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