I read the book "Inverse Spectral Theory" by J. Pöschel and E. Trubowitz. There are one lemma: if $|z-\pi n|\geq \pi/4$ for all $n\in\mathbb{Z}$, then $$e^{|\operatorname{Im} z|} \leq 4 |\sin z|$$

screenshot of proof from the book

First, I don't understand the beginning of the proof. I see that $|\sin z|$ is even and periodic, but how does it follow that we can consider $0 \leq x \leq \pi/2$ and $|z|\geq\pi/4$.

Second, are there any books that explore similar inequalities from complex analysis?


1 Answer 1


The left hand side does not depend on the real part of $z$. The fact that $|\sin z|$ has period of $\pi$ (real), it means that if the lemma is true for any interval with length $\pi$, it will be true in any other interval, shifted by integer multiples of $\pi$. So let's choose $x\in[-\pi/2,\pi/2]$. Now we know that $|\sin(z)|=|\sin(-z)|$. So if the lemma is valid for positive $z$, it will be valid for negative $z$ as well. Since we chose the real part to be in the $[-\pi/2,\pi/2]$ interval at the previous point, we can restrict our proof even more, to the $[0,\pi/2]$ interval.

Now how about the second condition? When we write $z=x+iy$, you have $|z-n\pi|=|(x-n\pi)+iy|=\sqrt{(x-n\pi)^2+y^2}$. This has to be greater than $\pi/4$ for any $n$. Since $x$ is in the $[0,\pi/2]$ interval, if you choose $n=0$ or $n=1$ you will get $(x-n\pi)^2\in[0,\pi^2/4]$. If you choose any other $n$ then $(x-n\pi)^2>x^2$. So it is enough to show for the minimum value. For $n=0$ your $|z-n\pi|>\pi/4$ is written as $|z|>\pi/4$

  • $\begingroup$ Can you explain the last 3 sentences in more detail, please? Estimation $(x-n\pi)^2 > x^2$ and about the minimum value. Minimum value of what? Why it is enough? $\endgroup$
    – Alex
    Mar 7, 2019 at 15:11
  • $\begingroup$ Ok, I understood. We have to show that $(x-n\pi)^2 \geq x^2$ for all $n \in \mathbb{Z}, x \in [0,\pi/2]$. It's pretty easy. Then we conclude that $|z-n \pi | \geq |z|$. And then all is clear. $\endgroup$
    – Alex
    Mar 7, 2019 at 17:02
  • $\begingroup$ Exactly. You got the point $\endgroup$
    – Andrei
    Mar 7, 2019 at 17:04

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