<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation
Our Terms of Use (click here to view) and Privacy Policy (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use and Privacy Policy.

6.14: Equations Using DeMoivre's Theorem

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
[object Object]%
Progress
Try again
Progress
[object Object]%
Retry

You are given an equation in math class:

\begin{align*}x^4 = 16\end{align*}

and asked to solve for "x". "Excellent!" you say. "This should be easy. The answer is 2."

"Not quite so fast," says your instructor.

"I want you to find the complex roots as well!"

Can you do this? Read on, and by the end of this Concept, you'll be able to solve equations to find complex roots.

Watch This

DeMoivres Theorem: Find all roots. Real and Imaginary.

Guidance

We've already seen equations that we would like to solve. However, up until now, these equations have involved solutions that were real numbers. However, there is no reason that solutions need to be limited to the real number line. In fact, some equations cannot be solved completely without the use of complex numbers. Here we'll explore a little more about complex numbers as solutions to equations.

The roots of a complex number are cyclic in nature. This means that when the roots are plotted on the complex plane, the \begin{align*}n^{th}\end{align*} roots are equally spaced on the circumference of a circle.

Since you began Algebra, solving equations has been an extensive topic. Now we will extend the rules to include complex numbers. The easiest way to explore the process is to actually solve an equation. The solution can be obtained by using De Moivre’s Theorem.

Example A

Consider the equation \begin{align*}x^5 - 32 = 0\end{align*}. The solution is the same as the solution of \begin{align*}x^5 = 32\end{align*}. In other words, we must determine the fifth roots of 32.

Solution:

\begin{align*}x^5-32 &= 0 \ \text{and} \ x^5=32.\\ r &= \sqrt{x^2+y^2}\\ r &= \sqrt{(32)^2+(0)^2}\\ r &= 32\\ \theta &= \tan^{-1} \left(\frac{0}{32}\right)=0\end{align*}

Write an expression for determining the fifth roots of \begin{align*}32 = 32 + 0i\end{align*}

\begin{align*}32^{\frac{1}{5}} &= [32( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{5}}\\ &= 2\left(\cos \frac{2\pi k}{5}+i \sin \frac{2\pi k}{5}\right)k=0, 1, 2, 3, 4\\ x_1 &= 2\left(\cos \frac{0}{5}+i \sin \frac{0}{5}\right) \rightarrow 2(\cos 0+i \sin 0)=2 && for \ k=0\\ x_2 &= 2\left(\cos \frac{2\pi}{5}+i \sin \frac{2\pi}{5}\right) \approx 0.62 + 1.9i && for \ k=1\\ x_3 &= 2\left(\cos \frac{4\pi}{5}+i \sin \frac{4\pi}{5}\right) \approx -1.62 + 1.18i && for \ k=2\\ x_4 &= 2\left(\cos \frac{6\pi}{5}+i \sin \frac{6\pi}{5}\right) \approx -1.62-1.18i && for \ k=3\\ x_5 &= 2\left(\cos \frac{8\pi}{5}+i \sin \frac{8\pi}{5}\right) \approx 0.62-1.9i && for \ k=4\end{align*}

Example B

Solve the equation \begin{align*}x^3- 27 = 0\end{align*}. This is the same as the equation \begin{align*}x^3 = 27\end{align*}.

Solution:

\begin{align*}x^3 = 27\\ r &= \sqrt{x^2+y^2}\\ r &= \sqrt{(27)^2+(0)^2}\\ r &= 27\\ \theta &= \tan^{-1} \left(\frac{0}{27}\right)=0\end{align*}

Write an expression for determining the cube roots of \begin{align*}27 = 27 + 0i\end{align*}

\begin{align*}27^{\frac{1}{3}} &= [27( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{3}}\\ &= 3\left(\cos \frac{2\pi k}{3}+i \sin \frac{2\pi k}{3}\right)k=0, 1, 2\\ x_1 &= 3\left(\cos \frac{0}{3}+i \sin \frac{0}{3}\right) \rightarrow 3(\cos 0 +i \sin 0)=3 && for \ k=0\\ x_2 &= 3\left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right) \approx -1.5 + 2.6i && for \ k=1\\ x_3 &= 3\left(\cos \frac{4\pi}{3}+i \sin \frac{4\pi}{3}\right) \approx -1.5 - 2.6i && for \ k=2\\ \end{align*}

Example C

Solve the equation \begin{align*}x^4 = 1\end{align*}

Solution:

\begin{align*}x^4 = 1\\ r &= \sqrt{x^2+y^2}\\ r &= \sqrt{(1)^2+(0)^2}\\ r &= 1\\ \theta &= \tan^{-1} \left(\frac{0}{1}\right)=0\end{align*}

Write an expression for determining the cube roots of \begin{align*}1 = 1 + 0i\end{align*}

\begin{align*}1^{\frac{1}{4}} &= [1( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{4}}\\ &= 1\left(\cos \frac{2\pi k}{4}+i \sin \frac{2\pi k}{4}\right)k=0, 1, 2, 3\\ x_1 &= 1\left(\cos \frac{0}{4}+i \sin \frac{0}{4}\right) \rightarrow 3(\cos 0 +i \sin 0)= 1 && for \ k=0\\ x_2 &= 1\left(\cos \frac{2\pi}{4}+i \sin \frac{2\pi}{4}\right) = 0 + i = i && for \ k=1\\ x_3 &= 1\left(\cos \frac{4\pi}{4}+i \sin \frac{4\pi}{4}\right) = -1 - 0i = -1 && for \ k=2\\ x_4 &= 1\left(\cos \frac{6\pi}{4}+i \sin \frac{6\pi}{4}\right) = 0 - i = -i && for \ k=3\\ \end{align*}

Vocabulary

DeMoivres Theorem: DeMoivres theorem relates a complex number raised to a power to a set of trigonometric functions by stating that the complex number raised to a power is equal to the trigonometric representation of the number with the power times the angle under consideration as the argument for the trigonometric form.

Guided Practice

1. Rewrite the following in rectangular form: \begin{align*}[2(\cos 315^\circ + i \sin 315^\circ)]^3\end{align*}

2. Solve the equation \begin{align*}x^4 + 1 = 0\end{align*}. What shape do the roots make?

3. Solve the equation \begin{align*}x^3-64=0\end{align*}. What shape do the roots make?

Solutions:

1. \begin{align*}r &= 2 \ \text{and} \ \theta=315^\circ \ \text{or} \ \frac{7\pi}{4}.\\ z^n &= [r(\cos \theta+i \sin \theta)]^n=r^n(\cos n\theta+i \sin n\theta)\\ z^3 &= 2^3 \left[(\cos 3 \left(\frac{7\pi}{4}\right)+i \sin 3 \left(\frac{7\pi}{4}\right)\right]\\ z^3 &= 8 \left(\cos \frac{21\pi}{4}+i \sin \frac{21\pi}{4}\right)\\ z^3 &= 8 \left(-\frac{\sqrt{2}}{2}-i \frac{\sqrt{2}}{2}\right)\\ z^3 &= -4\sqrt{2}-4i\sqrt{2}\end{align*}

\begin{align*}\frac{21\pi}{4}\end{align*} is in the third quadrant so both are negative.

2.

\begin{align*}& x^4+1=0 && r=\sqrt{x^2+y^2}\\ & x^4=-1 && r=\sqrt{(-1)^2+(0)^2}\\ & x^4=-1+0i && r=1\\ & \theta = \tan^{-1} \left(\frac{0}{-1}\right)+\pi=\pi\end{align*}

Write an expression for determining the fourth roots of \begin{align*}x^4 = -1 + 0i\end{align*}

\begin{align*}(-1+0i)^{\frac{1}{4}} &= [1(\cos (\pi+2\pi k)+i \sin (\pi+2\pi k))]^{\frac{1}{4}}\\ (-1+0i)^{\frac{1}{4}} &= 1^{\frac{1}{4}} \left(\cos \frac{\pi+2\pi k}{4}+i \sin \frac{\pi+2\pi k}{4}\right)\\ x_1 &= 1 \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} && \text{for} \ k=0\\ x_2 &= 1 \left(\cos \frac{3\pi}{4}+i \sin \frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} && \text{for} \ k=1\\ x_3 &= 1 \left(\cos \frac{5\pi}{4}+i \sin \frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} && \text{for} \ k=2\\ x_4 &= 1 \left(\cos \frac{7\pi}{4}+i \sin \frac{7\pi}{4}\right)=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} && \text{for} \ k=3\end{align*}

If a line segment is drawn from each root on the polar plane to its adjacent roots, the four roots will form the corners of a square.

3.

\begin{align*}x^3-64=0 \rightarrow x^3 = 64+0i\end{align*}

\begin{align*}64 + 0i = 64(\cos (0 + 2 \pi k) + i \sin (0 + 2 \pi k))\end{align*}

\begin{align*}x = (x^3)^{\frac{1}{3}} &= (64 + 0i)^{\frac{1}{3}}\\ & = \sqrt[3]{64} \left(\cos \left(\frac{0+2\pi k}{3}\right)+i \sin \left(\frac{0+2\pi k}{3}\right)\right)\end{align*}

\begin{align*}z_1 &=4 \left(\cos \left(\frac{0+2\pi 0}{3}\right)+i \sin \left(\frac{0+2\pi 0}{3}\right)\right)\\ &=4 \cos 0 + 4i \sin 0 \\ &= 4 \text{ for } \ k=0\end{align*}

\begin{align*}z_2 &=4 \left(\cos \left(\frac{0+2\pi}{3}\right)+i \sin \left(\frac{0+2\pi}{3}\right)\right)\\ &=4 \cos \frac{2\pi}{3}+4i \sin \frac{2\pi}{3}=-2 + 2i \sqrt{3} \text{ for } \ k=1\end{align*}

\begin{align*}z_3 &=4 \left(\cos \left(\frac{0+4\pi}{3}\right)+i \sin \left(\frac{0+4\pi}{3}\right)\right)\\ &=4 \cos \frac{4\pi}{3}+4i \sin \frac{4\pi}{3}\\ &=-2-2i \sqrt{3} \text{ for } \ k=2\end{align*}

If a line segment is drawn from each root on the polar plane to its adjacent roots, the three roots will form the vertices of an equilateral triangle.

Concept Problem Solution

Since you want to find the fourth root of 16, there will be four solutions in all.

\begin{align*}x^4=16.\\ r &= \sqrt{x^2+y^2}\\ r &= \sqrt{(16)^2+(0)^2}\\ r &= 16\\ \theta &= \tan^{-1} \left(\frac{0}{16}\right)=0\end{align*}

Write an expression for determining the fourth roots of \begin{align*}16 = 16 + 0i\end{align*}

\begin{align*}16^{\frac{1}{4}} &= [16( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{4}}\\ &= 2\left(\cos \frac{2\pi k}{4}+i \sin \frac{2\pi k}{4}\right)k=0, 1, 2, 3, 4\\ x_1 &= 2\left(\cos \frac{0}{4}+i \sin \frac{0}{4}\right) \rightarrow 2(\cos 0+i \sin 0)=2 && for \ k=0\\ x_2 &= 2\left(\cos \frac{2\pi}{4}+i \sin \frac{2\pi}{4}\right) = 2i && for \ k=1\\ x_3 &= 2\left(\cos \frac{4\pi}{4}+i \sin \frac{4\pi}{4}\right) = -2 && for \ k=2\\ x_4 &= 2\left(\cos \frac{6\pi}{4}+i \sin \frac{6\pi}{4}\right) = -2i && for \ k=3\\ \end{align*}

Therefore, the four roots of 16 are \begin{align*}2, -2, 2i, -2i\end{align*}. Notice how you could find the two real roots if you seen complex numbers. The addition of the complex roots completes our search for the roots of equations.

Practice

Solve each equation.

  1. \begin{align*}x^3=1\end{align*}
  2. \begin{align*}x^5=1\end{align*}
  3. \begin{align*}x^8=1\end{align*}
  4. \begin{align*}x^5=-32\end{align*}
  5. \begin{align*}x^4+5=86\end{align*}
  6. \begin{align*}x^5=-1\end{align*}
  7. \begin{align*}x^4=-1\end{align*}
  8. \begin{align*}x^3=8\end{align*}
  9. \begin{align*}x^6=-64\end{align*}
  10. \begin{align*}x^3=-64\end{align*}
  11. \begin{align*}x^5=243\end{align*}
  12. \begin{align*}x^3=343\end{align*}
  13. \begin{align*}x^7=-128\end{align*}
  14. \begin{align*}x^{12}=1\end{align*}
  15. \begin{align*}x^6=1\end{align*}

Image Attributions

Show Hide Details
Description
Difficulty Level:
At Grade
Subjects:
Grades:
Date Created:
Sep 26, 2012
Last Modified:
Mar 23, 2016
Files can only be attached to the latest version of Modality
Reviews
100 % of people thought this content was helpful.
0
Loading reviews...
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
MAT.TRG.673.L.1