De Moivre's Theorem and Roots of Unity

Multiplying in polar form

Here is where polar form earns its keep. When you multiply two complex numbers in polar form, the moduli multiply and the arguments add. Geometrically, multiplying by a number of modulus r and angle θ scales by r and rotates by θ — multiplication is rotation-and-scaling.

z1 = r1(cos α + i sin α),  z2 = r2(cos β + i sin β)

z1 · z2 = r1·r2 [ cos(α + β) + i sin(α + β) ]

example: 2(cos 30° + i sin 30°) · 3(cos 40° + i sin 40°)
       = 6 (cos 70° + i sin 70°)

De Moivre's theorem for powers

Multiply z by itself n times and the rule compounds: the modulus is raised to the n-th power, the argument is multiplied by n. That is De Moivre's theorem. It turns a fearsome power like (1 + i)^10 into a one-line computation.

[ r(cos θ + i sin θ) ]^n = r^n ( cos nθ + i sin nθ )

example: (1 + i)^10
  polar:  1 + i = sqrt(2)(cos 45° + i sin 45°)
  power:  (sqrt(2))^10 ( cos 450° + i sin 450° )
        = 32 (cos 90° + i sin 90°)     [450° = 360° + 90°]
        = 32 (0 + i·1) = 32i

Roots of unity

Run De Moivre backward to take roots. The equation z^n = 1 has exactly n solutions, the nth roots of unity. Each has modulus 1 (so they lie on the unit circle) and arguments spaced evenly by 360°/n. They are the corners of a regular n-gon, one of them always at z = 1.

solve z^3 = 1   (cube roots of unity)

  1 = cos(0° + 360°k) + i sin(0° + 360°k),  k = 0,1,2
  z = cos(120°k) + i sin(120°k)

  k=0: cos 0°   + i sin 0°   = 1
  k=1: cos 120° + i sin 120° = -1/2 + (sqrt(3)/2) i
  k=2: cos 240° + i sin 240° = -1/2 - (sqrt(3)/2) i

  three points, 120° apart, on the unit circle ✓

1. Write the target in polar form, including +360°k for the argument.
2. Take the n-th root of the modulus and divide the argument by n.
3. Let k = 0, 1, …, n-1 to list all n distinct roots; further k just repeats them.