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# 6.13: DeMoivre's Theorem and nth Roots

Difficulty Level: At Grade Created by: CK-12
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Practice DeMoivre's Theorem and nth Roots

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You are in math class one day when your teacher asks you to find 3i\begin{align*}\sqrt{3i}\end{align*}. Are you able to find roots of complex numbers? By the end of this Concept, you'll be able to perform this calculation.

### Guidance

Other Concepts in this course have explored all of the basic operations of arithmetic as they apply to complex numbers in standard form and in polar form. The last discovery is that of taking roots of complex numbers in polar form. Using De Moivre’s Theorem we can develop another general rule – one for finding the nth\begin{align*}n^{th}\end{align*} root of a complex number written in polar form.

As before, let z=r(cosθ+isinθ)\begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} and let the nth\begin{align*}n^{th}\end{align*} root of z\begin{align*}z\end{align*} be v=s(cosα+isinα)\begin{align*}v = s (\cos \alpha + i \sin \alpha)\end{align*}. So, in general, zn=v\begin{align*}\sqrt[n]{z}=v\end{align*} and vn=z\begin{align*}v^n=z\end{align*}.

znr(cosθ+isinθ)n[r(cosθ+isinθ)]1nr1n(cos1nθ+isin1nθ)r1n(cosθn+isinθn)=v=s(cosα+isinα)=s(cosα+isinα)=s(cosα+isinα)=s(cosα+isinα)\begin{align*}\sqrt[n]{z} &= v\\ \sqrt[n]{r(\cos \theta+i \sin \theta)} &=s(\cos \alpha + i \sin \alpha)\\ [r(\cos \theta+i \sin \theta)]^{\frac{1}{n}} &= s(\cos \alpha +i \sin \alpha)\\ r^{\frac{1}{n}}\left(\cos \frac{1}{n} \theta+i \sin \frac{1}{n}\theta \right) &= s(\cos \alpha+i \sin \alpha)\\ r^{\frac{1}{n}}\left(\cos \frac{\theta}{n}+i \sin \frac{\theta}{n} \right) &= s(\cos \alpha+i \sin \alpha)\end{align*}

From this derivation, we can conclude that r1n=s\begin{align*}r^{\frac{1}{n}}=s\end{align*} or sn=r\begin{align*}s^n=r\end{align*} and α=θn\begin{align*}\alpha=\frac{\theta}{n}\end{align*}. Therefore, for any integer k(0,1,2,n1)\begin{align*}k (0, 1, 2, \ldots n -1)\end{align*}, v\begin{align*}v\end{align*} is an nth\begin{align*}n^{th}\end{align*} root of z\begin{align*}z\end{align*} if s=rn\begin{align*}s=\sqrt[n]{r}\end{align*} and α=θ+2πkn\begin{align*}\alpha=\frac{\theta+2\pi k}{n}\end{align*}. Therefore, the general rule for finding the nth\begin{align*}n^{th}\end{align*} roots of a complex number if z=r(cosθ+isinθ)\begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} is: rn(cosθ+2πkn+isinθ+2πkn)\begin{align*}\sqrt[n]{r} \left(\cos \frac{\theta+2\pi k}{n}+i \sin \frac{\theta+2\pi k}{n}\right)\end{align*}. Let’s begin with a simple example and we will leave θ\begin{align*}\theta\end{align*} in degrees.

#### Example A

Find the two square roots of 2i\begin{align*}2i\end{align*}.

Solution: Express 2i\begin{align*}2i\end{align*} in polar form.

r=x2+y2r=(0)2+(2)2r=4=2cosθ=0θ=90\begin{align*}& r=\sqrt{x^2+y^2} && \cos \theta=0\\ & r=\sqrt{(0)^2+(2)^2} && \qquad \theta=90^\circ\\ & r=\sqrt{4}=2\end{align*}

(2i)12=212(cos902+isin902)=2(cos45+isin45)=1+i\begin{align*}(2i)^{\frac{1}{2}}=2^{\frac{1}{2}} \left(\cos \frac{90^\circ}{2}+i \sin \frac{90^\circ}{2}\right)=\sqrt{2}(\cos 45^\circ +i \sin 45^\circ)=1+i\end{align*}

To find the other root, add 360\begin{align*}360^\circ\end{align*} to θ\begin{align*}\theta\end{align*}.

(2i)12=212(cos4502+isin4502)=2(cos225+isin225)=1i\begin{align*}(2i)^{\frac{1}{2}}=2^{\frac{1}{2}} \left(\cos \frac{450^\circ}{2}+i \sin \frac{450^\circ}{2}\right)=\sqrt{2}(\cos 225^\circ +i \sin 225^\circ)=-1-i\end{align*}

#### Example B

Find the three cube roots of 22i3\begin{align*}-2-2i \sqrt{3}\end{align*}

Solution: Express 22i3\begin{align*}-2-2i \sqrt{3}\end{align*} in polar form:

rrr=x2+y2=(2)2+(23)2=16=4θ=tan1(232)=4π3\begin{align*}r &=\sqrt{x^2+y^2}\\ r &= \sqrt{(-2)^2+(-2\sqrt{3})^2}\\ r &= \sqrt{16}=4 && \theta = \tan^{-1} \left(\frac{-2\sqrt{3}}{-2}\right)=\frac{4\pi}{3}\end{align*}

\begin{align*}& \sqrt[n]{r}\left( \cos \frac{\theta + 2\pi k}{n}+i \sin \frac{\theta + 2\pi k}{n}\right)\\ \sqrt[3]{-2-2i \sqrt{3}} &= \sqrt[3]{4} \left(\cos \frac{\frac{4 \pi}{3} + 2\pi k}{3}+i \sin \frac{\frac{4\pi}{3} +2\pi k}{3}\right) k=0, 1, 2\end{align*}

\begin{align*}z_1 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{0}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{0}{3}\right)\right] && k=0\\ &= \sqrt[3]{4}\left[\cos \frac{4\pi}{9}+i \sin \frac{4\pi}{9}\right]\\ z_2 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{2\pi}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{2\pi}{3}\right)\right] && k=1\\ &= \sqrt[3]{4}\left[\cos \frac{10\pi}{9}+i \sin \frac{10\pi}{9}\right]\\ z_3 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{4\pi}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{4\pi}{3}\right)\right] && k=2\\ &= \sqrt[3]{4}\left[\cos \frac{16\pi}{9}+i \sin \frac{16\pi}{9}\right]\end{align*}

In standard form: \begin{align*}z_1=0.276+1.563i, z_2=-1.492-0.543i, z_3=1.216-1.02i\end{align*}.

#### Example C

Calculate \begin{align*}\left(\cos \frac{\pi}{4} + i\sin\frac{\pi}{4}\right)^{1/3}\end{align*}

Using the for of DeMoivres Theorem for fractional powers, we get:

\begin{align*}\left(\cos \frac{\pi}{4} + i\sin\frac{\pi}{4}\right)^{1/3}\\ = \cos \left(\frac{1}{3} \times \frac{\pi}{4} \right) + i\sin \left(\frac{1}{3} \times \frac{\pi}{4} \right)\\ = \left(\cos \frac{\pi}{12} + i\sin \frac{\pi}{12} \right) \end{align*}

### Vocabulary

DeMoivres Theorem: DeMoivres theorem relates a complex number raised to a power to a set of trigonometric functions.

### Guided Practice

1. Find \begin{align*}\sqrt[3]{27i}\end{align*}.

2. Find the principal root of \begin{align*}(1 + i)^{\frac{1}{5}}\end{align*}. Remember the principal root is the positive root i.e. \begin{align*}\sqrt{9}=\pm 3\end{align*} so the principal root is +3.

3. Find the fourth roots of \begin{align*}81i\end{align*}.

Solutions:

1.

\begin{align*}&&& a=0 \ and \ b=27\\ & \sqrt[3]{27i} =(0+27i)^{\frac{1}{3}} && x=0 \ and \ y=27\\ & \text{Polar Form} && r=\sqrt{x^2+y^2} \qquad \qquad \theta=\frac{\pi}{2}\\ &&& r=\sqrt{(0)^2+(27)^2}\\ &&& r=27\\ &&& \sqrt[3]{27i} = \left[27 \left(\cos (\frac{\pi}{2} + 2 \pi k) +i \sin (\frac{\pi}{2} + 2 \pi k) \right)\right]^{\frac{1}{3}} \text{for } k = 0, 1, 2\\ &&& \sqrt[3]{27i} = \sqrt[3]{27} \left[\cos \left(\frac{1}{3}\right) \left(\frac{\pi}{2} + 2 \pi k\right)+i \sin \left(\frac{1}{3}\right) \left(\frac{\pi}{2} + 2 \pi k\right)\right] \text{for } k = 0, 1, 2\\ &&& \sqrt[3]{27i} = 3 \left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right) \text{for } k = 0\\ &&& \sqrt[3]{27i} = 3 \left(\cos \frac{5\pi}{6}+i \sin \frac{5\pi}{6}\right) \text{for } k = 1\\ &&& \sqrt[3]{27i} = 3 \left(\cos \frac{9\pi}{6}+i \sin \frac{9\pi}{6}\right) \text{for } k = 2\\ &&& \sqrt[3]{27i} = 3\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right), 3\left(\frac{-\sqrt{3}}{2}+\frac{1}{2}i \right), -3i\end{align*}}}

2.

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

\begin{align*}(1+i)^{\frac{1}{5}} &= \left[\sqrt{2} \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right]^{\frac{1}{5}}\\ (1+i)^{\frac{1}{5}} &= \sqrt{2}^{\frac{1}{5}}\left[\cos \left(\frac{1}{5}\right) \left(\frac{\pi}{4}\right)+i \sin \left(\frac{1}{5}\right) \left(\frac{\pi}{4}\right)\right]\\ (1+i)^{\frac{1}{5}} &= \sqrt[10]{2} \left(\cos \frac{\pi}{20}+i \sin \frac{\pi}{20} \right)\end{align*}

In standard form \begin{align*}(1+i)^{\frac{1}{5}}=(1.06+1.06i)\end{align*} and this is the principal root of \begin{align*}(1+i)^{\frac{1}{5}}\end{align*}.

3.

\begin{align*}81i\end{align*} in polar form is:

\begin{align*}& r=\sqrt{0^2+81^2}=81, \tan \theta =\frac{81}{0}= und \rightarrow \theta=\frac{\pi}{2} \quad 81 \left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\\ & \left[81 \left(\cos \left(\frac{\pi}{2}+2 \pi k \right)+i \sin \left(\frac{\pi}{2}+2\pi k\right)\right)\right]^{\frac{1}{4}}\\ & 3 \left(\cos \left(\frac{\frac{\pi}{2}+2\pi k}{4}\right)+i \sin \left(\frac{\frac{\pi}{2}+2 \pi k}{4}\right)\right)\\ & 3 \left(\cos \left(\frac{\pi}{8}+\frac{\pi k}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{\pi k}{2}\right)\right)\\ & z_1 =3 \left(\cos \left(\frac{\pi}{8}+\frac{0 \pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{0 \pi}{2}\right)\right)=3 \cos \frac{\pi}{8}+3i \sin \frac{\pi}{8}=2.77+1.15i\\ & z_2 =3 \left(\cos \left(\frac{\pi}{8}+\frac{\pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{\pi}{2}\right)\right)=3 \cos \frac{5 \pi}{8}+3i \sin \frac{5 \pi}{8}=-1.15+2.77i\\ & z_3 =3 \left(\cos \left(\frac{\pi}{8}+\frac{2\pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{2\pi}{2}\right)\right)=3 \cos \frac{9\pi}{8}+3i \sin \frac{9 \pi}{8}=-2.77-1.15i\\ & z_4 =3 \left(\cos \left(\frac{\pi}{8}+\frac{3\pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{3\pi}{2}\right)\right)=3 \cos \frac{13 \pi}{8}+3i \sin \frac{13 \pi}{8}=1.15-2.77i\end{align*}

### Concept Problem Solution

Finding the two square roots of \begin{align*}3i\end{align*} involves first converting the number to polar form:

\begin{align*}r=\sqrt{x^2+y^2} \\ r=\sqrt{(0)^2+(3)^2} \\ r=\sqrt{9}=3\end{align*}

And the angle:

\begin{align*} \cos \theta=0\\ \theta=90^\circ\\\end{align*}

\begin{align*}(3i)^{\frac{1}{2}}=3^{\frac{1}{2}} \left(\cos \frac{90^\circ}{2}+i \sin \frac{90^\circ}{2} \right)=\sqrt{3}(\cos 45^\circ +i \sin 45^\circ)= \frac{\sqrt{6}}{2} \left( 1+i \right)\end{align*}

To find the other root, add \begin{align*}360^\circ\end{align*} to \begin{align*}\theta\end{align*}.

\begin{align*}(3i)^{\frac{1}{2}}=3^{\frac{1}{2}} \left(\cos \frac{450^\circ}{2}+i \sin \frac{450^\circ}{2}\right)=\sqrt{3}(\cos 225^\circ +i \sin 225^\circ)= \frac{\sqrt{6}}{2} \left( -1-i \right)\end{align*}

### Practice

Find the cube roots of each complex number. Write your answers in standard form.

1. \begin{align*}8(\cos 2\pi+i\sin 2\pi)\end{align*}
2. \begin{align*}3(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\end{align*}
3. \begin{align*}2(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})\end{align*}
4. \begin{align*}(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3})\end{align*}
5. \begin{align*}(3+4i)\end{align*}
6. \begin{align*}(2+2i)\end{align*}

Find the principal fifth roots of each complex number. Write your answers in standard form.

1. \begin{align*}2(\cos \frac{\pi}{6}+i\sin \frac{\pi}{6})\end{align*}
2. \begin{align*}4(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2})\end{align*}
3. \begin{align*}32(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\end{align*}
4. \begin{align*}2(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3})\end{align*}
5. \begin{align*}32i\end{align*}
6. \begin{align*}(1+\sqrt{5}i)\end{align*}
7. Find the sixth roots of -64 and plot them on the complex plane.
8. How many solutions could the equation \begin{align*}x^6+64=0\end{align*} have? Explain.
9. Solve \begin{align*}x^6+64=0\end{align*}. Use your answer to #13 to help you.

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### Vocabulary Language: English

$n^{th}$ roots of unity

The $n^{th}$ roots of unity are the $n^{th}$ roots of the number 1.

complex number

A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$.

complex plane

The complex plane is the graphical representation of the set of all complex numbers.

De Moivre's Theorem

De Moivre's theorem is the only practical manual method for identifying the powers or roots of complex numbers. The theorem states that if $z= r(\cos \theta + i \sin \theta)$ is a complex number in $r cis \theta$ form and $n$ is a positive integer, then $z^n=r^n (\cos (n\theta ) + i\sin (n\theta ))$.

trigonometric polar form

To write a complex number in trigonometric form means to write it in the form $r\cos\theta+ri\sin\theta$. $rcis\theta$ is shorthand for this expression.

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