$\mathbb{C}/\mathbb{Z}$ 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.

The proof, as written, is definitely wrong -- the inverse map you define does not vanish on $\mathbb{Z}$, so it cannot be a map out of $\mathbb{C}/\mathbb{Z}$.
I think the idea is to use the complex exponential function. Notice that $\alpha \mapsto \exp(2\pi i \alpha)$ defines a group homomorphism from $\mathbb{C}$ to $\mathbb{C} \setminus \{0\}$ that vanishes on $\mathbb{Z}$. Now prove it's an isomorphism (I suggest doing this geometrically -- you can get anything of norm $1$ by taking $\alpha \in \mathbb{R}$, and then scale)...
