How to compute the $(2n-1)$th complex roots of 1? I have to solve
$$
(e^{i\theta})^{2n-1}=1,
$$
i.e. to find the $(2n-1)$th roots on the unit circle.
How can I solve this?

I've read that I can use Euler's formula to write the solutions as
$$
e^{2\pi i\frac{k}{(2n-1)}}, ~~0\leq k< n.
$$
How do I get there?
 A: You want $z=re^{i\theta}$ such that $z^{2n-1} = 1$. In other words: $$(re^{i\theta})^{2n-1} = e^{0} \implies r^{2n-1}e^{(2n-1)i\theta} = 1\cdot e^0$$ From here $r=1$, since $r>0$. And the exponential has period $2\pi i $, so: $$(2n-1)i\theta = 0 + 2k\pi i, \quad k \in \Bbb Z \implies \theta = \frac{2k\pi}{2n-1},\quad k\in \Bbb Z. $$ Values of $k$ from $0$ to $2n-2$ will give you different roots. You'll get the same roots again for $k \geq 2n-1$.
A: One way to understand Euler's solution is by looking at the unit circle:
Multiplying a complex number with $e^{i\theta}$ is essentially rotating it in $\theta$ radians
if we define g as:
$g\equiv e^{\frac{i2\pi}{(2n-1)}}$
g is equivalent to a rotation of $\frac{360}{2n-1}$ degrees
Your question could be reformulated as:
I am going to make 2n-1 turns of angle $\theta$, and I want to be facing the same orientation as I started, what turn should I make ?
The answer is $\theta\in\{\frac{2\pi}{2n-1},\frac{4\pi}{2n-1},\dots,\frac{(2n-2)2\pi}{2n-1},2\pi\}$
Which is equivalent to $\{g^0,g^1,\dots,g^{2n-2},g^{2n-1}\}$
