# Suppose $|f|$ is constant on $\delta D$. Show that $f$ has at least one zero in $D$.

I'm doing some exercises for an exam and I've come across this one from a past comprehensive that I can't solve, can anyone give me any tips/hints?

Suppose that $f$ is analytic in a domain $G$ in the complex plane and not constant. Let $D$ be a disc whose closure is contained in $G$. Suppose $|f|$ is constant on $\delta D$. Show that $f$ has at least one zero in $D$.

I've tried showing that since $|f|$ is constant on $\delta D$ $f$ is constant on $\delta D$ (via CR equations), but I am not sure that is right (can anyone tell me if it is or not?).

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Hint: Assume for contradiction that $f$ has no zeroes and apply the maximum modulus principle to $1/f$.
Hey, so from this approach I can get that $1/f$ cannot be holomorphic in $D$, but is that a sufficient condition to say that $f$ has at least on zero in $D$. What I mean is does the faliure of holomorphicity only occur because we are dividing by 0 or could it happen due to something else? – user62245 Feb 14 '13 at 5:08
@user62245 Presumably you showed that if $f$ has no zeroes then $f$ is constant on $D$ and so constant on $G$. Which contradicts that $f$ was not constant, thereby $f$ has a zero in $D$. But also, in answer to your question $1/f$ is holomorphic on the domain of $f$ minus the zeroes of $f$. – JSchlather Feb 14 '13 at 5:10