Question: $f(z)$ is analytic in $C$ and $Im(f(z))\leq 0$. I want to show that $f(z)$ is a constant.
Approach: I know that if $f$ is analytic on a closed curve then the line integral along that curve of $f$ is $0$. But I how do I use the fact that the $f$ is analytic on the lower half plane? And how exactly does it imply that $f$ is constant?
Any help would be greatly appreciated.
Thankyou
Hint: Is $\exp(-i f(z))$ bounded?
Let $f(z) = u(z) + i v(z)$ where $u$ and $v$ are real functions. We have:
$$ \exp(-if(z)) = \exp(-iu(z))\exp(v(z)) $$
Hence:
$$ \left|\exp(-if(z))\right| = \exp(v(z)) $$
Since $v(z) \le 0$. It follows that $\exp(-if(z))$ is bounded. By Liouville's theorem, it's constant. Thus $f$ is constant.