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 ${\displaystyle 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 $\bar{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 ${\displaystyle z_1{z}_2}$$\displaystyle z_1z_2$ is defined as

$\begin{array}{c}{z}_1{z}_2=(a_1+b_{1}i)(a_2+b_{2}i)\\ \phantom{xxx}=a_1{a}_2+a_1{b}_{2}i+a_2{b}_{1}i+(b_{1}i)(b_{2}i)\\ \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+(a_1{b}_2+a_2{b}_1)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

${\displaystyle (3-2i)(1+4i)=3+(-8i^2)+(12i-2i)=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 ${\displaystyle \frac{z_1}{z_2}}$$\displaystyle\frac{z_1}{z_2}$ is defined as ${\displaystyle \frac{z_1}{z_2}\left(\frac{\overline{z_2}}{\overline{z_2}}\right)}$$\displaystyle\frac{z_1}{z_2}\left(\frac{\overline{z_2}}{\overline{z_2}}\right)$

Example 06

${\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}}$$\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

${\displaystyle \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}=\cos (x)+i\sin (x)$$e^{i x}=\cos{(x)} +i\sin{(x)}$.

Investigation 02

Prove Euler's Formula using the following Power Series:

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

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

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

${\displaystyle {[\cos (x)+i\sin (x)]}^n=\cos (nx)+i\sin (nx)}$$\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: ${\displaystyle \sin (2x)}$$\displaystyle \sin{(2x)}$.

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

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

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

5. Evaluate: ${\displaystyle \int {\cos }^3(x)dx}$$\displaystyle \int{\cos^3{(x)}\,dx}$.

Investigation 06

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

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

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

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

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

6. Evaluate: ${\displaystyle \int \sin (4x)\cos (5x)dx}$$\displaystyle \int{\sin{(4x)}\cos{(5x)}\,dx}$

Investigation 07

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

2. Use the above to evaluate: ${\displaystyle \int e^x\cos (2x)dx}$$\displaystyle \int{e^x\cos{(2x)}\,dx}$

Hyperbolic Functions

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

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

• ${\displaystyle \tanh (x)=\frac{\sinh (x)}{\cosh (x)}=\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}}$

• ${\displaystyle \coth (x)=\frac{\cosh (x)}{\sinh (x)}=\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$

• ${\displaystyle \text{csch}(x)=\frac{1}{\sinh (x)}=\frac{2}{e^x-e^{-x}}}$$\displaystyle \text{csch}\,{(x)}=\frac{1}{\sinh{(x)}}=\frac{2}{e^x-e^{-x}}$

• ${\displaystyle \text{sech}(x)=\frac{1}{\cosh (x)}=\frac{2}{e^x+e^{-x}}}$$\displaystyle \text{sech}\,{(x)}=\frac{1}{\cosh{(x)}}=\frac{2}{e^x+e^{-x}}$

Investigation 08

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

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

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

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

5. Show ${\displaystyle \frac{d}{dx}\tanh (x)={\text{sech}}^2(x)}$$\displaystyle \frac{d}{dx}\tanh{(x)}=\text{sech}^2\,{(x)}$.

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