I am trying to solve some problems in complex analysis, but I am not succeeding in the following problem. Suppose that $f$ is a function with the following properties:

  • $f$ is non-constant;
  • $f$ is meromorphic on $\mathbb{C}$;
  • $f(z)=f(z+\sqrt{2})=f(z+i\sqrt{2})$ for $z\not\in P$, where $P$ is the set of poles of $f$;
  • $f$ has at most one pole in $\{z\in\mathbb{C}:|z|\leq 1\}$.

Then $f$ has exactly one non-simple pole in $\{z\in\mathbb{C}:|z|\leq 1\}$.

I guess I can solve it if I change two things:

  • the subset where $f$ has at most one pole, I assume that $f$ has at most one pole in $P_0:=\{z\in\mathbb{C}:z=a+b\tau$, $0\leq a<1$ and $0\leq b<1\}$, the fundamental parallelogram, where $\tau:=\frac{\sqrt{2}i}{\sqrt{2}}=i$;
  • the periodic condition is valid for all $z\in\mathbb{C}$.

In this case, all the data about $f$ is in $P_0$. Hence, if $f$ does not have poles in $P_0$, then $f$ is analytic in $P_0$ (I'm not sure about this), implying that $f$ is constant, contradiction. Thus, $f$ has exactly one pole. Finally, since the total number of poles of an elliptic function in $P_0$ is always $\geq 2$, we have that the pole is not simple.

So, my question is how to solve this problem without the changing in the assumptions, or how to prove that my assumptions are equivalent to the original hypotheses of the problem.

Thank you!

  • 2
    $\begingroup$ I'm not sure I believe this: what's stopping the Weierstrass elliptic function $\wp(z)$ with periods $\sqrt{2}$ and $i\sqrt{2}$ satisfying the conditions while not having a non-simple pole at $0$? Either way, it's worth asking why this is not a counterexample. $\endgroup$ – Chappers May 2 '15 at 1:59
  • $\begingroup$ Sorry! I typed wrong. Now I believe that is correct. $\endgroup$ – Pryscilla Silva May 2 '15 at 2:27
  1. Denote the only pole inside $|z|\leq 1$ by $a$ and show that it is also the only pole inside $|z-a|\leq 1$.

  2. Consider the square $C$ with vertices $a\pm\frac{\sqrt2}{2}$, $a\pm\frac{i\sqrt2}{2}$ inscribed inside the circle $|z-a|\leq 1$, and oriented counterclockwise.

  3. Integrate $f$ along $C$: if the pole is simple, the result should be non-zero by residue theorem. However, this contradicts the periodicity properties of $f$.

Remark: The case with no pole inside $|z|\leq 1$ is trivial (use a similar square and periodicity to show that $f$ is then analytic in the entire complex plane, hence constant).

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