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?

$f$ 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 $D.$
