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Suppose $f$ is entire and, for some rectangle $R$, $f(R)$ is a rectangle. Prove $f$ is linear.

I have tried some ways,

Let the first rectangle be $R_1$ and the second one be $R_2$. $ f : R_1 \longmapsto R_2 $

And, I construct one linear map $h_1 : R_1 \longmapsto S_1$, $S_1$ is a unit square. Similarly a $h_2 : R_2 \longmapsto S_2$.

At the beginning, I want to say there should be a identical map $I : S_1 \longmapsto S_2$, but I failed.
Second, I want to use extended Liouville's Theorem to say there is a linear mapping between $S_1$, $S_2$, it doesn't satisfy the requirement of the theorem.

Since the map, say $h$, satisfying $|h(z)| \leq |z| + \sqrt{2} $.

What should I try? How to construct a linear map between $S_1$ and $S_2$ ?

Richard
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1 Answers1

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Step 1: prove that $f$ sends vertices to vertices. Indeed, if a point on a side of $R$ goes to a vertex, then interior angle $\pi$ becomes $\pi/2$, which is impossible for a holomorphic function. (Under a holomorphic map, angles are either preserved or multiplied by an integer $n\ge 2$.)

Step 2: use Schwarz reflection about the sides of the rectangle to show that $f$ is symmetric about each side. Let's say $R$ is $0<x<a$, $0<y<b$ and $f(R)$ is $0<x<A$, $0<y<B$, with $f$ preserving the orientation of each edge (you can always bring to this form by composing $f$ with linear functions). Then $$f(-x+iy)=(-\operatorname{Re}f(x+iy), \operatorname{Im}f(x+iy))$$ and similarly for four other sides. Doing this repeatedly, you will find that $$ f(x+2a+iy)=f(x+iy)+2A,\\ f(x+i(y+2b)) = f(x+iy)+2Bi $$

Step 3. Conclude that the function $f$ has at most linear growth at infinity: $ |f(z)|\le \alpha |z|+\beta$.

Step 4. Finish as in Entire function bounded by a polynomial is a polynomial

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