homomorphism $f: \mathbb{C}^* \rightarrow \mathbb{R}^*$ with multiplicative groups, prove that kernel of $f$ is infinite. Let $f: \mathbb{C}^* \rightarrow \mathbb{R}^*$ be a homomorphism of the multiplicative group of complex numbers to the multiplicative group of real numbers. I need to show that the kernel of $f$ must be infinite.
I do know that $\mathbb{C}^*$ and $\mathbb{R}^*$ are not isomorphic to each other from here. So does that mean $f$ is not onto? But how will I be able to show that the kernel is infinite?
Thanks in advance.
 A: Hint: There are infinitely many complex numbers $z$ that satisfy the equation $z^n=1$ for some odd integer $n$. What can you say about their homomorphic images?
A: I'm not terribly sure if this is the same thing Jyrki is hinting at, and I wouldn't like to give the game away, but at any rate I believe that if this is not a different solution, it at least approaches it from a slightly different angle:
We know that $\ker f$ is a subgroup of $\mathbb{C}^\ast$, so let's ask what are the finite subgroups of $\mathbb{C}^\ast$?
The rest I hide in a spoiler tag below to let you think about this a bit more.

First observation: if $G < \mathbb{C}^\ast$ is a finite subgroup then all $z \in G$ must have norm $1$, i.e. $z \in S^1$.
(For otherwise $z, z^2, z^3, \dotsc$ is an infinite sequence of distinct elements in $G$.)

Suppose then that $G$ is finite and is the kernel of $f:\mathbb{C}^\ast \rightarrow \mathbb{R}^\ast$.
Then let $z \in G$, and then

$$f(2z) = f(2)f(z) = 2\cdot 0 = 0,$$

so $2z \in \ker f = G$. But this is a contradiction since $|2z| = 2 \neq 1$.

In other words, no finite subgroup $G$ of $\mathbb{C}^\ast$ can be a kernel.
