Factorization of a Complex Polynomial

Question

My book (Complex Analysis by Gamelin) states that a complex polynomial $$p(z) = a_nz^n + a_{n-1}+\cdots +a_1z + a_0$$, where $z$ is an element of the complex numbers, can be factored as a product of linear factors $$p(z) = c(z-z_1)^{m_1} \ldots (z-z_k)^{m_k}$$ where the $ z_j $'s are distinct and $ m_j \ge 1$. I don't understand how they arrived at this factorization. Can someone explain this to me? Thanks!

Answer 1

This follows from

1. Every complex polynomial $p$ with degree $\deg p > 0$ has a zero, i.e. for at least one $\alpha \in \mathbb{C}$ you have $p(\alpha) = 0$.

2. If a polynomial $p$ has zero $\alpha$, i.e. if $p(\alpha) = 0$, then the polynomial $(z - \alpha)$ divides the polynomial $p(z)$. In other words, if $\alpha$ is a zero of $p$, then there is a polynomial $q$ such that $p(z) = q(z)(z - \alpha)$. Note that it follows that $\deg q = \deg p - 1$.

Thus, if you start from $p$, fact (1) guarantees that there is some zero $\alpha_1$, and using (2) we can rewrite $p$ as $p(z) = (z - \alpha_1)q_1(z)$. Now we apply the same idea to $q_1(z)$ to further rewrite to $p(z) = (z - \alpha_1)(z - \alpha_2)q_2(z)$. This continues until $q_{n+1}(z)$ has degree $0$, because then (1) fails so we can no longer be certain that $q_{n+1}$ has a zero. But since degree $0$ implies that $q_{n+1}(z)=c$ for some $c$, we overall get that $$ p(z) = (z - \alpha_1)(z-\alpha_2) \ldots (z - \alpha_n)c \text{.} $$

Note that some of the $\alpha_i$ might be equal. If that is relevant, we can group the equal $\alpha_i$ together, and call the number of times a particular $\alpha_i$ appears $r_i$. This, then, yields $$ p(z) = c(z - \alpha_1)^{r_1} \ldots (z - \alpha_k)^{r_k} $$

Answer 2

This is a direct result of the fundamental theorem of algebra established by Gauss and which states that every non constant complex polynomial of the form $$\mathcal O(z) = a_nz^n + a_{n-1}+...+a_1z + a_0$$ can be factored into the product of polynomials of order $1$ and/or $2$.