Polar and Exponential Form, De Moivre's Theorem, and Roots

Modulus-argument (polar) form

z = r(cos θ + i sin θ) = r cis θ = re^(iθ)

where r = abs(z) and θ = arg z, usually with −π < θ ≤ π.

Example. 1 + i: r = √2, θ = π/4, so 1 + i = √2 e^(iπ/4).

Multiplying and dividing

• Multiply: multiply the moduli, add the arguments.
• Divide: divide the moduli, subtract the arguments.

(2 cis π/3)(3 cis π/6) = 6 cis π/2 = 6i.

De Moivre's theorem

(r cis θ)ⁿ = rⁿ cis(nθ)

Example. (1 + i)⁸ = (√2)⁸ cis(2π) = 16.

Roots

The n nth roots of z = r cis θ are

r^(1/n) cis((θ + 2kπ)/n), k = 0, 1, ..., n − 1.

They form a regular n-gon on a circle of radius r^(1/n).

Example. Cube roots of 8: 2 cis 0, 2 cis(2π/3), 2 cis(4π/3), that is 2, −1 + √3 i, −1 − √3 i.

The nth roots of unity add up to 0.

Trigonometric identities

De Moivre can derive identities, for example cos 3θ = 4cos³θ − 3cos θ, by expanding (cos θ + i sin θ)³ and equating real parts.

Quick summary

• z = re^(iθ); multiply moduli, add arguments.
• De Moivre: (r cis θ)ⁿ = rⁿ cis nθ.
• n roots, equally spaced by 2π/n.