 at 02:45

score 1 · Answer 1 · answered Nov 26 '20 at 02:44

1

Yes, there are infinitely many complex numbers $z$ with $|z|=1$, including $e^{2 \pi i/n}$ for all $n\in \Bbb N$.

answered Nov 26 '20 at 02:44

J. W. Tanner

Is there a particular reason to focus on the roots of unity in this case? After all, it works for $e^{i \theta}$ for $\theta \in [0, 2\pi)$ in general. – PrincessEev Nov 26 '20 at 02:45
@EeveeTrainer: not particularly, though the ones I focused on did answer the question – J. W. Tanner Nov 26 '20 at 02:47

score 1 · Answer 2 · answered Nov 26 '20 at 02:46

1

Indeed, all complex numbers on the circle $S^1$: namely all numbers of the form $e^{i\theta}$ for $\theta\in \mathbb{R}$

answered Nov 26 '20 at 02:46

Rachid Atmai

score 0 · Answer 3 · answered Nov 26 '20 at 03:02

0

Ummm... sure. Here's a graph of them:

answered Nov 26 '20 at 03:02

David G. Stork

score -1 · Answer 4 · answered Nov 26 '20 at 02:47

-1

$$|z|=1$$ $$\sqrt{x^2+y^2}=1$$ $${x^2+y^2}=1.$$

answered Nov 26 '20 at 02:47

Unknown

4 Answers4