# Are all endomorphisms of C^* just power maps?

If an endomorphism of $\mathbb{C}^*$ is assumed to be holomorphic, then it is not difficult to show that it must be of the form $z^n$ for some $n\in \mathbb Z$. I was wondering if this is true without any topological or analytic assumptions on the endomorphism. My intuition says that there should be lots of them, but I'm usually working with endomorphisms which are at least holomorphic if not a true morphism of varieties, so I'm not sure.

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$C^\times$ is isomorphic as a group to $S^1\times\mathbb R$, and $\mathbb R$, as any infinite dimensional vector space over $\mathbb Q$, has lots of endomorphisms as an abelian group! –  Mariano Suárez-Alvarez Feb 4 '13 at 21:16
Although I think you mean $R_{>0}$, I see what you're saying, so thanks for the help! –  HNuer Feb 4 '13 at 21:27
No, I mean $\mathbb R$, just because the multiplicative group $\mathbb R_{>0}$ and the additive group $\mathbb R$ are isomorphic. –  Mariano Suárez-Alvarez Feb 4 '13 at 21:27
@HNuer: Well, you did not specify yourselves the kind of structure on $\mathbf C^*$ for which you wanted to determine the endomorphisms. –  ACL Feb 4 '13 at 22:45
All continuous endomorphisms are of the form $z\mapsto z^n\bar z^m$, for some $n,m \in \mathbb Z$. –  Andreas Thom Feb 9 '13 at 10:44