If $f(0) = \pi i$ and $e^{Re f(z)} \leq 1$ for all $|z| < 1,$ then $f(z) = \pi i$ for all $z \in D.$

Question

Let $f$ be analytic on $D = \{z, |z| < 1\}.$ Assume that $f(0) = \pi i$ and $e^{Re f(z)} \leq 1$ for all $|z| < 1.$ Show that $f(z) = \pi i$ for all $z \in D.$

My idea is to show that $f(z)$ is constant on the region, which will give the answer since we know $f(0) = \pi i.$

I first tried using Liouville's theorem on $e^{Re f(z)}$ since it is bounded, however, we don't know that $e^{Re f(z)}$ is analytic. If there is a way to use Liouville's theorem on $e^{Re f(z)},$ then I know how to get the answer from there.

I then thought about using Liouville's theorem on $f,$ but even though we know it is analytic, I don't know how to show that $f$ is bounded.

Answer 1

Hint: the condition $e^{\Re f(z)} \le 1$, which is the same as $\Re f(z)\le 0$, says that the image of $f$ is contained in the left half plane. And conformally, half planes and disks are the same type of object (because lines and circles are the same on the Riemann sphere).

Can you find an injective function $g$ (perhaps a Möbius transformation) that maps the left half plane into the unit disk, and then consider the function $g\circ f$?

Answer 2

If $f(z)$ were not constant, the open mapping theorem would imply that the image of $D$ under $f(z)$ would be an open set containing $f(0) = \pi i$. Thus there would be some $z$ in $D$ with $Re f(z) > 0$, which contradicts the condition that $e^{Re f(z)} \leq 1$ for all $z \in D$.