$f:\mathbb{C}\rightarrow\mathbb{C}\setminus\{z\in\mathbb{C}:\Im(z)=0\text{ and }\Re(z)\ge 0\}$ is constant

Question

I want to show that if $$f:\mathbb{C}\rightarrow\mathbb{C}\setminus\{z\in\mathbb{C}:\Im(z)=0\text{ and }\Re(z)\ge 0\}$$ is holomorphic, then f is constant.

I tried to show that $f$ is bounded, then I can use Liouville's theorem.

Is this the right way and how can I see that $f$ is bounded?

Answer 1

Hint- you have a holomorphic branch of the logarithm in the given region. Now compose maps.

Alternatively, use the square root function to map the slit plane to the upper half plane, and then use the standard transformation that takes the UHP to the unit disk.

Answer 2

The function $-f$ does not assume nonpositive real values, hence the principal value $g:={\rm pv}\sqrt{-f}$ is well defined. The values of $g$ lie in the right half-plane, and this implies that the entire function $h:=e^{-g}$ is bounded. By Liouville's theorem therefore all three functions $h$, $g$, and $f$ are constant.

Answer 3

Hint :

Use Picard's little theorem.

$f$ is entire and it omits all non-negative real numbers.