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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?

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    $\begingroup$ is there any assumption about $f$ being non-constant? $\endgroup$ May 3, 2013 at 4:04
  • $\begingroup$ @JohnMartin Thank you, I missed one line. $\endgroup$
    – Myshkin
    May 3, 2013 at 4:07
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    $\begingroup$ 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. $\endgroup$ May 3, 2013 at 4:12

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$f$ 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 $D.$

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$|f|$ 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.

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  • $\begingroup$ 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? $\endgroup$ May 3, 2013 at 5:40
  • $\begingroup$ not |f| but f. Since |f| = c , image of open disc will be contained in circle of radius c which is contradiction to open mappping theorem. Identity maps open disk to itself and not in any circle. $\endgroup$
    – rohit
    May 3, 2013 at 7:52

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