I am reading Bak and Newman's Complex Analysis and I can't figure out how to do the following exercise:

Prove that an entire function which maps a parallelogram onto another parallelogram, and maps each side of the original parallelogram onto a side of its image, must be a linear polynomial.

There is no solution, but a hint: to use the previous exercise, in which I proved the following, using Schwarz reflection principle:

If an entire funtion maps two horizontal lines onto two other horizontal lines, then its derivative is periodic.

I tried to show that the derivative must be periodic in two directions, hence be bounded, thereby constant, so that we could conclude $f (z)=az+b $. But I couldn't figure out how to do this, because we are not explicitely given the hypotheses of the previous exercise.

Thank you for your help.

  • $\begingroup$ Use the Schwarz reflection principle. It must be in the same book. $\endgroup$ – Alexandre Eremenko May 31 '18 at 20:41
  • $\begingroup$ I don't see how to apply it, should I apply it several times? $\endgroup$ – Friedrich May 31 '18 at 22:02
  • $\begingroup$ Infinitely many times. $\endgroup$ – Alexandre Eremenko Jun 1 '18 at 13:52
  • $\begingroup$ If it were a rectangle I could tile the plane by repeating the argument but it is a parallelogram... $\endgroup$ – Friedrich Jun 1 '18 at 13:57

Let us start with the hint: Suppose that an entire function $f$ maps a line $A$ into a line $A'$ and a line $B$ parallel to $A$ into a line $B'$. Reflection Principle says that $a'\circ f=f\circ a$, where small letters denote reflections with respect to the corresponding lines. Similarly we have $b'\circ f=f\circ b'$. From these two relations follow that $$f\circ a\circ b=a'\circ f\circ b=a'\circ b'\circ f.$$ But composition of two reflections in parallel lines is a shift, so what we really obtained is $f(z+h)=f(z)+h'$, with some constants $h\neq 0,h'$. Differentiating this we obtain that $f'$ has a period $h$.

Now consider the parallelogram. One parallelogram (in the domain of $f$) is bounded by parallel lines $A,B$ and parallel lines $C,D$, and it is mapped onto some parallelogram. Then each line $A,B,C,D$ must be mapped onto some line and we have 4 lines $A',B',C',D'$. These lines $A',B',C',D'$ must be pairwise distinct. This follows from the Schwarz symmetry and uniqueness: if $f$ maps a little piece of a line $A$ into a line $A'$ then it maps the whole $A$ into $A'$. Then by considering all possibilities, it is easy to conclude that $A'$ must be parallel to $B'$ and $C'$ must be parallel to $D'$. So $f$ maps two pairs of parallel lines into two pairs of parallel lines.

From the hint exercise we conclude that $f'$ has two periods which are not collinear. Therefore, by Liouville's theorem $f'$ must be constant.

Uniqueness theorem. Suppose that $f$ maps a little piece of a line to a little piece of a line. WLOG, both lines are the real line, and the little piece contains $0$. Let $\sigma(z)=\overline{z}$ be the reflection with respect to the real line. Then near $0$ we have $\sigma\circ f=f\circ\sigma$. This implies that all Taylor coefficients are real numbers. As $f$ is entire, it is real on the whole real line.

  • $\begingroup$ The part I'm having trouble with is what you refer to as Schwartz symmetry and uniqueness: how can we conclude that the image of A is a line? $\endgroup$ – Friedrich Jun 1 '18 at 14:58

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