True or False $Re f(z)>0$ [duplicate]

Question

Make an examples $f:\Bbb{C}\to\Bbb{C}$ holomorph (If it exists) with the property:(Justify your answer)

$Re f(z)>0$ for all $z\in\Bbb{C}$ and $f$ is not constant.

My attemp: I find a question in Conway that is discussed in $f : D \rightarrow \mathbb{C} $ analytic. Show $\operatorname{Re}(f(z)) \geq 0$.

But my question is difference. Because we have not the hypothesis $Re f(z)\geq 0$ and the domain is $\Bbb{C}$ instead of $D$. So we can not use the open mapping theorem and I think there is not such function!

Answer 1

No such $f$ exists. Suppose a non constant function $f(z)$ and let $f(z)=u+iv$ and consider the function $F(z)=e^{-f(z)}$. Then, $$|F(z)|=|e^{-u-iv}|=e^{-u}$$ Note that since $\Re f(z)= u>0$, it follows that $|F(z)|<1$ and by Liouville's Theorem $F(z)$ is constant. This implies that $F'(z)=-f'(z)e^{-f(z)}=0$ and this further implies that $f'(z)=0$. Hence, $f(z)$ must be a constant.