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DeMoivre's Theorem and nth Roots

Raise complex numbers to powers or find their roots.

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DeMoivre's Theorem and nth Roots

You are in math class one day when your teacher asks you to find \begin{align*}\sqrt{3i}\end{align*}. Are you able to find roots of complex numbers

De Moivre's Theorem and nth Roots

Other sections 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 \begin{align*}n^{th}\end{align*} root of a complex number written in polar form.

As before, let

\begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*}

and let the


root of



\begin{align*}v = s (\cos \alpha + i \sin \alpha)\end{align*}

. So, in general,





\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 \begin{align*}r^{\frac{1}{n}}=s\end{align*} or \begin{align*}s^n=r\end{align*} and \begin{align*}\alpha=\frac{\theta}{n}\end{align*}. Therefore, for any integer \begin{align*}k (0, 1, 2, \ldots n -1)\end{align*}, \begin{align*}v\end{align*} is an \begin{align*}n^{th}\end{align*} root of \begin{align*}z\end{align*} if \begin{align*}s=\sqrt[n]{r}\end{align*} and \begin{align*}\alpha=\frac{\theta+2\pi k}{n}\end{align*}. Therefore, the general rule for finding the \begin{align*}n^{th}\end{align*} roots of a complex number if \begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} is: \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 solve the following problems and leave \begin{align*}\theta\end{align*} in degrees.

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

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

\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*}

\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 \begin{align*}360^\circ\end{align*} to \begin{align*}\theta\end{align*}.

\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*}

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

\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*}.

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

Using 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*}


Example 1

Earlier, you were asked to solve \begin{align*}\sqrt{3i}\end{align*}.

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

For the radius:

\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*}

Example 2

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

\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*}

Example 3

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.

\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*}.

Example 4

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

\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*}


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.

Review (Answers)

To see the Review answers, open this PDF file and look for section 6.13. 

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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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