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I am stuck on the following problem:

Suppose an entire function maps two horizontal lines onto two other horizontal lines. Prove that its derivative is periodic.

The author supplies a hint: Assume $f = u+iv$ maps the lines $y=y_1$ and $y=y_2$ onto $v=v_1$ and $v=v_2$ with $y_2-y_1 = c$ and $v_2-v_1 = d$. Show then that $f(z+2ci)+f(z)+2di$ for all $z$.

I am trying to apply the Schwarz reflection principle to solve this problem.

What I've done so far:

If we let $\gamma$ be the analytic arc given by the first line, $x+iy_1$. Then, the reflection of $x+iy_2$ over $\gamma$ is $x+i(y_1-c)$, or

$$w = \gamma(x+ic) = x+i(y_1+c) = x+iy_2 \\ w^* = \gamma(x-ic) = x+i(y_1-c).$$

Now, what we want is that $f(z+2ci) = f(z)+2di$. Take $z = w^*$, so $z+2ci = w$.

Then, we have $f(z+2ci) = f(w) = f(w^*) + 2di$.

Then, we need to compute $f(w^*)$ and show that it is equal to the reflection of the image of $w$ under $f$ over $\lambda := f(\gamma) = u+iv_1$.

But I'm stuck on where to go from here.

Is this even the right approach? I'm taking it for a specific $z$... clearly this can't work for all $z$...

Martin
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Emily
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1 Answers1

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Suppose the source horizontal lines are $\mathrm{Im}(z)=a$ and $\mathrm{Im}(z)=b$ and the destination horizontal lines are $\mathrm{Im}(z)=c$ and $\mathrm{Im}(z)=d$.

The functions $g(z)=f(z+ia)-ic$ and $h(z)=f(z+ib)-id$ are entire functions that map the real line to the real line. By the Schwarz Reflection Principle, we get $$ \begin{align} f(z+ia)-ic=g(z)=\overline{g(\bar{z})}=\overline{f(\bar{z}+ia)}+ic\tag{1a}\\ f(z+ib)-id=h(z)=\overline{h(\bar{z})}=\overline{f(\bar{z}+ib)}+id\tag{1b} \end{align} $$ Using $(1a)$ and $(1b)$, we get $$ \begin{align} f(x+iy) &=f(x+i(y-a)+ia)\\ &=\overline{f(x-i(y-a)+ia)}+2ic\tag*{by $(1a)$}\\ &=\overline{f(x-i(y-2a+b)+ib)}+2ic\\ &=f(x+i(y-2a+b)+ib)+2ic-2id\tag*{by $(1b)$}\\ &=f(x+iy+2i(b-a))+2i(c-d)\tag{2} \end{align} $$ That is $$ f(z+2i(a-b))=f(z)+2i(c-d)\tag{3} $$ Taking the derivative of $(3)$ yields $$ f'(z+2i(a-b))=f'(z)\tag{4} $$ Therefore, $f'$ is periodic.

robjohn
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  • Is the $c,d$ you are using in your destination horizontal lines the same as the distance between the source horizontal lines? – Emily Feb 25 '13 at 15:10
  • No. In my answer, I do not use the hint. $f$ sends $\mathrm{Im}(z)=a$ to $\mathrm{Im}(z)=c$ and sends $\mathrm{Im}(z)=b$ to $\mathrm{Im}(z)=d$. – robjohn Feb 25 '13 at 18:52