Quotient group of $C^*$ by Circle group is isomorphic to $R^+$

Question

$C^*$ is the set of complex numbers except $0$ and $S'$ is the circle group. I have to show the quotient group $C^*/S'$ is isomorphic to $R^+$. Let's define $f:C^*→ R^+$, $z→|z|$. Why is $Kerf$ $S'$? If I show that, the map is clearly surjective and by the first isomorphism theorem, $C^*/S'$ would be isomorphic to $R^+$. Is my reasoning right? Is there a better way of doing this?

Answer 1

$\ker f=\{z\in\mathbb{C}^*\,|\,\lvert z\rvert=1\}=S^1$.