Let $C^*$ denotes the multiplicative group of complex numbers different from zero and $T^*$ be the subgroup of all $z\in C^*$ such that $|z|=1$. Prove that if $G$ is a subgroup of $C^*$ which is compact (as a set of complex numbers), then $G\subset T^*.$
I started to prove it with contradiction but couldn't reach in the result. How can I solve it?
Any help is appreciated.
Suppose that it is not contained in $T^*$, there exists $z\in G$ such that $|z|\neq 1$ you have $|z|$ or $|z|^{-1}>1$, if $|z|>1$, $|z|^n$ is not bounded this implies that $G$ is not bounded contradiction since a compact subset of $\mathbb{C}$ is bounded. If $|z|^{-1}>1$, apply the same argument.