Chapter 4 Polynomials

Section 4.A Polynomials

4.12 fundamental theorem of algebra, first version

Every nonconstant polynomial with complex coefficients has a zero in 𝐂.

Proof:

Note, this is just a summary of proof, and I added some steps that are not very clear to me.

We will need the De Moivre’s theorem

$$(\cos \theta + 𝑖 \sin \theta)^𝑘 = \cos 𝑘\theta + 𝑖 \sin 𝑘\theta.$$

We will need another fact: a continuous real-valued function on a closed disk in $𝐑^2$ attains a minimum value.

Assume we have

$$p(z) = c_0 + c_1 z + \cdots + c_m z^m$$

Consider the function $|p(z)|$ when $|z| \rightarrow +\infty$.

$$\begin{align*} |c_m z^m + c_{m-1} z^{m-1} + \cdots + c_1 z^1 + c_0| &= |z^m| | c_m + c_{m-1} z^{-1} + \cdots + c_1 z^{-(m-1)} + c_0 z^{-m}| \\ \end{align*}$$

Note that from we have triangle inequality

$$\begin{align*} &|c_{m-1} z^{-1} + \cdots + c_1 z^{-(m-1)} + z^{-m}| \\ & \leq \frac{|c_{m-1}|}{|z|} + \cdots \frac{|c_1|}{|z^{m-1}|} + \cdots \frac{|c_0|}{|z^{m}|} \\ & \leq |c_m/2| \text{ when |z| is big enough } \end{align*}$$

So when $|z|$ is big enough

$$\begin{align*} & | c_m + c_{m-1} z^{-1} + \cdots + c_1 z^{-(m-1)} + c_0 z^{-m}| \\ & \geq |c_m| - | c_{m-1} z^{-1} + \cdots + c_1 z^{-(m-1)} + z^{-m} | \\ & \geq |c_m| - |c_m / 2| \\ & = |c_m/2| \end{align*}$$

So when $|z| > r$, $|p(z)| \geq |c_m/2||z^m| > |c_m/2| r^m$. Then $|p(z)|$ achieves its minimum in $B(0, r)$.

Normalize $p(z)$ to $q(z) = 1 + a_k z^k + \cdots + a_m z^m$, where $|q(z)|$ achieves its minimum $1$ at $z = 0$.

Let $\beta^k = -1/a_k$, let $t \in (0,1)$ such that $|\beta t| < 1$, then check the value when $z = \beta t$,

$$\begin{align*} & |q(z)| = 1 + a_k z^k + \cdots + a_m z^m \\ & \leq | 1 + a_k z^k| + |a_{k+1} z^{k+1}| + \cdots + |a_m z^m| \\ & \leq | 1 - t^k| + |a_{k+1} z^{k+1}| + \cdots + |a_m z^m| \\ & \leq (1 - t^k) + |a_{k+1} \beta^{k+1}| t^{k+1} + \cdots + |a_m \beta^m|t^m \\ & \leq (1 - t^k) + c t^{k+1} \\ & = 1 - t^k(1 - ct) \\ \end{align*}$$

So if $t < \min \{1/c, 1/|\beta |\}$, where $c = |a_{k+1} \beta^{k+1}| + \cdots + |a_m \beta^m|$, $|q(z)| < 1$.

Then we achive a contradiction, so $\min |p(z)| = 0$.

$\square$

4.13 fundamental theorem of algebra, second version

If $𝑝 ∈ 𝒫(𝐂)$ is a nonconstant polynomial, then $𝑝$ has a unique factorization (except for the order of the factors) of the form

$$p(z) = c (z-\lambda_1) \cdots (z-\lambda_m),$$

where $c, \lambda_1, \cdots, \lambda_m \in C$.

Proof:

The existence part can be derived from induction, 4.6 and 4.12.

The uniqueness is as follows:

Assume

$$(z-\lambda_1) \cdots (z-\lambda_m) = (z-\tau_1) \cdots (z-\tau_m)$$

Since $\lambda_1$ is a zero, then we can find $\tau_i = \lambda_1$. Let $\tau_1 = \lambda_1$.

For all $z \neq \lambda_1$, we have

$$(z-\lambda_2) \cdots (z-\lambda_m) = (z-\tau_2) \cdots (z-\tau_m)$$

If when $z = \lambda_1$, the left and the right side is not equal, then assume

$$\begin{align*} p_1(z) &= (z-\lambda_2) \cdots (z-\lambda_m) \\ p_2(z) &= (z-\tau_2) \cdots (z-\tau_m) \\ \end{align*}$$

$p_1(z) - p_2(z)$ has infinite zeros. This contradicts with 4.8. So $p_1(z) = p_2(z)$ for all $z \in C$.

Then we can continue with induction.

$\square$

4.16 factorization of a polynomial over 𝐑

Suppose $𝑝 ∈ 𝒫(𝐑)$ is a nonconstant polynomial. Then $𝑝$ has a unique factorization (except for the order of the factors) of the form

$$𝑝(𝑥) = 𝑐(𝑥 − \lambda_1)⋯(𝑥 − \lambda_𝑚) (𝑥^2 + 𝑏_1𝑥 + 𝑐_1)⋯(𝑥^2 + 𝑏_𝑀𝑥 + 𝑐_𝑀),$$

Proof summary:

If $\lambda$ is a zero then $\bar{\lambda}$ is also a root, assume $(x - \lambda)(x - \bar{\lambda}) = x^2 + bx + c$, then $b, c \in \mathbb{R}$.

Let $p(x) = (x^2 + bx + c) q(x)$, so $q(x) = \frac{p(x)}{(x^2 + bx + c)}$.

For any $x \in \mathbb{R}$, $q(x) \in \mathbb{R}$.

Let

$$q(x) = a_0 + a_1 x + \cdots + a_{m} x^{m}$$

Then

$$0 = \Im q(x) = \Im a_0 + \Im a_1 x + \cdots + \Im a_{m} x^{m}$$

have infinite zeros. So $\Im a_0 = \Im a_1 = \cdots = \Im a_{m} = 0$

$\square$