This question was asked in comprehensive which I am doing .
Let $\Omega$ be a bounded region and g $\in C(\bar\Omega) \cap H(\Omega)$. Assume that |g| is a non-zero constant on $d\Omega$ . If g is not constant on $\Omega $ then show that g has atleast 1 one zero in $\Omega$ .
Maximum modulus principle can be used and using hypothesis |g| is constant let c on $d\Omega$. So, max g $\leq$ c . But I am not able to proceed from this.
Note: This question has an answer here:Suppose $|f|$ is constant on $\delta D$. Show that $f$ has at least one zero in $D$.
but I couldn't deduce how Question asker wrote that 1/f will not be holomorohic using Maximum principleprinciple and user 622… was last seen 7years ago. So, I don't think I should ask him.
Kindly tell how should I approach this question.
Best Answer
Assume $|g|=c\;\,$on $d\Omega$.
By the maximum modulus principle, we have $|g(z)|\le c$ for all $z\in\overline\Omega$.
Since $g$ is holomorphic and non-constant on $\Omega$, $g(\Omega)$ must contain a nonempty open subset of $\mathbb{C}$.
It follows that $|g|$ is non-constant on $\Omega$, hence if $b$ is the minimum value of $|g|$ on $\overline\Omega$, we must have $b < c$.
Suppose $b > 0$.
Then $g(z)\ne 0$ for all $z\in\Omega$.
Hence by the minimum modulus principle, the minimum value of $|g|$ on $\overline\Omega$ must occur at some $z\in d\Omega$, contrary to $b < c$.
Therefore $b=0$.