# Proof of inequality (complex analysis)

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 $$e^{|\operatorname{Im} z|} \leq 4 |\sin z|$$

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?

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 $$|z|>\pi/4$$
• 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?
• 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.