5.3.7: Equations Using DeMoivre's Theorem

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Complex roots of an equation.

You are given an equation in math class:

\(x^4=16\)

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!"

Equations Using De Moivre's Theorem

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 nth 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.

Using De Moivre's Theorem

1. Consider the equation \(x^5−32=0\). The solution is the same as the solution of \(x^5=32\). In other words, we must determine the fifth roots of 32.

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

Write an expression for determining the fifth roots of 32=32+0i

\(\begin{array}{rlr}
32^{1/5} & =\left[32(\cos (0+2 \pi k)+i \sin (0+2 \pi k)]^{\dfrac{1}{5}}\right. & \\
& =2\left(\cos \dfrac{2 \pi k}{5}+i \sin \dfrac{2 \pi k}{5}\right) k=0,1,2,3,4 & \\
x_{1} & =2\left(\cos \dfrac{0}{5}+i \sin \dfrac{0}{5}\right) \rightarrow 2(\cos 0+i \sin 0)=2 & & \text { for } k=0 \\
x_{2} & =2\left(\cos \dfrac{2 \pi}{5}+i \sin \dfrac{2 \pi}{5}\right) \approx 0.62+1.9 i & & \text { for } k=1 \\
x_{3} & =2\left(\cos \dfrac{4 \pi}{5}+i \sin \dfrac{4 \pi}{5}\right) \approx-1.62+1.18 i & & \text { for } k=2 \\
x_{4} & =2\left(\cos \dfrac{6 \pi}{5}+i \sin \dfrac{6 \pi}{5}\right) \approx-1.62-1.18 i & & \text { for } k=3 \\
x_{5} & =2\left(\cos \dfrac{8 \pi}{5}+i \sin \dfrac{8 \pi}{5}\right) \approx 0.62-1.9 i & & \text { for } k=4
\end{array}\)

This is the same as the equation \(x^3=27\).

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

Write an expression for determining the cube roots of \(27=27+0i\)

\(\begin{aligned}
27^{1/3} &=\left[27(\cos (0+2 \pi k)+i \sin (0+2 \pi k)\right]^{1/3}& \\
&=3\left(\cos \dfrac{2 \pi k}{3}+i \sin \dfrac{2 \pi k}{3}\right) k=0,1,2 & \\
x_{1} &=3\left(\cos \dfrac{0}{3}+i \sin \dfrac{0}{3}\right) \rightarrow 3(\cos 0+i \sin 0)=3 & \text { for } k=0 \\
x_{2} &=3\left(\cos \dfrac{2 \pi}{3}+i \sin \dfrac{2 \pi}{3}\right) \approx-1.5+2.6 i & \text { for } k=1 \\
x_{3} &=3\left(\cos \dfrac{4 \pi}{3}+i \sin \dfrac{4 \pi}{3}\right) \approx-1.5-2.6 i & \text { for } k=2
\end{aligned}\)

3. Solve the equation \(x^4=1\)

\(\begin{array}{l}
x^{4}=1 \\
r=\sqrt{x^{2}+y^{2}} \\
r=\sqrt{(1)^{2}+(0)^{2}} \\
r=1 \\
\theta=\tan ^{-1}\left(\dfrac{0}{1}\right)=0
\end{array}\)

Write an expression for determining the cube roots of \(1=1+0i\)

\(\begin{aligned}
1^{1/4} &=\left[1(\cos (0+2 \pi k)+i \sin (0+2 \pi k)\right]^{1/4}& \\
&=1\left(\cos \dfrac{2 \pi k}{4}+i \sin \dfrac{2 \pi k}{4}\right) k=0,1,2,3 & \\
x_{1} &=1\left(\cos \dfrac{0}{4}+i \sin \dfrac{0}{4}\right) \rightarrow 3(\cos 0+i \sin 0)=1 & \text { for } k=0 \\
x_{2} &=1\left(\cos \dfrac{2 \pi}{4}+i \sin \dfrac{2 \pi}{4}\right)=0+i=i & \text { for } k=1 \\
x_{3} &=1\left(\cos \dfrac{4 \pi}{4}+i \sin \dfrac{4 \pi}{4}\right)=-1-0 i=-1 & \text { for } k=2 \\
x_{4} &=1\left(\cos \dfrac{6 \pi}{4}+i \sin \dfrac{6 \pi}{4}\right)=0-i=-i & \text { for } k=3
\end{aligned}\)

Example \(\PageIndex{1}\)

Earlier, you were asked to find the complex root of \(x^4=16\).

Solution

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

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

Write an expression for determining the fourth roots of \(16=16+0i\)

\(\begin{aligned}
16^{1/4} &=\left[16(\cos (0+2 \pi k)+i \sin (0+2 \pi k)\right]^{1/4}& & \\
&=2\left(\cos \dfrac{2 \pi k}{4}+i \sin \dfrac{2 \pi k}{4}\right) k=0,1,2,3,4 & & \\
x_{1} &=2\left(\cos \dfrac{0}{4}+i \sin \dfrac{0}{4}\right) \rightarrow 2(\cos 0+i \sin 0)=2 & & \text { for } k=0 \\
x_{2} &=2\left(\cos \dfrac{2 \pi}{4}+i \sin \dfrac{2 \pi}{4}\right)=2 i & & \text { for } k=1 \\
x_{3} &=2\left(\cos \dfrac{4 \pi}{4}+i \sin \dfrac{4 \pi}{4}\right)=-2 & & \text { for } k=2 \\
x_{4} &=2\left(\cos \dfrac{6 \pi}{4}+i \sin \dfrac{6 \pi}{4}\right)=-2 i & & \text { for } k=3
\end{aligned}\)

Therefore, the four roots of 16 are \(2,\; −2,\; 2i,\; −2i\). 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.

Example \(\PageIndex{2}\)

Rewrite the following in rectangular form: \([2(\cos 315^{\circ} +i\sin 315^{\circ} )]^3\)

Solution

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

\(\dfrac{21\pi }{4}\) is in the third quadrant so both are negative.

Example \(\PageIndex{3}\)

Solve the equation \(x^4+1=0\). What shape do the roots make?

Solution

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

Write an expression for determining the fourth roots of \(x^4=−1+0i\)

\(\begin{array}{rlr}
(-1+0 i)^{1/4} & =[1(\cos (\pi+2 \pi k)+i \sin (\pi+2 \pi k))]^{1/4} & \\
(-1+0 i)^{1/4} & =1^{\dfrac{1}{4}}\left(\cos \dfrac{\pi+2 \pi k}{4}+i \sin \dfrac{\pi+2 \pi k}{4}\right) & \\
x_{1} & =1\left(\cos \dfrac{\pi}{4}+i \sin \dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}+i \dfrac{\sqrt{2}}{2} & & \text { for } k=0 \\
x_{2} & =1\left(\cos \dfrac{3 \pi}{4}+i \sin \dfrac{3 \pi}{4}\right)=-\dfrac{\sqrt{2}}{2}+i \dfrac{\sqrt{2}}{2} & & \text { for } k=1 \\
x_{3} & =1\left(\cos \dfrac{5 \pi}{4}+i \sin \dfrac{5 \pi}{4}\right)=-\dfrac{\sqrt{2}}{2}-i \dfrac{\sqrt{2}}{2} & & \text { for } k=2 \\
x_{4} & =1\left(\cos \dfrac{7 \pi}{4}+i \sin \dfrac{7 \pi}{4}\right)=\dfrac{\sqrt{2}}{2}-i \dfrac{\sqrt{2}}{2} & & \text { for } k=3
\end{array}\)

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.

Example \(\PageIndex{4}\)

Solve the equation \(x^3−64=0\). What shape do the roots make?

Solution

\(\begin{aligned} x^3−64&=0\rightarrow x^3=64+0i \\ 64+0i&=64(cos(0+2\pi k)+isin(0+2\pi k)) \end{aligned}\)

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

\(\begin{aligned}
z_{1} &=4\left(\cos \left(\dfrac{0+2 \pi 0}{3}\right)+i \sin \left(\dfrac{0+2 \pi 0}{3}\right)\right) \\
&=4 \cos 0+4 i \sin 0 \\
&=4 \text { for } k=0
\end{aligned}\)

\(\begin{aligned}
z_{2} &=4\left(\cos \left(\dfrac{0+2 \pi}{3}\right)+i \sin \left(\dfrac{0+2 \pi}{3}\right)\right) \\
&=4 \cos \dfrac{2 \pi}{3}+4 i \sin \dfrac{2 \pi}{3}=-2+2 i \sqrt{3} \text { for } k=1
\end{aligned}\)

\(\begin{aligned}
z_{3} &=4\left(\cos \left(\dfrac{0+4 \pi}{3}\right)+i \sin \left(\dfrac{0+4 \pi}{3}\right)\right) \\
&=4 \cos \dfrac{4 \pi}{3}+4 i \sin \dfrac{4 \pi}{3} \\
&=-2-2 i \sqrt{3} \text { for } k=2
\end{aligned}\)

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.