= Solution
A one-dimensional continuous <group representation> is a continuous <group homomorphism> $\rho:S^1\to\mathbb C^\times$. Its image is compact. If some $|\rho(z)|$ differed from one, its positive or negative powers would be unbounded, so the image lies in $S^1$. Lift $t\mapsto\rho(e^{it})$ to a continuous real argument $a(t)$ with $a(0)=0$. Then $a(s+t)-a(s)-a(t)$ is a continuous integer multiple of $2\pi$ and vanishes at $(0,0)$, hence is always zero. The continuous <Cauchy functional equation> gives $a(t)=ct$. Periodicity forces $e^{2\pi ic}=1$, so $c=n\in\mathbb Z$. Consequently $\boxed{\rho(z)=z^n}$; conversely every such map is a continuous one-dimensional <group representation>.
Back to article page
