# On an analytic function $f$ on $\{z:\lvert z\rvert \leq 3\}$ such that $\{1,i,-1,-i\}\subset f^{-1}(0)$

How to prove the following inequality with the hypotheses in the title of the question?$$\lvert f(0)\rvert\le\frac1{80}\max_{\lvert z\rvert=3}\lvert f(z)\rvert$$ By maximum principle, clearly it's true without the division by $80$. I am not able to get through this one. Any ideas?

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Let $$g(z) = \frac{f(z)}{(z-1)(z-i)(z+1)(z+i)} = \frac{f(z)}{z^4-1}.$$
Then $g$ is also analytic on $D_3 = \{ z : |z| \le 3 \}$ and now we can use the maximum prinicple on $g$:
$$|f(0)| = |g(0)| \le \max_{|z|=3} \left| \frac{f(z)}{z^4-1} \right| \le \frac1{80} \max_{|z|=3} |f(z)|.$$
Thanks very much! This construction for $g$ did flash in my mind! I only hadn't put it together. Thanks a lot! –  Host-website-on-iPage Oct 28 '12 at 8:44
Very ingenious, mrf (this construction for $g$ never entered my mind!) : +1 –  Georges Elencwajg Oct 28 '12 at 8:49