Properties of Polynomials with Complex Coefficients

# Properties of Polynomials with Complex Coefficients

Proposition 1: Let $p(x) \in \wp ( \mathbb{C} )$ be a nonconstant polynomial. Then $p(x)$ has a unique factorization $p(x) = c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m)$ where $\lambda_1, \lambda_2, ..., \lambda_m \in \mathbb{C}$ are the roots of $p$ and $c \in \mathbb{C}$. |

**Proof:**We will carry out this proof by induction to first show that such a factorization exists. Let $p(x) \in \wp (\mathbb{C})$ be a nonconstant polynomial. Then $\mathrm{deg} (p) = m ≥ 1$. For $m ≥ 1$, let $S(m)$ be the statement that every polynomial $p(x) \in \wp (\mathbb{C})$ with degree $m$ has a factorization in the form $c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m)$ where $\lambda_1, \lambda_2, ..., \lambda_m \in \mathbb{C}$ are the roots of $p(x)$ and $c \in \mathbb{C}$.

- Suppose that $m = 1$. Then $p(x) = a_0 + a_1x$ where $a_0, a_1 \in \mathbb{C}$ and $a_1 \neq 0$. Clearly $p(x)$ can be written in the form $p(x) = a_1 \left ( x - \left ( \frac{-a_0}{a_1} \right ) \right )$, and so $S(1)$ is true.

- Now suppose that $m > 1$ and that all polynomials of degree $m - 1$ can be factored in the form above. We know by The Fundamental Theorem of Algebra that since $\mathrm{deg} (p) = m ≥ 1$ then $p(x)$ has a root, say $\lambda_1$. Therefore for some polynomial $q(x) \in \wp ( \mathbb{C} )$ where $\mathrm{deg} (q) = m - 1$ we have that:

\begin{align} \quad p(x) = (x - \lambda_1) q(x) \end{align}

- Since $\mathrm{deg} (q) = m - 1$, then by our induction hypothesis it follows that $q(x)$ can be factored in the form specified earlier, so let $q(x) = c(x - \lambda_2)...(x - \lambda_m)$ and substitute this into the equation above so that we get:

\begin{align} \quad p(x) = c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) \end{align}

- So $S(m)$ is true. Thus by the Principle of Mathematical, $S(m)$ is true for all $m ≥ 1$. We now only need to show that this factorization is unique.

- Suppose that $p(x)$ can be factorized in this form in two different ways, that is for $d, \tau_1, \tau_2, ..., \tau_m \in \mathbb{C}$ we have that:

\begin{align} \quad c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) = d(x - \tau_1)(x - \tau_2)...(x - \tau_m) \end{align}

- We note that $c = d$ as when we expand both sides of these polynomials we get that $c$ and $d$ will both be the coefficient of the term $x^n$ and hence must be equal, therefore:

\begin{align} \quad (x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) = (x - \tau_1)(x - \tau_2)...(x - \tau_m) \end{align}

- Now since $\lambda_1$ makes the lefthand side of the equation equal to zero, then it must also make the righthand side of this equation equal to zero, and so $\lambda_1$ is equal to one of the $\tau$'s. Without loss of generality, suppose that $\lambda_1 = \tau_1$. Thus the term $(x - \lambda_1)$ and $(x - \tau_1)$ are the same, and dividing both sides by this term yields:

\begin{align} \quad (x - \lambda_2)...(x - \lambda_m) = (x - \tau_2)...(x - \tau_m) \end{align}

- We can continue this process with pairing $\lambda_2, \lambda_3, ..., \lambda_m$ with $\tau_2, \tau_3, ..., \tau_m$ which shows that the factorization is unique. $\blacksquare$