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I have to show that, if the title is true, then f is constant.

I can't find the way to approach this. I thought using the fact that, since f is whole, $f(z)=\sum_{n=0}^\infty a_nz^n$ for some $a_n\in\mathbb{C}$. Then $f(0)=a_0=f(z_0)=f(z_1)$. I wanted to show that $g(z) = f(z)-a_0$ trying to build a convergent sequence of zeros of $g(z)$ using the $z_0$ and $z_1$ so I could use the identity principle.

I tried it less, because I couldn't find a way to use it, but tried showing that using the $g$ defined above, $g^{(n)}(0)=0$ $\forall n\in\mathbb{N}$

Thanks in advance

Silkking
  • 991

1 Answers1

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If $z_0$ and $z_1$ are $\mathbb{R}$-independent, for any $z \in \mathbb{C}$ you can write $z = (n + r_0) z_0 + (m + r_1) z_1$ for some $n,m \in \mathbb{Z}$ and $r_1, r_2 \in [0,1)$. Then we have: $$\begin{align*} f(z) & = f((n + r_0) z_0 + (m + r_1) z_1) \\ & = f(r_0z_0 + r_1z_1) \end{align*} $$ Thus, we have that$f(\mathbb{C}) \subseteq f(\Omega)$ where $\Omega$ is the set of all linear combinations of $z_1$ and $z_2$ with coefficients in $[0,1]$. That is, $\Omega$ is a closed parallelogram. $\Omega$ is clearly compact, and thus the image $f(\Omega)$ is bounded. Then by the above inclusion, $f$ is a bounded entire function, and hence is constant by Liouville's theorem.

Joe
  • 2,963