# Show that an analytic function $f$ has a zero on the unit disc given that $|f(z)|>2$ on the boundary and $f(0) = 1$

Question: Suppose $f$ is analytic in a region A containing the unit disc $D = \{ z: |z| \leq 1 \}$ and such that $|f(z)|>2$ whenever $|z| = 1$. If $f(0) = 1$, show that $f$ has a zero in D.

Thoughts: I am studying for an exam in complex analysis and this is one of the more theoretical questions from an old exam. Since my knowledge of complex analysis theory is small, I'm not quite sure how to proceed. I supposed it has something to do with that $f$ must be negative somewhere on $D$ and since the function is analytic in the region, $f$ must have a zero somewhere according the mean-value theorem or an analogous version of it. All input is very much appreciated.

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Do you know the maximum modulus principle? Suppose $f$ has no zeros in $D$, and consider the function $1/f$...
Thank you for the hint. I have not heard about this before I read this article, but now I think I can construct a proof: Assume $f$ has no zeros in $D$. Then $1/f$ is analytic on the whole region $D$. We also have that $|1/f|<1/2$ on the boundary of D and $|1/f|=1$ in $z=0$. But this must mean that $|1/f|$ takes a maximum value somewhere on the disc! This violates the principle you showed me and hence the assumption was incorrect, so $f$ must have a zero on D. Was this what you had in mind? –  Sid Aug 29 '13 at 0:35