Question

state and prove liouville's theorem​

Answer 1

Step-by-step explanation:

Liouville's Theorem states that a bounded holomorphic function on the entire complex plane must be constant. It is named after Joseph Liouville. Picard's Little Theorem is a stronger result .

Statement

Let $f : \mathbb{C} \to \mathbb{C}$ be a holomorphic function. Suppose there exists some real number $M \ge 0$ such that $\lvert f(z) \rvert \le M$ for all $z \in \mathbb{C}$. Then $f$ is a constant function.

Proof

We use Cauchy's Integral Formula.

Pick some $z_0 \in \mathbb{C}$; let $C_R$ denote the simple counterclockwise circle of radius $R$ centered at $z_0$. Then\[\lvert f'(z_0) \rvert = \biggl\lvert \frac{1}{2\pi i} \int_C -\frac{f(z)}{(z-z_0)^2}dz \biggr\rvert \le \frac{M}{R^2} .\]Since $f$ is holomorphic on the entire complex plane, $R$ can be arbitrarily large. It follows that $f'(z) = 0$, for every point $z \in \mathbb{C}$. Now for any two complex numbers $A$ and $B$,\[f(B) - f(A) = \int_{A}^B f'(z)dz = 0 ,\]so $f$ is constant, as desired. $\blacksquare$

Extensions

It follows from Liouville's theorem if $f$ is a non-constant entire function, then the image of $f$ is dense in $\mathbb{C}$; that is, for every $z_0 \in \mathbb{C}$, there exists some $z \in \text{Im}\,f$ that is arbitrarily close to $z_0$.

Proof

Suppose on the other hand that there is some $z_0$ not in the image of $f$, and that there is a positive real $\epsilon$ such that $\text{Im}\,f$ has no point within $\epsilon$ of $z_0$. Then the function\[g(z) = \frac{1}{f(z) - z_0}\]is holomorphic on the entire complex plane, and it is bounded by $1/\epsilon$. It is therefore constant. Therefore $f$ is constant. $\blacksquare$

Picard's Little Theorem offers the stronger result that if $f$ avoids two points in the plane, then it is constant. It is possible for an entire function to avoid a single point, as $\exp(z)$ avoids 0.