Let $f: \mathbb{C}- \{0\} \rightarrow \mathbb{C}$ be a holomorphic function on the punctured complex plane, and suppose that $f(2z)=f(z)$ for all $z \neq 0$. Prove that $f$ is constant.
Proof: Consider the annulus $A=\{ z: \frac{1}{2}\leq z\leq 1 \}$. Since this set is compact the modulus attains a maximum $M$. Given arbitrary $z$ we can by either successively multiplying or dividing by 2, find $w \in A$ such that $f(w)=f(2^{k}z)=f(z)$ Hence the values in the punctured complex plane are bounded. By Riemann's theorem, since the modulus is bounded in the vicinity of 0, it has to be a removable singularity. Hence, $f$ is an entire function in the complex plane which is bounded, and by Liouville's theorem is constant.
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