Answer: An Entire Function with Non-Negative Real Part

Geometric Intuition

Think about what the condition says about the image of $f$: the entire output of $f$ is confined to the right half-plane — a closed, convex half of $\mathbb{C}$.

Now recall Liouville's theorem needs boundedness. The right half-plane is not bounded, so we can't apply Liouville directly. But here's the key geometric picture:

The right half-plane and the unit disk are conformally equivalent — one is just a "bent" version of the other.

The Möbius map $g(w) = (w-1)/(w+1)$ literally folds the right half-plane into the unit disk: points near the origin map near $-1$, points far out toward $+\infty$ map near $+1$, and the imaginary axis (boundary of the half-plane) wraps onto the unit circle.

So if $f(\mathbb{C})$ is trapped in the right half-plane, then $g \circ f$ is trapped in the unit disk — and that is bounded. Liouville then says $g \circ f$ is constant, and unwrapping gives $f$ is constant.

The insight: you don't need the image to be bounded, just conformal-equivalent to something bounded.

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Solution

Define the Möbius transformation (https://www.youtube.com/watch?v=0z1fIsUNhO4&t=7s)

$g(w) = \frac{w - 1}{w + 1}.$

This map sends the right half-plane $\{w \in \mathbb{C} : \operatorname{Re}(w) \geq 0\}$ into the closed unit disk $\{|g| \leq 1\}$. Indeed, for $\operatorname{Re}(w) \geq 0$:

$|w - 1|^2 = (\operatorname{Re}(w) - 1)^2 + \operatorname{Im}(w)^2,$ $|w + 1|^2 = (\operatorname{Re}(w) + 1)^2 + \operatorname{Im}(w)^2.$

Their difference is $|w-1|^2 - |w+1|^2 = -4\operatorname{Re}(w) \leq 0$, so $|g(w)| \leq 1$.

Now consider the composition

$h(z) = g(f(z)) = \frac{f(z) - 1}{f(z) + 1}.$

• $h$ is entire: $f$ is entire, and $g$ is holomorphic away from $w = -1$. Since $\operatorname{Re}(f(z)) \geq 0$, we have $f(z) \neq -1$ for all $z$, so $h$ is holomorphic everywhere.
• $h$ is bounded: $|h(z)| = |g(f(z))| \leq 1$ for all $z \in \mathbb{C}$.

By Liouville's theorem, a bounded entire function is constant. So $h \equiv c$ for some $|c| \leq 1$.

Inverting, $f(z) = g^{-1}(c) = \frac{1+c}{1-c}$, which is a constant (assuming $c \neq 1$; but $c = 1$ would require $f(z) \to \infty$, impossible for an entire function).

Therefore $f$ is constant. $\blacksquare$

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Why the Trick Works

The Möbius map $g(w) = (w-1)/(w+1)$ is the unique (up to rotation) conformal map of the right half-plane to the unit disk fixing the real axis. Composing it with $f$ converts a geometric constraint on the image (image ⊂ right half-plane) into analytic boundedness, which is exactly what Liouville needs.

This composition trick is a template: whenever you know the image of an entire function avoids or is contained in some region, try to compose with a conformal map that compresses that region into a disk.

Generalization

The same argument shows: if $f$ is entire and its image omits any open half-plane (or any disk, or any simply-connected proper open subset of $\mathbb{C}$), then $f$ is constant. This is essentially one direction of Picard's little theorem — an entire function that omits even two points must be constant.