# Proof $\mathbb{C}^* \cong \mathbb{C} / \mathbb{Z}$

I am trying to prove the isomorphism between $$\mathbb{C}^*$$ and $$\mathbb{C} / \mathbb{Z}$$. I already established the way to do it:

1. find a surjective homomorphism $$f: \mathbb{C} \to \mathbb{C}^*$$, such that $$Ker(f)=\mathbb{Z}$$
2. take the homomorphism $$\phi: \mathbb{C} \to \mathbb{C}/\mathbb{Z}$$
3. Then there exists a homomorphism $$g: \mathbb{C}/\mathbb{Z} \to \mathbb{C}^*$$, and then we have to prove that $$g$$ is a isomorphism.

My problem mostly is in finding a surjective homomorphism $$f$$ such that the $$Ker(f)=\mathbb{Z}$$. Anyone that can help me out?

• You probably mean $(\mathbb C^*, \times)$ and $(\mathbb C/\mathbb Z, +)$. It helps to make this clear. – Trebor Jun 19 '20 at 7:12
• In general, when looking for a homomorphism that goes from "addition" to "multiplication", a good first thing to try is an exponential function, i.e. something like $f(x) = e^x$. – Ben Grossmann Jun 19 '20 at 7:17

$$e^{z_1 + z_2} = e^{z_1} \cdot e^{z_2}$$. But we know $$e^{n 2\pi i} = 1$$ for each $$n \in \mathbb Z$$...
• Ok yes I had something like that in mind as well. So the homomorphism would be: $f: z\mapsto e^{2i\pi z}$. However, is that surjective? These are only the answers for the unit circle, right? (so the solutions for $z=re^{ni2\pi}$, where $r=1$) – dikkemaatjes Jun 19 '20 at 7:22