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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.

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    Sorry, fixed the typo.2012-08-05

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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.

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    Also use $|z^4 - 1| \geq |z|^4 - 1 = 80$ on $|z|=3$.2012-08-05