Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$f$ 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?

share|cite|improve this question
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
up vote 1 down vote accepted

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

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

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

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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.