Liouville's Theorem and the Fundamental Theorem of Algebra

Two of the most elegant applications of Cauchy's theory: every bounded entire function is constant, and every non-constant polynomial has a complex root.

Liouville's Theorem

Theorem4.4Liouville's theorem

If $f: \mathbb{C} \to \mathbb{C}$ is entire (holomorphic on all of $\mathbb{C}$ ) and bounded, then $f$ is constant.

Proof

For any $z_0 \in \mathbb{C}$ and $R > 0$ , Cauchy's inequality gives

$|f'(z_0)| \leq \frac{M}{R}$

where $M = \sup_{z \in \mathbb{C}} |f(z)|$ . Since $R$ is arbitrary, letting $R \to \infty$ yields $f'(z_0) = 0$ for all $z_0$ . Hence $f$ is constant. $\blacksquare$

■

RemarkSharpness and generalizations

Liouville's theorem is sharp: $e^z$ is entire and satisfies $|e^z| = e^{\text{Re}(z)}$ , which is unbounded. Generalizations include:

An entire function bounded by $C|z|^n$ is a polynomial of degree at most $n$ .
Picard's little theorem: a non-constant entire function takes every complex value with at most one exception.

Fundamental Theorem of Algebra

Theorem4.5Fundamental theorem of algebra

Every non-constant polynomial $p(z) = a_nz^n + \cdots + a_0$ with $a_n \neq 0$ and $n \geq 1$ has at least one root in $\mathbb{C}$ .

Consequently, $p(z)$ factors completely as $p(z) = a_n(z - z_1)(z - z_2)\cdots(z - z_n)$ where $z_1, \ldots, z_n$ are the roots (counted with multiplicity).

ExampleFactoring polynomials

The polynomial $p(z) = z^4 + 1$ has no real roots but factors over $\mathbb{C}$ :

$z^4 + 1 = (z - e^{i\pi/4})(z - e^{3i\pi/4})(z - e^{5i\pi/4})(z - e^{7i\pi/4}).$

The roots are the eighth roots of unity with odd index: $e^{i(2k+1)\pi/4}$ for $k = 0,1,2,3$ .

Applications of Liouville's Theorem

ExampleDoubly periodic functions

An entire function $f$ that is doubly periodic — $f(z + \omega_1) = f(z + \omega_2) = f(z)$ for two $\mathbb{R}$ -linearly independent periods $\omega_1, \omega_2$ — must be constant. The values of $f$ are determined by its values on the compact fundamental parallelogram $\{s\omega_1 + t\omega_2 : 0 \leq s, t \leq 1\}$ . By continuity on a compact set, $f$ is bounded, hence constant by Liouville's theorem.

This motivates the study of elliptic functions: meromorphic doubly periodic functions (which must have poles).

RemarkNon-existence results

Liouville's theorem is a powerful tool for non-existence proofs:

There is no entire function satisfying $|f(z)| \geq 1$ for all $z$ (otherwise $1/f$ would be bounded entire, hence constant, so $f$ is constant, contradicting $|f| \geq 1$ unless $f$ is a constant of modulus $\geq 1$ , which is consistent only if $f$ is constant).
There is no holomorphic square root function on $\mathbb{C} \setminus \{0\}$ : if $g(z)^2 = z$ for all $z \neq 0$ , then $g$ extends to an entire function (by boundedness near $0$ ) with $g(0) = 0$ , contradicting $g(z)^2 = z$ .

The Open Mapping Theorem

Theorem4.6Open mapping theorem

If $f$ is a non-constant holomorphic function on a domain $D$ , then $f$ is an open map: it sends open sets to open sets.

RemarkConsequences

The open mapping theorem has several important consequences:

Maximum modulus principle follows immediately: if $|f|$ attains a maximum at an interior point $z_0$ , then $f(D)$ is open and contains $f(z_0)$ , so there exist nearby values with larger modulus — contradiction.
Inverse function theorem: if $f'(z_0) \neq 0$ , then $f$ is locally injective near $z_0$ and the inverse is also holomorphic.
Combined with the identity theorem, it shows that non-constant holomorphic maps are "locally $n$ -to- $1$ near a zero of order $n$ ."

ExampleImage of a disk

The function $f(z) = z^2$ maps the open disk $D(0,1)$ onto $D(0,1)$ (each point $w \in D(0,1)$ has two preimages $\pm\sqrt{w}$ ). The boundary circle $|z| = 1$ maps to $|w| = 1$ (each point covered twice, since $z$ and $-z$ give the same $w$ ). This illustrates the open mapping property.