What are element with finite order at $\mathbb{C}^*/U$? I need to find the elements with finite order at the group - $\mathbb{C}^*/U$.
$U$ - is the Circle Uint.
$\mathbb{C}^*$ - is $(\mathbb{C}/0,\cdot)$.
I need to write also the proof, and I'll be glad to get help with this...
I think that only $|z|=1,z\in\mathbb{C}$ is finite, I'm right?
I'm little bit confuse so I'd like to get help with the explanation about the finite order elements and the proof.
Thank you!
 A: Every non-zero complex number $z\in\Bbb C^\times$ can be uniquely written as
$$
z=re^{2\pi it}
$$
where $r$ is a positive real number. Moreover $zz^\prime=(rr^\prime)e^{2\pi i(t+t^\prime)}$ so that the (norm) map
$$
\Bbb C^\times\longrightarrow\Bbb R^{>0},\qquad
z\mapsto r
$$
is a surjective homomorphism with kernel $U$. Therefore
$$
\frac{\Bbb C^\times}{U}\simeq\Bbb R^{>0}.
$$
So the question is: what are the torsion elements (i.e. the finite order elements) in the multiplicative group $\Bbb R^{>0}$?
A: We now that every nonzero complex number $z$ can be uniquely expressed as $z=r\omega$, where $r=|z|>0$ and $\omega$ belongs to the unit circle. If $z_1=r_1\omega_1$ and $z_2=r_1\omega_2$,
then $|z_1z_2|=r_1r_2$ and $\omega_1\omega_2$ belong to the unit circle.
This allows us to write the multiplicative group $\mathbb C^*$ as a product of two multiplicative groups $\mathbb R^+$ and $U$, the unit circle, i.e.,
$$
(r,\omega)\cdot (r',\omega')=(rr',\omega\omega').
$$
As $\mathbb C^*=\mathbb R^+\times U$, then 
$$
\mathbb C^*/U=\frac{\{r\omega:r\in\mathbb R^+\,\&\,\omega\in U\}}{
\{ r\omega_1=r\omega_2:\text{for all $r,\omega_1,\omega_2$} \} }\cong \mathbb R^+.
$$
