If $f$ is an entire function (ie analytic in the complex plane) and is bounded, then f must be a constant

! Not hard to prove using Cauchy's Estimate. For example, since $\sin z$ is not a constant, by Lioville's Theorem we know that there must be some points in the complex plane that $\sin z$ will go off to infinity.

A rather counter-intuitive example is when $f$ is an entire function with $u(z) \le 0 \, \forall z \in \mathbb{C}$, then $f$ must be constant ! The proof starts with considering the function $g(z) = e^{f(z)}$.

Furthermore, Liouville's Theorem can be used to prove (via contradiction) the Fundamental Theorem of Algebra, which states that any polynomial

$$\begin{aligned} p(z) = a_0 + a_1z + \cdots + a_n z^n \end{aligned}$$

with $a_0, \cdots, a_n \in \mathbb{C}$ with $n \ge 1$ and $a_n \not = 0$ has a zero in $\mathbb{C}$. In other words, there exists $z_0 \in \mathbb{C}$ such that $p(z_0) = 0$. As a consequence, polynomials can always be factored in $\mathbb{C}$ (but this is not the case in $\mathbb{R}$) ! So

$$\begin{aligned} p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n) \end{aligned}$$

where $z_1, \cdots , z_n \in \mathbb{C}$ are the zeros of $p$. A typical example would be $p(z) = z^2 + 1$.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 3:40 PM