# Holomorphic function - inequalities

$f$ is holomorphic on some neighborhood of $\lbrace z\in\mathbb{C}: \frac{3\pi}{2}\leq |z|\leq\frac{5\pi}{2}\rbrace$. On both $\lbrace z\in\mathbb{C}:|z|=\frac{3\pi}{2}\rbrace$ and $\lbrace z\in\mathbb{C}:|z|=\frac{5\pi}{2}\rbrace$ it satisfies $$|f(z)|\leq|\frac{\sin z}{(z-2\pi)(z+2\pi)}|$$ Show that $$|f(2\pi i)|\leq\frac{e^{2\pi}-e^{-2\pi}}{16\pi^2}$$

Is it a coincidence that if we plug $2\pi i$ into the first inequality (although the inequality is assumed to be true for numbers from a different set) we get what asked for? I tried to apply the maximum principle to get somewhere, but it doesn't seem to work. Any hints?

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This feels strange, $f$ might not even be defined there! I mean, you know a bound on f between two circles of radius 1.5 and 2.5, so it seems strange to be able to find a bound on its value on a circle of radius 1. Maybe you seek a bound on the analytic continuation of f... ? – Per Alexandersson Jan 31 '13 at 21:05
@Paxinum The cricles are of radius $\frac{3\pi}{2}$ and $\frac{5\pi}{2}$, not $\frac{3}{2}$ and $\frac{5}{2}$, so $2\pi i$ is within the stated region. – Brett Frankel Jan 31 '13 at 21:09
Ah, I just completely misread $2\pi i$ as an exponent... $e^{2\pi i}$ usually has absolute value 1, that was what I read it as. Silly habits... – Per Alexandersson Jan 31 '13 at 21:11

Apply the maximum principle to $$\frac{|f(z)(z-2\pi)(z+2\pi)|}{|\sin(z)|}$$

Then $$\frac{|f(2\pi i)(2\pi i-2\pi)(2\pi i+2\pi)|}{|\sin(2\pi i)|}\leq1$$

$$|f(2\pi i)(2\pi i-2\pi)(2\pi i+2\pi)|\leq|\sin(2\pi i)|$$

$$|f(2\pi i)|(2\pi\sqrt2)^2\leq\frac{|e^{i\cdot 2\pi i}-e^{-i\cdot 2\pi i}|}{|2i|}$$

$$|f(2\pi i)|\leq\frac{e^{2\pi}-e^{-2\pi}}{16\pi^2}$$

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If I understand it correctly, the above reasoning works if our newly-defined function is holomorphic on the same region as $f$? It's not a problem that the denominator disappears at $2\pi$ and $-2\pi$, since we can have $\sin z$ in the form of $(z-2\pi)(z+2\pi)h(z)$ for some holomorphic $h$, right? – czachur Feb 1 '13 at 8:06
@czachur That's correct. The function on the right can be extended to a holomorphic function just by plugging in the holes. – Brett Frankel Feb 1 '13 at 18:04