Answer: The Holomorphic Function That Misses Two Points

Key Idea / Intuition

If $f$ avoids two values $a$ and $b$, we can normalize so that $f$ avoids $0$ and $1$. Then $\log f$ can be defined globally (since $f$ never hits $0$), giving an entire function $g = \log f$ that avoids all real multiples of $2\pi i$. One more layer of cleverness — a classical trick using $\sin$ or the exponential — produces a bounded entire function, which by Liouville must be constant.

The cleanest route: use the fact that $\mathbb{C} \setminus \{0,1\}$ has the upper half-plane as its universal cover, combined with the monodromy theorem, to lift $f$ to a bounded holomorphic function.

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Formal Proof / Solution

Step 1: Normalize.

Since $a \neq b$, consider $h(z) = \frac{f(z) - a}{b - a}$. Then $h$ is entire, and $h$ omits both $0$ and $1$. So without loss of generality, assume $f$ itself is entire and satisfies $f(z) \notin \{0, 1\}$ for all $z$.

Step 2: Define a global logarithm.

Since $f$ is entire and never zero, $f$ has a global holomorphic logarithm: $g(z) = \log f(z),$ which is entire. Here we use that $\mathbb{C}$ is simply connected, so the logarithm can be defined without branch cuts. Thus $e^{g(z)} = f(z)$ and $g$ is entire.

Step 3: The image constraint on $g$.

Since $f(z) \neq 1$, we have $e^{g(z)} \neq 1$, which means: $g(z) \notin 2\pi i \mathbb{Z} \quad \text{for all } z.$

Step 4: Use the universal covering of $\mathbb{C} \setminus \{0,1\}$.

Here is the key classical fact: the upper half-plane $\mathbb{H}$ is the universal cover of $\mathbb{C} \setminus \{0,1\}$, via the modular lambda function $\lambda : \mathbb{H} \to \mathbb{C} \setminus \{0,1\}$.

Since $f : \mathbb{C} \to \mathbb{C} \setminus \{0,1\}$ and $\mathbb{C}$ is simply connected, by the monodromy theorem (lifting criterion for covering spaces), $f$ lifts to a holomorphic map: $\tilde{f} : \mathbb{C} \to \mathbb{H}$ satisfying $\lambda \circ \tilde{f} = f$.

Step 5: Apply Liouville's theorem.

The upper half-plane $\mathbb{H}$ is conformally equivalent to the open unit disk $\mathbb{D}$ via the Möbius transformation: $\phi(w) = \frac{w - i}{w + i}.$

Define $F = \phi \circ \tilde{f} : \mathbb{C} \to \mathbb{D}$. Then $F$ is an entire function with $|F(z)| < 1$ for all $z$.

By Liouville's theorem, $F$ is constant. Hence $\tilde{f}$ is constant, and therefore: $f = \lambda \circ \tilde{f}$ is constant. $\blacksquare$

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Why This Is Surprising

The result is sharp: an entire function can omit one value (e.g., $e^z$ omits $0$). Omitting two values is too much — the function is forced to be constant. The proof is conceptually beautiful: the "two missing values" rigidify the target space so much that its universal cover becomes the disk, and any entire map into the disk must be constant by Liouville.