Analyticity of Polynomial and Rational Complex Functions

# Analyticity of Polynomial and Rational Complex Functions

Recall from the Analytic Complex Functions page that if $A \subseteq \mathbb{C}$ is open, $z_0 \in A$, and $f : A \to \mathbb{C}$ then $f$ is said to be analytic at $z_0$ if there exists an $r > 0$ such that $f$ is complex differentiable on the open disk $D(z_0, r)$. Furthermore, we said that $f$ is analytic on $A$ if $f$ is analytic at every point $z_0 \in A$.

On the Analyticity of Sums, Differences, Products, and Quotients of Analytic Functions page we looked at a bunch of properties regarding the sums, differences, products, and quotients of analytic functions - all of such results are extensions from introductory calculus of a single variable.

We will now show that every complex polynomial is analytic on all of $\mathbb{C}$ and that every complex rational function is analytic on all $\mathbb{C}$ except at the points where it is undefined.

 Theorem 1: If $f$ is a polynomial then $f$ is analytic on all of $\mathbb{C}$.
• Proof: We will carry this proof out by induction. We first show that any constant function is analytic on all of $\mathbb{C}$. Let $c \in \mathbb{C}$ and let $f(z) = c$. Then for all $z_0 \in \mathbb{C}$ we have that:
(1)
\begin{align} \quad f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} = \lim_{z \to z_0} \frac{c - c}{z - z_0} = \lim_{z \to z_0} 0 = 0 \end{align}
• So $f'(z_0)$ exists (and is equal to $0$) for every $z_0 \in \mathbb{C}$ which shows that any constant function is analytic on all of $\mathbb{C}$.
• Now if $f(z) = z$ then for all $z_0 \in \mathbb{C}$ we have that:
(2)
\begin{align} \quad f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} = \lim_{z \to z_0} \frac{z - z_0}{z - z_0} = 1 \end{align}
• So $f'(z_0)$ exists (and is equal to $1$) for every $z_0 \in \mathbb{C}$ which shows that the function $f$ is analytic on all of $\mathbb{C}$.
• Now for any polynomial of order $n$, say $f(z) = a_0 + a_1z + ... + a_nz^n$ where $a_0, a_1, ..., a_n \in \mathbb{C}$ we have by the theorems regarding products of analytic functions that $f$ is analytic on all of $\mathbb{C}$ which completes the proof. $\blacksquare$
 Theorem 2: If $f$ is a rational function then $f$ is analytic on all of $\mathbb{C}$ except at the points where $f$ is undefined.
• Proof: Every rational function $f$ is a quotient of polynomials, say for some $n, m \in \{ 0, 1, ... \}$:
(3)
\begin{align} \quad f(z) = \frac{a_0 + a_1z + ... + a_nz^n}{b_0 + b_1z + ... + b_mz^m} \end{align}
• By Theorem 1, the numerator and denominator individually are analytic on all of $\mathbb{C}$, and by the theorem regarding quotients of analytic functions we have that $f$ is analytic on all of $\mathbb{C}$ except at the points where the denominator is $0$ (i.e., where $f$ is undefined). $\blacksquare$