De Moivre’s Theorem

A power is repeated multiplying

Multiplying two complex numbers multiplies their lengths and adds their angles. Squaring z is multiplying z by itself, so if z is r at $\theta$θ, then $z^2$z² is r × r at $\theta +\theta$θ + θ, which is $r^2$r² at $2\theta$2θ.

One more factor of z multiplies the length by r again and adds $\theta$θ again: $z^3=r^3$z³ = r³ at $3\theta$3θ. After n factors the length is $r^n$rⁿ and the angle is $n\theta$nθ.

The powers of 1 + i

1 + i has modulus $\sqrt{2}$√2 and argument 45°. Each power is the one before it stretched by $\sqrt{2}$√2 and turned a further 45°: ${\left(1+i\right)}^2=2$(1 + i)² = 2 at 90° = 2i, ${\left(1+i\right)}^3=2\sqrt{2}$(1 + i)³ = 2√2 at 135° = −2 + 2i, and ${\left(1+i\right)}^4=4$(1 + i)⁴ = 4 at 180° = −4.

Check by multiplying out: ${\left(1+i\right)}^2=1+2i+i^2=2i$(1 + i)² = 1 + 2i + i² = 2i. Then ${\left(1+i\right)}^3=2i\left(1+i\right)=2i+2i^2=-2+2i$(1 + i)³ = 2i(1 + i) = 2i + 2i² = −2 + 2i, and ${\left(1+i\right)}^4={\left(2i\right)}^2=4i^2=-4$(1 + i)⁴ = (2i)² = 4i² = −4.

z = 1 + i and its next three powers. The arrows are $\sqrt{2}$√2, 2, $2\sqrt{2}$2√2 and 4 long, at 45°, 90°, 135° and 180°: each is $\sqrt{2}$√2 times the length of the one before and 45° further round.

The theorem

De Moivre’s theorem: (r at $\theta {)}^n=r^n$θ)ⁿ = rⁿ at $n\theta$nθ. In the full notation, ${\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n=r^n\left(\cos n\theta +i\sin n\theta \right)$[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ).

For z = 2(cos 30° + i sin 30°), $z^3=2^3\left(\cos 90^{\circ }+i\sin 90^{\circ }\right)=8i$z³ = 2³(cos 90° + i sin 90°) = 8i. In parts, $z=\sqrt{3}+i$z = √3 + i, $z^2=3+2\sqrt{3}i+i^2=2+2\sqrt{3}i$z² = 3 + 2√3 i + i² = 2 + 2√3 i, and $z^3=\left(2+2\sqrt{3}i\right)\left(\sqrt{3}+i\right)=2\sqrt{3}+2i+6i+2\sqrt{3}{i}^2=8i$z³ = (2 + 2√3 i)(√3 + i) = 2√3 + 2i + 6i + 2√3 i² = 8i.

It holds for every positive whole number n. It holds for n = 1. If it holds for some n, then multiplying $z^n=r^n$zⁿ = rⁿ at $n\theta$nθ by z = r at $\theta$θ gives $r^{n+1}$rⁿ⁺¹ at $\left(n+1\right)\theta$(n + 1)θ, so it holds for n + 1 as well. By induction it holds for every n.

Large powers

${\left(1+i\right)}^8={\left(\sqrt{2}\right)}^8$(1 + i)⁸ = (√2)⁸ at 8 × 45° = 16 at 360°. A full turn points the same way as 0°, so ${\left(1+i\right)}^8=16$(1 + i)⁸ = 16. Multiplying out would take seven multiplications.

$1+\sqrt{3}i$1 + √3 i has modulus $\sqrt{1+3}=2$√(1 + 3) = 2 and argument 60°, since $\tan \theta =\sqrt{3}$tan θ = √3 in the first quadrant. So ${\left(1+\sqrt{3}i\right)}^5=2^5$(1 + √3 i)⁵ = 2⁵ at 5 × 60° = 32 at 300°. 300° is past 180°, so take off a full turn: 300° − 360° = −60°. Then ${\left(1+\sqrt{3}i\right)}^5=32\left(\cos \left(-60^{\circ }\right)+i\sin \left(-60^{\circ }\right)\right)=16-16\sqrt{3}i$(1 + √3 i)⁵ = 32(cos(−60°) + i sin(−60°)) = 16 − 16√3 i, about 16 − 27.713i.

$\sqrt{3}+i$√3 + i is 2 at 30°, so ${\left(\sqrt{3}+i\right)}^6=2^6$(√3 + i)⁶ = 2⁶ at 180° = 64 at 180° = −64.

On the unit circle

When r = 1, every power has length 1 too, and the theorem reads ${\left(\cos \theta +i\sin \theta \right)}^n=\cos n\theta +i\sin n\theta$(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. So ${\left(\cos 20^{\circ }+i\sin 20^{\circ }\right)}^9=\cos 180^{\circ }+i\sin 180^{\circ }=-1$(cos 20° + i sin 20°)⁹ = cos 180° + i sin 180° = −1.

Raising $\cos \theta +i\sin \theta$cos θ + i sin θ to a power multiplies the angle inside the cosine and the sine. Multiple Angles by De Moivre expands the left side to find formulas for $\cos n\theta$cos nθ and $\sin n\theta$sin nθ.

Negative and zero powers

$z^{-1}=\frac{1}{z}$z⁻¹ = 1/z, and dividing 1 by r at $\theta$θ gives $\frac{1}{r}$1/r at $-\theta$−θ. So $z^{-n}={\left(\frac{1}{z}\right)}^n=r^{-n}$z⁻ⁿ = (1/z)ⁿ = r⁻ⁿ at $-n\theta$−nθ, and the theorem holds for negative whole numbers too.

${\left(1+i\right)}^{-4}={\left(\sqrt{2}\right)}^{-4}$(1 + i)⁻⁴ = (√2)⁻⁴ at $-4\times 45^{\circ }=\frac{1}{4}$−4 × 45° = 1/4 at −180°. −180° is outside the range, so add a full turn: $\frac{1}{4}$1/4 at 180°, which is $-\frac{1}{4}$−1/4. Check: ${\left(1+i\right)}^4=-4$(1 + i)⁴ = −4, and $1÷\left(-4\right)=-\frac{1}{4}$1 ÷ (−4) = −1/4.

With n = 0 the theorem gives $r^0$r⁰ at 0°, which is 1, as $z^0$z⁰ is for any z other than 0.

The usual mistakes

Multiplying the length by n. (2 at $30^{\circ }{)}^3$30°)³ is 8 at 90°, not 6 at 90°: each of the three factors multiplies the length by 2, so it is $2^3$2³.

Leaving the angle as it was. Each factor adds another 30°, so the cube is at 90°, not 30°.

Raising the cosine and the sine to the power. ${\left(\cos 30^{\circ }+i\sin 30^{\circ }\right)}^3$(cos 30° + i sin 30°)³ is cos 90° + i sin 90° = i, not ${\cos }^{3}30^{\circ }+i{\sin }^{3}30^{\circ }\approx 0.650+0.125i$cos³30° + i sin³30° ≈ 0.650 + 0.125i, a number about 0.66 long rather than 1.

Leaving the argument out of range. ${\left(1+\sqrt{3}i\right)}^5$(1 + √3 i)⁵ has argument −60°; 300° names the same direction but is outside $-180^{\circ }<\theta \le 180^{\circ }$−180° < θ ≤ 180°.

A Ferris wheel that stops every twelfth of a turn

The application below measures turns in radians: $\frac{\pi }{6}$π/6 is 30°. Each stop multiplies the car’s position by 1 at 30°, so n stops multiply it by 1 at 30n°. Five stops make 150°. Sixteen make 480°, which is 120° once a full turn is taken off.