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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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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

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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.

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

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