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

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!

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.