# Analytic $f$ with $|f|$ constant on $\partial D$, I need to show that it has atleast one $0$ inside $D$

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 $G$,$|f|$ constant on $\partial D$, I need to show that it has atleast one $0$ inside $D$.

Suppose it has no zero inside $D$, then I consider $\frac{1}{f}$ which is analytic and non vanishing inside $D$ and must attain its minima on $\partial D$, minima of $1\over f$ is maxima of $f$ that is also attained on $\partial D$, so minima and maxima of $f$ are attained on $\partial D$, but does that create any contradiction? am I in the right path?

• is there any assumption about $f$ being non-constant? May 3, 2013 at 4:04
• @JohnMartin Thank you, I missed one line. May 3, 2013 at 4:07
• I think somewhere you should use the hypothesis that $|f|$ is constant on $\partial D$ in order to conclude that the max and min of $f$ are in fact the same, which in turn implies that $f$ is constant, contrary to the hypothesis that $f$ is non-constant. May 3, 2013 at 4:12

$f$ is non-constant on $G\implies f$ is non-constant on $D.$ (follows as a silly corollary of the identity theorem)
As you have noticed your assumption that $f$ has no zero on $D$ tells that maximum and minimum of $|f|$ (not of $f$) are attained on $\partial D.$ $|f|$ being constant on $\partial D,|f|$ becomes constant on $D$ $(\max=\min);$ more clearly $|f|$ attains maximum at a point in $D,$ which is impossible unless $f$ is constant on $D.$
$|f|$ attains its minimum and maximum on boundary where it is constant = $c > 0$.
This implies $|f|$ is constant on the whole disk and so it is on its interior. Thus the image of the open disk under $f$ is contained in circle of radius $c$. A contradiction to open mapping theorem except if $f$ is constant.
• Are you saying that if $f$ is analytic and non-constant on a open disk $D$ then it's impossible for $|f|$ to get contained in any circle? If it be so, then what about the identity mapping on the unit open disk? May 3, 2013 at 5:40