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

You seem to be mainly confused about the nature of the additive group $\mathbb{C}/\mathbb{Z}$. By definition of the quotient, two vectors $z,w \in \mathbb{C}$ are equivalent in $\mathbb{C}/\mathbb{Z}$ if and only if $z - w \in \mathbb{Z}$. Geometrically, in the complex plane, this means that $z$ is to be identified with $w$ if and only if $w$ is at the same vertical level as $z$ and shifted an integral number of units to the left or to the right. This means that a fundamental polygon for the quotient, or a particularly nice choice of coset representatives, is given by the band $[0,1] \times \mathbb{R} \subset \mathbb{C}$ modulo the relation $(0,y) \sim (1,y)$. In other words, you take an infinitely long band and glue together the two sides to form a cylinder. Now try think of a way to interpret the punctured complex plane as an infinitely long cylinder.
