If $f(z)$ has positive real part, show that $f$ is a constant function.

Let $f$ be an entire function. If $f(z)$ has positive real part, show that $f$ is a constant function.

Proof: Let $f(z) = u + vi$. Consider the function $g(z) = e^{-f}$. We have

$$|g(z)|=|e^{-u - vi}|=|e^{-u}||\cos(v)-i\sin(v)|$$

Since $u$ is positive, $|g(z)|$ is bounded. Since $g(z)$ is entire, $g(z)$ has to be constant. Thus, $f$ has to be constant as well.

I am not sure about the last sentence though because $e^z = e^{z + 2k\pi i}$. I tried to argue that if $f$ is not a constant, then it has to be unbounded. But that doesn't mean the real part has to be unbounded so I don't have an obvious contradiction here.

Solution 1:

Here is a similar approach: Let $g(z) = {1 \over 1+f(z)}$ and note that $|1+f(z)| \ge 1$ for all $z$ and so $|g(z)| \le 1$ for all $z$ hence $g$ is constant (and non zero). Then $f(z) = {1 \over g(z)} -1$, and so $f$ is constant.