3.5: Complex Zeros

First, some notation. The symbol $\mathbb{C}$ means the set of all complex numbers, similar to $\mathbb{R}$.

Motivation

Question: Can we factor irreducibles?

Answer: yes, if you involve complex numbers.

Factor $P(x) = x^2 + 1$ into linear factors.

$c$ is a zero if

  • $(x - c)$ is a factor of $x^2 + 1$ and
  • $x = c$ is a solution to $x^2 + 1 = 0$.

Solving the equation: \begin{align} x^2 + 1 &= 0\\ x^2 &= -1\\ \sqrt{x^2} &= \pm \sqrt{-1}\\ x &= \pm i \end{align}

The solutions are $x = i$ and $x = -i$.

Therefore $x^2 + 1 = (x - i)(x - (-i)) = \boxed{(x - i)(x + i)}$

We just saw an irreducible $x^2 + 1$ break down into complex zeros.

Zeros must always exist if your resultant factors are allowed to have complex numbers (called a factorization over $\mathbb{C}$).

This is exactly what the next theorem says:

Fundamental Theorem of Algebra
Every polynomial \[P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \qquad a_n \neq 0\] with complex coefficients has at least one complex zero.

Recall the Real Factorization Theorem:\[P(x) = (\text{linear factors}) \cdot (\text{irreducible factors})\]

Breaking down the irreducibles into linear factors gives you a complete factorization over $\mathbb{C}$:

Complex Factorization Theorem
Suppose $P(x)$ is a degree $n$ polynomial with complex coefficients. Then:
  1. there exists complex numbers $a, c_1, c_2, \dots, c_n$ where \[P(x) = a(x-c_1)(x-c_2)\cdots(x-c_{n-1})(x-c_{n})\] and
  2. $P(x)$ has exactly $n$ zeros, provided a zero of multiplicity $k$ is counted $k$ times.
Suppose $P(x) = x^3 - 3x^2 + x - 3$.
  1. How many zeros are there, counting multiplicity?
  2. Find a complete factorization over $\mathbb{R}$.
  3. Find a complete factorization over $\mathbb{C}$.
Suppose $P(x) = x^3 - 2x + 4$, and you know that $P(-2) = 0$.
  1. Find a complete factorization over $\mathbb{R}$.
  2. Find a complete factorization over $\mathbb{C}$.

Next example looks at multiplicity.

Suppose $P(x) = 3x^5 + 12x^3 + 12x$.
  1. How many zeros are there, counting multiplicity?
    The degree is five, so five zeros.
  2. Find a complete factorization over $\mathbb{R}$.

    We're looking for linear factors and irreducibles.

    No zero was given, so let's use Chapter 1 methods. Use GCF first, pull out $3x$:

    \begin{align} 3x^5 + 12x^3 + 12x &= 3x (x^4 + 4x^2 + 4) && \text{GCF} \\&= 3x\bigg(\big(x^2\big)^2 + 4x^2 + 4\bigg) && \text{Rewriting to substitute $y = x^2$} \\&= 3x\bigg(y^2 + 4y + 4\bigg) && \text{Substitute $y = x^2$} \\&= 3x(y+2)^2 && \text{Factoring quadratic} \\&= 3x(x^2+2)^2 && \text{Backsubstitute } y = x^2 \end{align}

    Now we need to check if the quadratic $(x^2 + 2)$ is irreducible or not.

    We have $a = 1, b = 0, c = 2$, so \begin{align} b^2 - 4ac &= 0^2 - 4 \cdot 1 \cdot 2 \\&= -8 \\&< 0 \end{align} $x^2 + 2$ is irreducible. No zeros over $\mathbb{R}$.

    A complete factorization over $\mathbb{R}$ is $\boxed{P(x) = 3x(x^2+2)^2}$

  3. Find a complete factorization over $\mathbb{C}$.

    We know a CF over $\mathbb{C}$ just means to break down the irreducibles into linear factors from the CF over $\mathbb{R}$.

    From part 2, we saw $x^2 + 2$ was the only irreducible. So we need to factor it.

    Remember that $c$ is a zero if

    • $(x - c)$ is a factor of $x^2 + 2$ and
    • $x = c$ is a solution to $x^2 + 2 = 0$.

    Solving the equation: \begin{align} x^2 + 2 &= 0\\ x^2 &= -2\\ \sqrt{x^2} &= \pm \sqrt{-2}\\ x &= \pm i\sqrt{2} \end{align}

    The solutions are $x = i\sqrt{2}$ and $x = -i\sqrt{2}$.

    Therefore $x^2 + 2 = (x - i\sqrt{2})(x - (-i\sqrt{2})) = (x - i\sqrt{2})(x + i\sqrt{2})$

    Now substitute into the CF over $\mathbb{R}$ and simplify: \begin{align} P(x) &= 3x(x^2 + 2)^2 \\&= 3x\bigg[(x - i\sqrt{2})(x + i\sqrt{2})\bigg]^2 && \text{Substituting} \\&= \boxed{3x(x-i\sqrt{2})^2(x+i\sqrt{2})^2} && \text{Using } (ab)^2 = a^2b^2 \end{align}

    Notice that \[P(x) = \underbrace{3x}_{\text{multiplicity } 1} \cdot \underbrace{(x-i\sqrt{2})^2}_{\text{multiplicity } 2}\cdot \underbrace{(x+i\sqrt{2})^2}_{\text{multiplicity } 2}\]

    Adding up the multiplicities $1 + 2 + 2 = 5$, which is the degree of $P(x)$!

Complex Zeros Come in Pairs

complex conjugate
The complex conjugate of $a + bi$ is $a - bi$.

In other words, the imaginary part sign was flipped.

Find the complex conjugate of $-3 - 2i$.
Find the complex conjugate of $\dfrac{1 + i\sqrt{5}}{2}$.
Conjugate Pairs Theorem
If $P(x)$ has real coefficients and $a + bi$ is a zero, then $a - bi$ must also be a zero.

Why?

The irreducibles are the only source of complex zeros.

Irreducibles are quadratics with $b^2 - 4ac < 0$, so the only source of complex zeros are from the quadratic formula \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] in particular, the $\pm \sqrt{b^2 - 4ac}$.

This $\pm$ creates conjugate pairs!

Give an example of a degree 3 polynomial with zeros $\frac{1}{2}$ and $3 - i$. Leave factored.