De Moivre, powers and roots: VCE Specialist Mathematics Units 3 & 4

1. Powers become a geometric operation

Cartesian expansion works for small powers, but it quickly becomes inefficient. Polar form exposes the repeated action. If z = r cis θ, multiplication by z scales a nonzero point by r and rotates it by θ. Repeating that multiplication n times produces modulus rⁿ and angle nθ. De Moivre’s theorem formalises the angular part: (cis θ)ⁿ = cis(nθ) for every integer n.

The statement contains conditions worth separating. The angle θ is real, and cis θ always has modulus 1, so it is never zero. For a general nonzero complex number, [r cis θ]ⁿ = rⁿcis(nθ), with r > 0 and integer n. Positive powers of zero are also defined directly, but zero has no polar argument. Negative powers require a nonzero base, and 0⁰ is not supplied by this theorem.

The integer restriction matters. Replacing n with a fraction and expecting one unique complex value ignores the multiple roots discussed below. The root problem must account for every argument of the target number, including angles differing by 2π. This is why simply dividing a single principal argument often produces only one of several answers.

2. Prove the theorem for every integer power

First establish the multiplication identity cis α cis β = cis(α + β). Expand the left side and use i² = −1. Its real part is cos α cos β − sin α sin β = cos(α + β), and its imaginary part is sin α cos β + cos α sin β = sin(α + β). This angle-addition result is the ingredient for an induction proof.

For positive integers, the base case n = 1 is immediate: (cis θ)¹ = cis θ. Assume (cis θ)^k = cis(kθ) for an arbitrary positive integer k. Then (cis θ)^k+1 = (cis θ)^kcis θ = cis(kθ)cis θ = cis((k + 1)θ). The inductive hypothesis is used in the second equality. Therefore the theorem holds for all positive integers.

For n = 0, (cis θ)⁰ = 1 = cis 0, because the base is nonzero. For a negative integer n = −m with m positive, use cis(mθ)cis(−mθ) = cis 0 = 1. Thus (cis θ)^−m = 1/(cis θ)^m = 1/cis(mθ) = cis(−mθ). This extends the result to negative integers without pretending that induction from 1 already covered them.

Multiplying by rⁿ gives the general polar formula. Check both parts when applying it: the modulus is raised to the power, while the argument is multiplied by the power. The common incorrect expression r cis(nθ) leaves the distance unchanged even when repeated multiplication should scale it.

3. Derive all the nth roots

Let w = R cis Θ be nonzero, with R > 0, and let n be a positive integer. To solve zⁿ = w, write z = ρ cis φ with ρ > 0. Matching moduli gives ρⁿ = R, so ρ = R^1/n, the positive real nth root. Matching arguments gives nφ = Θ + 2kπ for some integer k, not merely nφ = Θ.

Why does this list contain exactly n distinct values? Consecutive arguments differ by 2π/n. Among k = 0 through n − 1, no two differ by a full turn, so their points are distinct. Increasing k by n adds 2π to the angle, returning to the same point. Every possible integer k therefore reproduces one of these n values. This establishes completeness as well as producing a formula.

Choosing a different argument Θ + 2jπ for w changes the order of the list, not the set of roots. You may leave roots in exact cis form unless Cartesian form is requested. If principal arguments are requested, adjust each angle individually. Do not delete a root merely because its initial angle exceeds π; it still represents a valid point.

If w = 0, solve separately: zⁿ = 0 has the unique distinct solution z = 0. It does not give n different points on a radius-zero circle. In polynomial language this root has multiplicity n, which is explained in C03. Distinguishing distinct roots from roots counted with multiplicity prevents an apparent contradiction.

4. Roots of unity organise the pattern

The nth roots of unity solve zⁿ = 1. Since 1 = cis 0, they are cis(2kπ/n) for k = 0, 1, …, n − 1. They lie on the unit circle, with one root at 1. For n ≥ 3 they are the vertices of a regular n-sided polygon centred at the origin. For n = 2 the two roots are opposite endpoints of a diameter; n = 1 gives just 1.

If ω = cis(2π/n), the roots can be listed as 1, ω, ω², …, ωⁿ⁻¹. Multiplying a root by ω rotates it one position around the set. After n steps, ωⁿ = 1 returns to the starting point. More generally, if a is one nth root of a nonzero w, multiplying a by each nth root of unity generates all the roots of w. This explains the shared radius and equally spaced directions.

For n > 1, the roots of unity sum to zero. One algebraic explanation uses S = 1 + ω + ⋯ + ωⁿ⁻¹. Multiplication gives ωS = ω + ⋯ + ωⁿ = S because ωⁿ = 1. Hence (ω − 1)S = 0. Our chosen ω is not 1 when n > 1, so S = 0. The argument is a finite-sum calculation, and it agrees with the balanced geometric arrangement.

5. Exact work, numerical checks and common traps

Before computing roots, record the expected count, common modulus and spacing. After computing, check each defining equation, either using exact polar multiplication or exact Cartesian expansion. In a technology-active task, plot the root set and evaluate zⁿ − w. Small decimal residuals are evidence of numerical accuracy, not proof that there are no omitted roots. The derivation and the count establish completeness.

Use brackets carefully when entering ((Θ + 2kπ)/n). Dividing only Θ by n while leaving 2kπ undivided generates the same point repeatedly. Similarly, 2π/n is the spacing between roots, not the argument of every first root. The target argument determines the rotation of the entire polygon. A calculator’s single complex-root output may use one branch; the equation-solving task still requires the complete set.

De Moivre’s theorem is especially efficient when the base has an exact polar form. If a question instead asks for a Cartesian factorisation, carry the exact roots back into linear factors. Keep rounded angles out of an exact polynomial calculation: a tiny angle error can create a nonzero remainder when the exact root should cancel it.