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DeMoivre's Theorem and nth 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 \sqrt{3i} . Are you able to find roots of complex numbers? By the end of this Concept, you'll be able to perform this calculation.

Watch This

James Sousa: Determining the Nth Roots of a Complex Number

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 n^{th} root of a complex number written in polar form.

As before, let z = r(\cos \theta + i \sin \theta) and let the n^{th} root of z be v = s (\cos \alpha + i \sin \alpha) . So, in general, \sqrt[n]{z}=v and v^n=z .

\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)

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

Example A

Find the two square roots of 2i .

Solution: Express 2i in polar form.

& r=\sqrt{x^2+y^2} && \cos \theta=0\\& r=\sqrt{(0)^2+(2)^2} && \qquad \theta=90^\circ\\& r=\sqrt{4}=2

(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

To find the other root, add 360^\circ to \theta .

(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

Example B

Find the three cube roots of -2-2i \sqrt{3}

Solution: Express -2-2i \sqrt{3} in polar form:

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}

& \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

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]

In standard form: z_1=0.276+1.563i, z_2=-1.492-0.543i, z_3=1.216-1.02i .

Example C

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

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

\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)

Vocabulary

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

Guided Practice

1. Find \sqrt[3]{27i} .

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

3. Find the fourth roots of 81i .

Solutions:

1.

&&& 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 }}

2.

& 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}

(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)

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

3.

81i in polar form is:

& 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

Concept Problem Solution

Finding the two square roots of 3i involves first converting the number to polar form:

For the radius:

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

And the angle:

\cos \theta=0\\\theta=90^\circ\\

(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)

To find the other root, add 360^\circ to \theta .

(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)

Practice

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

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

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

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

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