Characteristic Equations with Complex Roots

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 4.3 – Auxiliary Equations with Complex Roots

Expected Educational Results

• Objective 11–1: I can identify complex solutions to the characteristic equation for $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear ODEs.

• Objective 11–2: I understand the form of the solution to an $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear ODE for complex roots to the characteristic equation.

• Objective 11–3: I can find the most general solution to $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear ODEs.

• Objective 11–4: I can find the solution to $n^{\text{th}}$$n^{\text{th}}$-degree homogeneous linear IVPs.

Algebra

Theorem: Fundamental Theorem of Algebra

A polynomial of degree $n$$n$, $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_2{x}^2+a_{1}x+a_0$$P(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0$, $a_n\ne 0$$a_n\ne 0$, has exactly $n$$n$ factors, counting multiplicities.

Corollary

By Fundamental Theorem of Algebra, $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_2{x}^2+a_{1}x+a_0=0$$P(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0=0$, where $a_n\ne 0$$a_n\ne 0$, has exactly $n$$n$ solutions.

Solving Polynomial Equations

Distributive Property and Zero Product Property of Complex Numbers

*Example 01

$\begin{array}{c}{x}^3+x=0\Rightarrow x(x^2+1)=0\\ \Rightarrow x=0\text{ or }{x}^2+1=0\\ \Rightarrow x=0\text{ or }{x}^2=-1\\ \Rightarrow x=0\text{ or }x=-i\text{ or }x=i\end{array}$$x^3+x=0 \Rightarrow x(x^2+1)=0\\ \Rightarrow x=0 \text{ or } x^2+1=0\\ \Rightarrow x=0 \text{ or } x^2=-1\\ \Rightarrow x=0 \text{ or } x=-i \text{ or } x=i$

Quadratic Equations

Quadratic Formula

$a{x}^2+bx+c=0\Rightarrow x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$ax^2+bx+c=0 \Rightarrow x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Discriminant

The discriminant of a quadratic equation $a{x}^2+bx+c=0$$ax^2+bx+c=0$ is $b^2-4ac$$b^2-4ac$.

The discriminant is used to determine the types of roots to a quadratic equation:

• If ${\displaystyle b^2-4ac>0}$$\displaystyle b^2-4ac>0$, the the roots, $r_1,r_2$$r_1, r_2$ are distinct (i.e., different) real numbers;

• If ${\displaystyle b^2-4ac=0}$$\displaystyle b^2-4ac=0$, the the roots, $r_1=r_2$$r_1 = r_2$ is a repeated (i.e., multiplicity 2) real number;

• If ${\displaystyle b^2-4ac<0}$$\displaystyle b^2-4ac<0$, the the roots, $r_1,r_2$$r_1, r_2$ conjugate pair of complex numbers.

Example 02

$\begin{array}{c}{x}^2-x+1=0\Rightarrow x={\displaystyle \frac{-(-1)\pm \sqrt{(-1{)}^2-4(1)(1)}}{2(1)}}\\ \Rightarrow x={\displaystyle \frac{1\pm \sqrt{1-4}}{2}}\\ \Rightarrow x={\displaystyle \frac{1\pm \sqrt{-3}}{2}}\\ \Rightarrow x={\displaystyle \frac{1-i\sqrt{3}}{2}}\text{ or }x={\displaystyle \frac{1+i\sqrt{3}}{2}}\end{array}$$x^2-x+1=0 \Rightarrow x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(1)}}{2(1)}\\ \Rightarrow x=\dfrac{1\pm\sqrt{1-4}}{2}\\ \Rightarrow x=\dfrac{1\pm\sqrt{-3}}{2}\\ \Rightarrow x=\dfrac{1-i\sqrt{3}}{2} \text{ or } x=\dfrac{1+i\sqrt{3}}{2}$

Quadratic-Like Equations

A quadratic-like equation has the form: ${\displaystyle a{x}^{2n}+b{x}^n+c=0,n\in \mathbb{R}}$$\displaystyle ax^{2n}+bx^{n}+c=0, n\in\mathbb{R}$.

Method to Solve Quadratic-Like Equations

1. Let ${\displaystyle u=x^n}$$\displaystyle u=x^{n}$ and substitute into the quadratic-like equation;

2. Factor ${\displaystyle a{u}^2+bu+c=0\Rightarrow (u+s)(u+t)=0}$$\displaystyle au^2+bu+c=0 \Rightarrow (u+s)(u+t)=0$, where $(s)(t)=c$$(s)(t)=c$ and $s+t=b$$s+t=b$;

3. Solve in terms of $u$$u$: ${\displaystyle (u+s)(u+t)=0\Rightarrow (u+s)(u+t)=0\Rightarrow u=-s\text{ or }u=-t}$$\displaystyle (u+s)(u+t)=0 \Rightarrow (u+s)(u+t)=0 \Rightarrow u=-s \text{ or } u=-t$;

4. Solve in terms of $x$$x$: ${\displaystyle u=-s\Rightarrow x^n=-s\Rightarrow x=(-s{)}^{\frac{1}{n}}}$$\displaystyle u=-s \Rightarrow x^{n}=-s \Rightarrow x=(-s)^{\frac{1}{n}}$\newline ${\displaystyle u=-t\Rightarrow x^n=-t\Rightarrow x=(-t{)}^{\frac{1}{n}}}$$\displaystyle u=-t \Rightarrow x^{n}=-t \Rightarrow x=(-t)^{\frac{1}{n}}$

Example 03

$\begin{array}{c}{x}^4+3x^2+4=0\text{ Let }u=x^2\\ \Rightarrow u^2+3u+4=0\\ \Rightarrow u=\frac{-3\pm \sqrt{3^2-4(1)(4)}}{2(1)}\\ \Rightarrow u=\frac{-3\pm \sqrt{9-16}}{2(1)}\\ \Rightarrow u=\frac{-3-i\sqrt{5}}{2}\text{ or }u=\frac{-3+i\sqrt{5}}{2}\\ \Rightarrow x^2=\frac{-3-i\sqrt{5}}{2}\text{ or }{x}^2=\frac{-3+i\sqrt{5}}{2}\\ \Rightarrow x=\sqrt{\frac{-3-i\sqrt{5}}{2}}\text{ or }x=\sqrt{\frac{-3+i\sqrt{5}}{2}}\end{array}$$x^4+3x^2+4=0 \text{ Let } u=x^2 \\ \Rightarrow u^2+3u+4=0 \\ \Rightarrow u=\frac{-3\pm\sqrt{3^2-4(1)(4)}}{2(1)}\\ \Rightarrow u=\frac{-3\pm\sqrt{9-16}}{2(1)}\\ \Rightarrow u=\frac{-3-i\sqrt{5}}{2} \text{ or } u=\frac{-3+i\sqrt{5}}{2} \\ \Rightarrow x^2=\frac{-3-i\sqrt{5}}{2} \text{ or } x^2=\frac{-3+i\sqrt{5}}{2} \\ \Rightarrow x=\sqrt{\frac{-3-i\sqrt{5}}{2}} \text{ or } x=\sqrt{\frac{-3+i\sqrt{5}}{2}}$

In order to write the solutions in standard cartesian form, we will need to use technology. For example, in Mathematica:

​x
1
(* real part of sqrt((-3-i*sqrt(5))\2) *)
2
Re[Sqrt[(-3-I*Sqrt[5])\2]
3

4
(* imaginary part of sqrt((-3-i*sqrt(5))\2) *)
5
Im[Sqrt[(-3-I*Sqrt[5])\2]

Difference of Two Cubes

${\displaystyle a^3{x}^3-b^3=0\Rightarrow (ax-b)(a^2{x}^2+abx+b^2)=0}$$\displaystyle a^3x^3-b^3 = 0 \Rightarrow \left(ax-b\right)\left(a^2x^2+abx+b^2\right)=0$

Example 06

$\begin{array}{c}8x^3-1=0\\ \Rightarrow 2^3{x}^3-1^3=0\\ \Rightarrow (2x-1)(2^2{x}^2+(2)(1)x+1^2)=0\\ \Rightarrow (2x-1)(4x^2+2x+1)=0\\ \Rightarrow 2x-1=0\text{ or }4x^2+2x+1=0\\ \Rightarrow x={\displaystyle \frac{1}{2}}\text{ or }x={\displaystyle \frac{-1-i\sqrt{3}}{4}}\text{ or }x={\displaystyle \frac{-1+i\sqrt{3}}{4}}\end{array}$$8x^3-1=0 \\ \Rightarrow 2^3x^3-1^3=0 \\ \Rightarrow \left(2x-1\right)\left(2^2x^2+(2)(1)x+1^2\right)=0\\ \Rightarrow \left(2x-1\right)\left(4x^2+2x+1\right)=0\\ \Rightarrow 2x-1=0 \text{ or } 4x^2+2x+1=0 \\ \Rightarrow x=\dfrac{1}{2} \text{ or } x=\dfrac{-1-i\sqrt{3}}{4} \text{ or } x=\dfrac{-1+i\sqrt{3}}{4}$

Sum of Two Cubes

${\displaystyle a^3{x}^3+b^3=0\Rightarrow (ax+b)(a^2{x}^2-abx+b^2)=0}$$\displaystyle a^3x^3+b^3 = 0 \Rightarrow \left(ax+b\right)\left(a^2x^2-abx+b^2\right)=0$

Finding Zeros Using Rational Root Theorem and Polynomial or Synthetic Division

Given the polynomial $a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_2{x}^2+a_{1}x+a_0=0,a_n\ne 0$$a_nx^n+a_{n-1}x^{n-1}+\cdots + a_2x^2+a_1x+a_0=0,a_n\ne 0$

Let $p_1,p_2,p_3,\dots$$p_1,p_2,p_3,\ldots$ be the factors of $a_0$$a_0$. Let $q_1,q_2,q_3,\dots$$q_1,q_2,q_3,\ldots$ be the factors of $a_n$$a_n$. Then the set of potential rational roots of the polynomial are:

$\{{\displaystyle \frac{p_1}{q_1}},-{\displaystyle \frac{p_1}{q_1}},{\displaystyle \frac{p_1}{q_2}},-{\displaystyle \frac{p_1}{q_2}},{\displaystyle \frac{p_2}{q_1}},-{\displaystyle \frac{p_2}{q_1}},\dots \}$$\left\{\dfrac{p_1}{q_1},-\dfrac{p_1}{q_1},\dfrac{p_1}{q_2},-\dfrac{p_1}{q_2},\dfrac{p_2}{q_1},-\dfrac{p_2}{q_1},\ldots\right\}$

Example 04

Find all three real roots of: $x^4-2x^3+3x^2-2x+2=0$$x^4-2x^3+3x^2-2x+2=0$.

The factors of $a_0=2$$a_0=2$ are $p=-1,1,-2,2$$p=-1,1,-2,2$.

The factors of $a_4=1$$a_4=1$ are $q=-1,1$$q=-1,1$.

Then the set of distinct potential rational roots of the polynomial are:

$\{{\displaystyle \frac{-1}{-1}},{\displaystyle \frac{-1}{1}},{\displaystyle \frac{-2}{-1}},{\displaystyle \frac{-2}{1}}\}$$\left\{\dfrac{-1}{-1},\dfrac{-1}{1},\dfrac{-2}{-1},\dfrac{-2}{1}\right\}$

Now use polynomial or synthetic division to find the three factors.

Using Polynomial Division

a. Divide the polynomial by $x-\left(\text{ rational root}\right)$$x-\left(\text{ rational root}\right)$, i.e., divide by $x-(-1)$$x-(-1)$:

${\displaystyle \frac{x^4-2x^3+3x^2-2x+2}{x+1}}$$\dfrac{x^4-2x^3+3x^2-2x+2}{x+1}$

Since the above polynomial division has a non-zero remainder, then $x+1$$x+1$ is not a factor of $x^4-2x^3+3x^2-2x+2=0$$x^4-2x^3+3x^2-2x+2=0$.

The same is true for all the other potential rational roots. Thus, there are no real solutions to the polynomial equation.

Since all roots are non-real numbers, we will need a method for solving a quartic ($4^{\text{th}}$$4^{\text{th}}$-degree polynomial) equation with [all] four complex roots which you probably do not have experience.

NOTE: Potential methods for solving polynomial equations that have only complex roots:

1. Substitute $x=y-\frac{2}{4}$$x=y-\frac{2}{4}$, where the 2 is the coefficient of the cubic term to reduce the quartic equation to a cubic equation;

2. Muller's method;

3. Numerical methods to approximate the complex roots;

4. If one of the complex roots is known, then polynomial or synthetic division can be used to reduce the degree of the polynomial.

Thus, we will will use technology to compute all roots of polynomial equations.

Practice 01

Solve, i.e., find all solutions, to the following equations:

1. ${\displaystyle x^2+3x=0}$$\displaystyle x^2+3x=0$

2. ${\displaystyle x^2-8=0}$$\displaystyle x^2-8=0$

3. ${\displaystyle x^2+4=0}$$\displaystyle x^2+4=0$

4. ${\displaystyle x^2-5x+8=0}$$\displaystyle x^2-5x+8=0$

5. ${\displaystyle x^2+2x=-4}$$\displaystyle x^2+2x=-4$

6. ${\displaystyle x^3+3x^2+2=0}$$\displaystyle x^3+3x^2+2=0$

7. ${\displaystyle x^4-5x^3-6x^2=0}$$\displaystyle x^4-5x^3-6x^2=0$

8. ${\displaystyle x^3-8=0}$$\displaystyle x^3-8=0$

9. ${\displaystyle x^3=-8}$$\displaystyle x^3=-8$

10. ${\displaystyle x^4+12x=0}$$\displaystyle x^4+12x=0$

11. ${\displaystyle x^4-9=0}$$\displaystyle x^4-9=0$

12. ${\displaystyle 2x^3-x^2-5x=3}$$\displaystyle 2x^3-x^2-5x=3$

13. ${\displaystyle x^3-7x-6=0}$$\displaystyle x^3-7x-6=0$

14. ${\displaystyle x^4+4x^3+3x^2-4x-4=0}$$\displaystyle x^4+4x^3+3x^2-4x-4=0$

15. ${\displaystyle x^4+5x^3+9x^2+7x+2=0}$$\displaystyle x^4+5x^3+9x^2+7x+2=0$

Determinants

See: PRE_10_Homogeneous_Equations.html.

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Last Modified: Monday, 6 September 2020 13:33 EDT