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Fall 2017 - Problem 7

construction

Prove that there exists a meromorphic function ff on C\C with the following three properties:

  1. f(z)=0f\p{z} = 0 if and only if z∈Zz \in \Z.
  2. f(z)=∞f\p{z} = \infty if and only if z−13∈Zz - \frac{1}{3} \in \Z.
  3. ∣f(x+iy)∣≤1\abs{f\p{x + iy}} \leq 1 for all x∈Rx \in \R and all y∈Ry \in \R with ∣y∣≥1\abs{y} \geq 1.
Solution.

Let f(z)=sin⁡(πz)sin⁡π(z−13)f\p{z} = \frac{\sin\p{\pi z}}{\sin\pi\p{z - \frac{1}{3}}}. sin⁡πz\sin\pi z has zeroes only at z∈Zz \in \Z, and these are all simple zeroes. Hence, ff is meromorphic and satisfies (1) and (2).

To show (3), we have

∣f(z)∣=∣eiπz−e−iπzeiπze−iπ/3−e−πzeiπ/3∣≤eπ∣Im⁡z∣+e−π∣Im⁡z∣eπ∣Im⁡z∣−e−π∣Im⁡z∣→∣Im⁡z∣→∞1.\abs{f\p{z}} = \abs{\frac{e^{i\pi z} - e^{-i\pi z}}{e^{i\pi z}e^{-i\pi/3} - e^{-\pi z}e^{i\pi/3}}} \leq \frac{e^{\pi\abs{\Im{z}}} + e^{-\pi\abs{\Im{z}}}}{e^{\pi\abs{\Im{z}}} - e^{-\pi\abs{\Im{z}}}} \xrightarrow{\abs{\Im{z}}\to\infty} 1.

Thus, ff is bounded on ∣Im⁡z∣≥1\abs{\Im{z}} \geq 1, so if C>0C > 0 is large enough, fC\frac{f}{C} still satisfies (1) and (2), and

∣f(x+iy)∣C≤1\frac{\abs{f\p{x + iy}}}{C} \leq 1

on ∣Im⁡z∣≥1\abs{\Im{z}} \geq 1, as required.