Prove that a holomorphic function with postive real part is constant

Question

Suppose that $f$ is holomorphic on $\mathbb C$ and that $\Re(f(z))\ge 0$ for all $z$. Show that $f$ is constant. [Hint: consider $e^{−f(z)}$.]

My thoughts: If $\Re(f(z))\ge 0 $ holds, then $e^{−f(z)}$ is a bounded holomorphic function (do I need to prove this or is it obvious?) So then by Liouville's Theorem $e^{−f(z)}$ is constant.

But then I'm not sure how to rigorously go from this $\exp(−f(z))$ back to $f(z)$.

Could anyone help me piece this together please?

Thanks

Answer 1

If you can show that $e^{-f(z)}$ is constant, then: $\frac {d}{dz}e^{-f(z)}=e^{-f(z)}f'(z)=0$, so $f'(z)=0$ , and $f(z)$ is constant.

Answer 2

A different approach(without using $e^{-f(z)}$):

We know that the real and imaginary parts of an analytic function are harmonic functions. By Liouville's theorem, we know that the real part of $f(z)$ is constant. Furthermore, we know that by Cauchy-Riemann equations the imaginary part of $f(z)$ is also constant. Thus we conclude that $f(z)$ is constant.