$\mathbb{C}/\mathbb{Z}$ is isomorphic to multiplicative group $\mathbb{C}\setminus\{0\}$ I have to show that $\mathbb{C}/\mathbb{Z}$ is isomorphic to the multiplicative group $\mathbb{C} \setminus \{0\}$. 
Proof.
Let $f:\mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}/\mathbb{Z}$ be the map
$$ f(\alpha) = \alpha \mathbb{Z}.$$
This map has inverse $f^{-1}(\alpha \mathbb{Z})  = \alpha$ so it is bijective. Furthermore, $$f(\alpha  \beta) = (\alpha \beta)\mathbb{Z} = \alpha \mathbb{Z} \beta \mathbb{Z} =f(\alpha) f(\beta)$$ and $f(1) = \mathbb{Z}$, so the map $f$ is also a homomorphism.
Question.
Is this correct? It feels a bit like I am cheating.

Since $\Bbb C\simeq\Bbb R^2$, we have $\Bbb C/\Bbb Z\simeq\Bbb R^2/\Bbb Z\simeq\Bbb R\times(\Bbb R/\Bbb Z)$.
On the other hand, the polar expression $z=\rho e^{i\theta}$ shows that there is an isomorphism $\Bbb C^\times\simeq\Bbb R^{>0}\times\Bbb C^1$ where $\Bbb C^1$ denotes the (sub)group of complex numbers of norm $1$.
To conclude observe that:
1) The map $r\mapsto e^{2\pi i r}$ sets an isomorphism $\Bbb R/\Bbb Z\simeq\Bbb C^1$.
2) The logarithm defines an isomorphism $\Bbb R^{>0}\simeq\Bbb R$.
Just put everything together.
