$|f(z)|$ is constant but $f(z)$ is not constant.

Question

According to this post Show that if $f$ is analytic in a domain $D$ and $|f(z)|$ is constant in $D$, then the function $f(z)$ is constant in $D$ We have,

If $f$ is analytic in a domain $D$ and $|f(z)|$ is constant in $D$ then the function $f(z)$ is constant in $D$.

But, If I consider the function $f(z) =e^{iz}$ then clearly $f$ is analytic on any domain $D\subseteq\mathbb{C}$. Moreover, $|f(z) |=1$ for all $z\in D$ But, $f(z) =e^{iz}$ is Not constant in $D$.

I think, I made some mistake. But I am not getting where! Please help me.

Answer 1

But, If I consider the function $f(z) =e^{iz}$ then clearly $f$ is analytic on any domain $D\subseteq\mathbb{C}$.

Correct.

Moreover, $|f(z) |=1$ for all $z\in D$

No. $|f(z)|=1$ if and only if $z\in\mathbb{R}$. But for example $f(i)=e^{-1}$ and so $|f(i)|\neq 1$.

When we are dealing with complex analysis, the term "domain" almost always means at least open subset of $\mathbb{C}$. And therefore $D\subseteq \mathbb{R}$ is impossible, since $\mathbb{R}$ has empty interior in $\mathbb{C}$.

And this implies that $|f(z)|=1$ for all $z\in D$ is also impossible.

Answer 2

Consider $z=x+iy$, then $f(z)=e^{iz}=e^{ix}e^{-y}=e^{-y}(\cos{x}+i \sin{x})$, so $|f|$ is not constant when $y$ increases. You may ignore that $z\in \mathbb{C}$, not a real variable.

Answer 3

$f$ entire and $\mid f(z)\mid=c\implies f$ is bounded. Then by Liouville's theorem $f$ is constant.

It's interesting to note that, in the link you provided, we find that this is also true when $f$ is only analytic on a domain $D\subset \Bbb C.$