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

From Harvard qualification exam, 1990. Let $f$ be a holomorphic function on a domain contained the closed disc $$|z|\le 3$$ such that $$f(\pm 1)=f(\pm i)=0$$ Show that $$|f(0) |\le \frac{1}{80}\max |f(z)|_{|z|=3}$$

I am confused with this question because I do not know how to use the condition $|z|\le 3$ at all. I also do not know how this related to the four zeros (looks arbitrarily to me). This question feels really standard so I venture to ask in here.

share|cite|improve this question
What's the maximum of a set of a complex numbers? In any case, the conclusion seems unlikely based on the example $f(z) = z^5 - z$. – Sean Eberhard Aug 5 '12 at 22:25
Sorry, fixed the typo. – Bombyx mori Aug 5 '12 at 22:29
up vote 7 down vote accepted

The assumption about the zeros of $f$ implies that $g(z) = f(z)/(z^4-1)$ is a holomorphic function defined in the same region. Now use the mean value property of holomorphic functions: the average of a holomorphic function over a circle is equal to its value at the centre.

share|cite|improve this answer
Also use $|z^4 - 1| \geq |z|^4 - 1 = 80$ on $|z|=3$. – Sean Eberhard Aug 5 '12 at 22:36

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.