Zeros in Complex Analysis 📂Complex Anaylsis

Zeros in Complex Analysis

Definition ¹

$\alpha \in \mathbb{C}$ being a zero of order $n$ of the function $f : \mathbb{C} \to \mathbb{C}$ means that for some function $g$, where $\displaystyle \lim_{z \to \alpha} g(z) \ne 0$, it can be expressed as follows: $$ f(z) = (z-\alpha)^{n} g(z) $$

Theorem

Zeros are isolated:

For a zero $f$, we can take a radius such that no other zeros exist around it.
For the zero $\alpha$ of $f$, there exists a neighborhood $\mathcal{N} (\alpha)$ where $z \in \mathcal{N} (\alpha) \setminus \left\{ \alpha \right\}$ to $f (z) \ne 0$.

Proof

Without loss of generality, assume that $g$ is analytical at the zero of order $n$ $\alpha$ of $f$ and let’s denote it as $g(\alpha) = 2 \beta \ne 0$.

Since $g$ is continuous at $\alpha$, for all $\beta$, there must exist a $\delta > 0$ satisfying the following: $$ | z - \alpha | < \delta \implies \left| g(z) - g(\alpha) \right| < |\beta| $$ Since we previously set as $g(\alpha) = 2 \beta$, by the triangle inequality, we have: $$ | z - \alpha | < \delta \implies |g(z)| \ge \left| |g(\alpha)| - \left| g(z) - g(\alpha) \right| \right| > |\beta| $$ Since $|z-\alpha| < \delta$ leads to $|g(z)| > |\beta|$, $\alpha$ cannot be a zero of $g$. Given $f(z) = (z-\alpha)^{n} g(z)$, specifically within this open ball $B \left( \alpha , \delta \right)$, only $\alpha$ becomes the zero of $f$: $$ f(z) \begin{cases} = 0 & , \text{if } z = \alpha \\ \ne 0 & , \text{if } z \in B \left( \alpha , \delta \right) \setminus \left\{ \alpha \right\} \end{cases} $$

■