# 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.

## Complex Numbers

Definition: Complex Number

A complex number is a number that can be displayed in the form $a+bi$$\displaystyle a+bi$. The real part of complex number is $a\in \mathbb{R}$$a\in\mathbb{R}$ and the imaginary part of complex number is $b\in \mathbb{R}$$b\in\mathbb{R}$. The imaginary unit $i$$i$ satisfies ${i}^{2}=-1$$i^2=-1$.

### Conjugate Pairs of Complex Numbers

#### Definition: Complex Conjugate Pairs

Let $z=a+bi$$z=a+bi$, then the complex conjugate of $z$$z$ is $\stackrel{―}{z}={z}^{\ast }=a-bi$$\overline{z}=z^{\ast}=a-bi$.

#### Multiplication of Imaginary Units

• ${i}^{2}=-1$$i^2=-1$

• ${i}^{3}=-i$$i^3=-i$

• ${i}^{4}=1$$i^4=1$

• ${i}^{5}=i$$i^5=i$

• etc.

#### Multiplication of Complex Numbers

Let ${z}_{1}={a}_{1}+{b}_{1}i$$z_1=a_1+b_1i$ and ${z}_{2}={a}_{2}+{b}_{2}i$$z_2=a_2+b_2i$, then the product ${z}_{1}{z}_{2}$$\displaystyle z_1z_2$ is defined as

$\begin{array}{c}{z}_{1}{z}_{2}=\left({a}_{1}+{b}_{1}i\right)\left({a}_{2}+{b}_{2}i\right)\\ \phantom{xxx}={a}_{1}{a}_{2}+{a}_{1}{b}_{2}i+{a}_{2}{b}_{1}i+\left({b}_{1}i\right)\left({b}_{2}i\right)\\ \phantom{xxx}={a}_{1}{a}_{2}+{a}_{1}{b}_{2}i+{a}_{2}{b}_{1}i+{b}_{1}{b}_{2}{i}^{2}\\ \phantom{xxx}={a}_{1}{a}_{2}-{b}_{1}{b}_{2}+\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)i\end{array}$$z_1z_2=(a_1+b_1i)(a_2+b_2i) \\ \phantom{xxx} = a_1a_2+a_1b_2i+a_2b_1i+(b_1i)(b_2i) \\ \phantom{xxx} = a_1a_2+a_1b_2i+a_2b_1i+b_1b_2i^2 \\ \phantom{xxx} = a_1a_2-b_1b_2+(a_1b_2+a_2b_1)i$

Example 05

$\left(3-2i\right)\left(1+4i\right)=3+\left(-8{i}^{2}\right)+\left(12i-2i\right)=11+10i$$\displaystyle (3-2i)(1+4i)=3+(-8i^2)+(12i-2i)=11+10i$

#### Division of Complex Numbers

Let ${z}_{1}={a}_{1}+{b}_{1}i$$z_1=a_1+b_1i$ and ${z}_{2}={a}_{2}+{b}_{2}i$$z_2=a_2+b_2i$, then the division $\frac{{z}_{1}}{{z}_{2}}$$\displaystyle\frac{z_1}{z_2}$ is defined as $\frac{{z}_{1}}{{z}_{2}}\left(\frac{\stackrel{―}{{z}_{2}}}{\stackrel{―}{{z}_{2}}}\right)$$\displaystyle\frac{z_1}{z_2}\left(\frac{\overline{z_2}}{\overline{z_2}}\right)$

Example 06

$\frac{1}{1+2i}=\frac{1}{1+2i}\left(\frac{1-2i}{1-2i}\right)=\frac{1-2i}{5}=\frac{1}{5}-\frac{2i}{5}$$\displaystyle \frac{1}{1+2i}=\frac{1}{1+2i}\left(\frac{1-2i}{1-2i}\right)=\frac{1-2i}{5}=\frac{1}{5}-\frac{2i}{5}$

#### Reciprocals of Imaginary Units

$\frac{1}{i}=\frac{1}{i}\left(\frac{i}{i}\right)=\frac{i}{-1}=-i$$\displaystyle \frac{1}{i}=\frac{1}{i}\left(\frac{i}{i}\right)=\frac{i}{-1}=-i$

## Trigonometry

### Euler's Formula

Definition: Euler's Formula

${e}^{ix}=\mathrm{cos}\left(x\right)+i\mathrm{sin}\left(x\right)$$e^{i x}=\cos{(x)} +i\sin{(x)}$.

#### Investigation 02

Prove Euler's Formula using the following Power Series:

1. ${e}^{x}=\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{n}}{n!}$$\displaystyle e^{x}=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$

2. $\mathrm{cos}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!}$$\displaystyle \cos{(x)}=\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n}}{(2n)!}}$

3. $\mathrm{sin}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)!}$$\displaystyle \sin{(x)}=\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n+1}}{(2n+1)!}}$

#### Investigation 03

Evaluate ${e}^{i\pi }$$e^{i\pi}$.

Here is a poem written about ${e}^{i\pi }$$e^{i\pi}$: http://www.3blue1brown.com/poems/epii.

Here is the poem being read by the author of the poem: https://www.youtube.com/watch?v=zLzLxVeqdQg.

### de Moivre's Formula

#### Definition: de Moivre's Formula

${\left[\mathrm{cos}\left(x\right)+i\mathrm{sin}\left(x\right)\right]}^{n}=\mathrm{cos}\left(nx\right)+i\mathrm{sin}\left(nx\right)$$\displaystyle \left[\cos{(x)} +i\sin{(x)}\right]^n=\cos{(nx)}+i\sin{(nx)}$.

#### Investigation 04

Prove de Moivre's Formula.

#### Investigation 05

1. Use de Moivre's formula to find an equivalent expression for: $\mathrm{sin}\left(2x\right)$$\displaystyle \sin{(2x)}$.

2. Use de Moivre's formula to find an equivalent expression for: $\mathrm{cos}\left(2x\right)$$\displaystyle \cos{(2x)}$.

3. Use de Moivre's formula to find an equivalent expression, in terms of cosine only, for: $\mathrm{cos}\left(3x\right)$$\displaystyle \cos{(3x)}$.

4. Use de Moivre's formula to find an equivalent expression, in terms of cosine only, for: ${\mathrm{cos}}^{3}\left(x\right)$$\displaystyle \cos^3{(x)}$.

5. Evaluate: $\int {\mathrm{cos}}^{3}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$$\displaystyle \int{\cos^3{(x)}\,dx}$.

#### Investigation 06

1. Prove: $\mathrm{cos}\left(x\right)=\frac{{e}^{ix}+{e}^{-ix}}{2}$$\displaystyle \cos{(x)}=\frac{e^{ix}+e^{-ix}}{2}$.

2. Prove: $\mathrm{sin}\left(x\right)=\frac{{e}^{ix}-{e}^{-ix}}{2i}$$\displaystyle \sin{(x)}=\frac{e^{ix}-e^{-ix}}{2i}$.

3. Find an equivalent expression for: $\mathrm{sin}\left(x+y\right)$$\displaystyle \sin{(x+y)}$.

4. Find an equivalent expression for: $\mathrm{cos}\left(x+y\right)$$\displaystyle \cos{(x+y)}$.

5. Find an equivalent expression for: $\mathrm{sin}\left(x-y\right)$$\displaystyle \sin{(x-y)}$.

6. Evaluate: $\int \mathrm{sin}\left(4x\right)\mathrm{cos}\left(5x\right)\phantom{\rule{0.167em}{0ex}}dx$$\displaystyle \int{\sin{(4x)}\cos{(5x)}\,dx}$

#### Investigation 07

1. Rewrite ${e}^{\left(1+2i\right)x}$$\displaystyle e^{(1+2i)x}$ in terms of ${e}^{x}$$\displaystyle e^x$, $\mathrm{cos}\left(x\right)$$\displaystyle \cos{(x)}$, and $\mathrm{sin}\left(x\right)$$\displaystyle \sin{(x)}$.

2. Use the above to evaluate: $\int {e}^{x}\mathrm{cos}\left(2x\right)\phantom{\rule{0.167em}{0ex}}dx$$\displaystyle \int{e^x\cos{(2x)}\,dx}$

### Hyperbolic Functions

• $\mathrm{sinh}\left(x\right)=\frac{{e}^{x}-{e}^{-x}}{2}$$\displaystyle \sinh{(x)}=\frac{e^x-e^{-x}}{2}$

• $\mathrm{cosh}\left(x\right)=\frac{{e}^{x}+{e}^{-x}}{2}$$\displaystyle \cosh{(x)}=\frac{e^x+e^{-x}}{2}$

• $\mathrm{tanh}\left(x\right)=\frac{\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}=\frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}$$\displaystyle \tanh{(x)}=\frac{\sinh{(x)}}{\cosh{(x)}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

• $\mathrm{coth}\left(x\right)=\frac{\mathrm{cosh}\left(x\right)}{\mathrm{sinh}\left(x\right)}=\frac{{e}^{x}+{e}^{-x}}{{e}^{x}-{e}^{-x}}$$\displaystyle \coth{(x)}=\frac{\cosh{(x)}}{\sinh{(x)}}=\frac{e^x+e^{-x}}{e^x-e^{-x}}$, $x\ne 0$$x\ne 0$

• $\text{csch}\phantom{\rule{0.167em}{0ex}}\left(x\right)=\frac{1}{\mathrm{sinh}\left(x\right)}=\frac{2}{{e}^{x}-{e}^{-x}}$$\displaystyle \text{csch}\,{(x)}=\frac{1}{\sinh{(x)}}=\frac{2}{e^x-e^{-x}}$

• $\text{sech}\phantom{\rule{0.167em}{0ex}}\left(x\right)=\frac{1}{\mathrm{cosh}\left(x\right)}=\frac{2}{{e}^{x}+{e}^{-x}}$$\displaystyle \text{sech}\,{(x)}=\frac{1}{\cosh{(x)}}=\frac{2}{e^x+e^{-x}}$

#### Investigation 08

1. Show ${e}^{x}=\mathrm{cosh}\left(x\right)+\mathrm{sinh}\left(x\right)$$\displaystyle e^x=\cosh{(x)}+\sinh{(x)}$.

2. Show ${e}^{-x}=\mathrm{cosh}\left(x\right)-\mathrm{sinh}\left(x\right)$$\displaystyle e^{-x}=\cosh{(x)}-\sinh{(x)}$.

3. Show $\frac{d}{dx}\mathrm{cosh}\left(x\right)=\mathrm{sinh}\left(x\right)$$\displaystyle \frac{d}{dx}\cosh{(x)}=\sinh{(x)}$.

4. Show $\frac{d}{dx}\mathrm{sinh}\left(x\right)=\mathrm{cosh}\left(x\right)$$\displaystyle \frac{d}{dx}\sinh{(x)}=\cosh{(x)}$.

5. Show $\frac{d}{dx}\mathrm{tanh}\left(x\right)={\text{sech}}^{2}\phantom{\rule{0.167em}{0ex}}\left(x\right)$$\displaystyle \frac{d}{dx}\tanh{(x)}=\text{sech}^2\,{(x)}$.