Orders of Roots of Analytic Complex Functions

We will soon develop some nice results regarding analytic complex functions and their Taylor series representation. We will first examine the roots of analytic complex functions by classifying them by their "order" which we define precisely below.

Definition: Let $A \subseteq \mathbb{C}$ be open, $f : A \to \mathbb{C}$ be analytic on $A$, and let $z_0 \in A$ be a root of $f$. Let $\displaystyle{T_f^{z_0} = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n}$ be the Taylor series of $f$ centered at $z_0$. Then $z_0$ is said to be a Root (Zero) of Order $k$ if $f(z) = (z - z_0)^k (a_k + a_{k+1}(z - z_0) + ... ) = (z - z_0)^k g(z)$ where $g(z) \neq 0$ and $g$ is analytic on $A$.

Note that a root of order $k$ occurs when $f(z_0)$, $f'(z_0)$, …, $f^{(k-1)}(z_0)$ in the Taylor series of $f$ centered at $z_0$ are $0$.

For example, consider the function $f(z) = \cos (z^2)$. The number $z_0 = 0$ is a root of $f$. Let's determine the order of $z_0 = 0$. We know that the Taylor series of $\sin z$ is given (and valid for all $z \in \mathbb{C}$) by:

(1)

\begin{align} \quad \sin z = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n + 1)!} z^{2n+1} \end{align}

Therefore the Taylor series of $\sin z^2$ is given (and valid for all $z \in \mathbb{C}$) by:

(2)

\begin{align} \quad \sin z^2 &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n + 1)!} (z^2)^{2n+1} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{4n+2} \\ &= z^2 - \frac{z^6}{3!} + \frac{z^{10}}{5!} - ... \\ &= z^2 \underbrace{\left ( 1 - \frac{z^4}{3!} + \frac{z^8}{5!} - ... \right )}_{g(z), g(0) \neq 0} \\ &= z^2 g(z) \end{align}

Therefore the order of $z_0 = 0$ for the function $\sin z^2$ is $2$.

For another example, consider the function $f(z) = z(z - 1)^2$ which is analytic on all of $\mathbb{C}$. The points $z_0 = 0$ and $z_1 = 1$ are both roots of $f$ and intuitively, the order of $z_0$ should be $1$ and the order of $z_1$ should be $2$. In other words, the definition of the order of a root of an analytic complex function should coincide with the multiplicity of a root for a complex polynomial. This is indeed the case since:

(3)

\begin{align} \quad f(z) = z\underbrace{(z - 1)}_{g_1(z), g_1(0) \neq 0} = zg_1(z) \end{align}

(4)

\begin{align} \quad f(z) = (z - 1)^2 \underbrace{z}_{g_2(z), g_2(1) \neq 0} = (z - 1)^2 g_2(z) \end{align}

The formulas above show that $z_0 = 0$ is a root of order $1$ and $z_1 = 1$ is a root of order $2$.