De Moivre's Theorem

De Moivre's Theorem provides a formula to calculate powers and roots of complex numbers expressed in polar form.

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De Moivre's Theorem : Page 20

\left[r(\cos \theta +i \sin \theta)\right]^n=r^n(\cos n \theta+i \sin n \theta)

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Using De Moivre's Theorem to Find Powers of Complex Numbers

Convert the number to Polar form.
Apply De Moivre's Theorem
Simplify

Finding Powers

Example

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Calculate $(1+i)^4$ .

First, express $1+i$ in Polar form.

r=\sqrt{1^2+1^2}=\sqrt{2}

\theta=\tan^{-1}\left( \tfrac{1}{1}\right)=\tfrac{\pi}{4}

Polar form.

1+i=\sqrt{2}\left(\cos \tfrac{\pi}{4} +i \sin \tfrac{\pi}{4}\right)

Apply De Moivre's Theorem

(1+i)^4=\left(\sqrt{2}\left(\cos \tfrac{\pi}{4} +i \sin \tfrac{\pi}{4}\right) \right)^4

=\left(\sqrt{2}\right)^4\left(\cos (4 \cdot \tfrac{\pi}{4}) +i \sin (4 \cdot \tfrac{\pi}{4})\right)

=4\left(\cos (\pi) +i \sin (\pi)\right)

=4(-1+0i)

=-4

Finding Roots

Example

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Find the solutions of $z^3=8.$

First, express $8$ in Polar form.

r=\sqrt{8^2+0^2}=8

\theta=0

Polar form.

z=8(\cos 0 +i \sin0)

To solve for $z$ , we would take the cubed root of both sides, which is equivalent to raising both sides to a power of $\tfrac{1}{3}$ .

z^\tfrac{1}{3}=\left(8(\cos 0 +i \sin0) \right)^\tfrac{1}{3}

In the argument of the trigonometric identities, you also need to add $2 \pi n$ , to account for the different rotation of the complex numbers in order to get your solutions.

z^\tfrac{1}{3}=\left(8(\cos (0+2\pi n) +i \sin(0+2\pi n) \right)^\tfrac{1}{3}

Now, apply De Moivre's theorem

=8^\tfrac{1}{3}\left(\cos \frac{(0+2\pi n)}{3} +i \sin\frac{(0+2\pi n)}{3} \right)

=2\left(\cos \frac{2\pi n}{3} +i \sin\frac{2\pi n}{3} \right)

Since we took the cubed root, we except to get 3 roots. That means we need to substitute 0,1,2 for $n$ .

z_0=2\left(\cos \frac{2\pi (0)}{3} +i \sin\frac{2\pi (0)}{3} \right)

z_0 = 2 (\cos 0 + i \sin 0) = 2

z_1=2\left(\cos \frac{2\pi (1)}{3} +i \sin\frac{2\pi (1)}{3} \right)

z_1 = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -1 + i\sqrt{3}

z_2=2\left(\cos \frac{2\pi (2)}{3} +i \sin\frac{2\pi (2)}{3} \right)

z_2 = 2 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = 2 \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = -1 - i\sqrt{3}

Thus, the three cube roots of $8$ are 2, -1 + i√3, and -1 - i√3.

De Moivre's Theorem Simplified Revision Notes for Leaving Cert Mathematics

De Moivre's Theorem

Finding Powers

Calculate $(1+i)^4$ .

Finding Roots

Find the solutions of $z^3=8.$

500K+ Students Use These Powerful Tools to Master De Moivre's Theorem For their Leaving Cert Exams.

Other Revision Notes related to De Moivre's Theorem you should explore

De Moivre's Theorem Simplified Revision Notes for Leaving Cert Mathematics

De Moivre's Theorem

Finding Powers

Calculate (1+i)4(1+i)^4(1+i)4.

Finding Roots

Find the solutions of z3=8.z^3=8.z3=8.

500K+ Students Use These Powerful Tools to Master De Moivre's Theorem For their Leaving Cert Exams.

Other Revision Notes related to De Moivre's Theorem you should explore

Calculate $(1+i)^4$ .

Find the solutions of $z^3=8.$