Polar form and de Moivre’s theorem

Writing a complex number by its distance and direction, z = r(cos θ + i sin θ) = re^iθ, turns multiplication into rotation and scaling, and powers into multiples of an angle. This lesson converts between Cartesian and polar forms, multiplies and divides in polar form, proves and uses de Moivre’s theorem, finds all n distinct nth roots of a complex number, and uses the binomial expansion of (cos θ + i sin θ)ⁿ to derive multiple-angle identities.

Distance and direction describe a complex number

In H2 Mathematics you located z = x + iy by its coordinates. The same point can be described by its distance from the origin and its direction. With modulus r = |z| > 0 and argument θ,

We write e^iθ for cos θ + i sin θ, so the same number is z = re^iθ. This is not a new kind of number: it is shorthand for the trigonometric form. The next section shows why the exponent notation behaves exactly as exponents should.

For example, z = - square root of 3 + i is in the second quadrant, with |z| = square root of (3 + 1) = 2 and reference angle π/6. So arg z = π-π/6 = 5π/6 and

The zero complex number has no argument, so it has no polar form. Every other complex number has exactly one polar form with r > 0 and θ in the principal interval.

Multiplying rotates and scales

Take z₁ = r₁e^iθ₁ and z₂ = r₂e^iθ₂. Multiplying out the trigonometric forms and using the compound-angle formulae gives

So multiplying by z₂ scales every distance by r₂ and rotates anticlockwise through θ₂. Dividing reverses both: for z₂ ≠ 0,

The sum of two principal arguments can leave the interval (-π,π]. The product is still correct, but the principal argument needs a full turn added or subtracted.

De Moivre’s theorem turns powers into multiples

Multiplying e^iθ by itself n times adds θ to the argument n times. This is de Moivre’s theorem: for every integer n,

For positive n it follows by induction from the product rule above: if it holds for n = k, multiply both sides by cos θ + i sin θ and add the arguments. For n = 0 both sides equal 1. For a negative integer n = -m, the reciprocal of cos mθ + i sin mθ is cos mθ-i sin mθ = cos(-mθ) + i sin(-mθ).

So (re^iθ)ⁿ = rⁿe^inθ. To find a high power, convert to polar form first:

Expanding (1-i)¹⁰ term by term would need eleven binomial terms; the polar form needs one multiplication of the argument.

Every nonzero number has n distinct nth roots

To solve zⁿ = w with w = Re^iφ ≠ 0, write z = re^iθ. De Moivre gives rⁿe^inθ = Re^iφ, so rⁿ = R and nθ is some argument of w, not necessarily the principal one:

These n values are different, because their arguments differ by less than a full turn. Any other integer k repeats one of them. All the roots lie on the circle of radius R^(1/n), a turn of 2π/n apart, so they are the vertices of a regular n-gon.

Binomial expansion turns de Moivre into identities

Write c = cos θ and s = sin θ. Expanding (c + is)³ with the binomial theorem and comparing it with cos 3θ + i sin 3θ:

Equating real and imaginary parts is valid because both sides are the same complex number and c, s are real.

The reverse job, writing a power of cos θ in terms of multiple angles, uses z = e^iθ. Then z⁻¹ = e^(-iθ), and

Roots of unity carry exact values

The solutions of zⁿ = 1 are the nth roots of unity e^(2kπ i/n), k = 0,1,…,n-1. For n ≥ 2 they sum to zero: they form a geometric series with ratio ω = e^(2π i/n) ≠ 1 and ωⁿ = 1, so the sum is (1-ωⁿ)/(1-ω) = 0.

Common mistakes

• Giving an argument outside the principal interval, or in the wrong quadrant, when converting to polar form.
• Adding moduli instead of multiplying them when multiplying two complex numbers.
• Finding only one nth root; there are n, spaced 2π/n apart.
• Dividing the argument by n before adding the multiples of 2π, which gives the same root repeatedly.
• Expanding (cos θ + i sin θ)ⁿ and forgetting the powers of i when separating real and imaginary parts.

Before you move on

Complex loci, such as |z-c| ≤ r and arg(z-a) = α, are taught next in Loci in the Argand diagram.