1The idea in one paragraph

Two facts carry this subtopic. First, if a polynomial has real coefficients, its non-real roots come in conjugate pairs: if 2 + i is a root, so is 2 − i. That lets you find every root of a cubic or quartic from a single complex one. Second, De Moivre's theorem: multiplying in polar form multiplies moduli and adds arguments, so doing it n times gives (r cis θ)ⁿ = rⁿ cis nθ. Powers of complex numbers become one line of arithmetic. Run the theorem backwards and it finds roots: the equation zⁿ = w has exactly n solutions, spaced evenly round a circle, found by remembering that an argument can have any multiple of 2π added to it.

2Complex roots come in conjugate pairs

You saw in 1.12 that x² − 4x + 13 = 0 has roots 2 ± 3i: a pair of conjugates. This is always true of a polynomial with real coefficients, whatever its degree.

If p(z) is a polynomial with real coefficients and p(w) = 0, then p(w*) = 0. Non-real roots come in conjugate pairs.

Why. Conjugating respects sums and products: (z + w)* = z* + w* and (zw)* = z*w*, so (zⁿ)* = (z*)ⁿ. A real coefficient is its own conjugate. So taking the conjugate of every term of p(w) gives p(w*). If p(w) = 0, then p(w*) = 0* = 0.

The word real is essential. z² − iz + 2 = 0 has roots 2i and −i, which are not conjugates of each other, because the coefficient −i is not real.

Consequences. A real polynomial of odd degree has at least one real root, because the non-real ones use up an even number. A real cubic has either three real roots, or one real root and a conjugate pair. Figure 1 shows the second case: the graph of the cubic crosses the x-axis only once, and the two missing roots are mirror images in the real axis.

The quadratic factor of a conjugate pair. If w and w* are roots, then (z − w)(z − w*) is a factor, and it has real coefficients:

since w + w* = 2Re(w) and ww* = |w|². For w = 2 − i: z² − 4z + 5. You can write it down in one line, which is the fastest route in most questions.

Worked example 1. Paper 1. Given that 2 − i is a root of z³ − 5z² + 9z − 5 = 0, find the other roots.

The first line must say why 2 + i is a root: "the coefficients are real". That sentence is the reasoning mark.

Worked example 2. Paper 1. Given that 1 + 2i is a root of z⁴ − 2z³ + az² − 2z + 5 = 0, where a ∈ ℝ, find a and the other three roots.

The line "z: 5p − 2q = −2 ✓" is a free check: four equations for three unknowns, and the spare one confirms the others. AHL 2.12 gives another route: the sum of the roots of this quartic is 2 and the product is 5.

3De Moivre's theorem

In 1.13 you multiplied two numbers in polar form by multiplying the moduli and adding the arguments. Multiply z = r cis θ by itself and the rule gives z² = r² cis 2θ; again, z³ = r³ cis 3θ; and so on. That is the theorem.

De Moivre's theorem: [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) = rⁿ cis nθ = rⁿe^(inθ). It is in the formula booklet.

In Euler form it is simply (re^(iθ))ⁿ = rⁿe^(inθ), the law of indices. Figure 2 shows the powers of one number spiralling out: each step multiplies the length by the same factor and turns by the same angle.

Worked example 3. Paper 1. Find (1 + i)¹⁰ in the form a + bi.

Expanding (1 + i)¹⁰ with the binomial theorem would take eleven terms. De Moivre takes three lines.

When is a power real? zⁿ is real exactly when its argument nθ is a multiple of π. For z = 1 + i = √2 cis(π/4), zⁿ = (√2)ⁿ cis(nπ/4) is real when nπ/4 is a multiple of π, that is, when n is a multiple of 4. The smallest positive value is n = 4, giving (1 + i)⁴ = 4 cis π = −4.

4Proving De Moivre's theorem by induction

The guide requires the proof for positive integers n. It is the standard induction structure from 1.15, applied to complex numbers. Write P(n) for the statement (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. (The rⁿ factor follows at once, since (rz)ⁿ = rⁿzⁿ.)

Worked example 4. Prove by mathematical induction that (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for all n ∈ ℤ⁺.

Conclusion. P(1) is true, and P(k) true implies P(k + 1) true. So, by the principle of mathematical induction, P(n) is true for all n ∈ ℤ⁺.

Every line of that is worth marks: the base case shown (not just "true for n = 1"), the assumption stated with "for some k", the step using the assumption visibly, the compound angle formulae named, and the conclusion written out. Section 1.15 explains why each is required.

5Negative and rational exponents

Negative integers. Since 1/(r cis θ) = (1/r) cis(−θ), from 1.13, a negative power is a positive power of that reciprocal:

So De Moivre's theorem holds for every integer n. For example (√3 − i)⁻³: √3 − i = 2 cis(−π/6), so (√3 − i)⁻³ = 2⁻³ cis(π/2) = i/8.

Rational exponents. Put n = 1/q and the theorem says (r cis θ)^(1/q) = r^(1/q) cis(θ/q). That is a qth root: raise it to the power q and you get r cis θ back. But it is not the only one, because the argument of w is not unique. r cis θ = r cis(θ + 2π) = r cis(θ + 4π) = …, and dividing each of those arguments by q gives a different answer. This is the extension to rational exponents that the guide names, and it is exactly the method for finding roots.

6The nth roots of a complex number

To solve zⁿ = w, write w in polar form with every possible argument, then take the nth root. Figure 5, at the end of this section, sets out the four steps.

Worked example 5. Paper 1. Solve z³ = 8i, giving your answers in the form a + bi.

k = 3 gives 2 cis(13π/6) = 2 cis(π/6) again, so there are exactly three roots. Check one: (−2i)³ = −8i³ = −8(−i) = 8i.

Figure 3 shows what always happens. The n roots of zⁿ = w all have the same modulus, |w|^(1/n), so they lie on one circle; and consecutive arguments differ by 2π/n, so they are evenly spaced, forming a regular n-sided polygon centred on the origin.

zⁿ = R cis φ has exactly n roots: z = R^(1/n) cis((φ + 2πk)/n), for k = 0, 1, …, n − 1. They lie on a circle of radius R^(1/n), 2π/n apart.

Roots of unity. The roots of zⁿ = 1 are the nth roots of unity. Since 1 = cis(2πk):

Writing ω = cis(2π/n), the roots are 1, ω, ω², …, ωⁿ⁻¹, the corners of a regular n-gon inscribed in the unit circle, one corner at 1. Figure 4 draws the fifth roots of unity.

They add up to zero. 1 + ω + ω² + … + ωⁿ⁻¹ is a geometric series with ratio ω ≠ 1, so its sum is (ωⁿ − 1)/(ω − 1) = (1 − 1)/(ω − 1) = 0. Geometrically, the arrows from the centre of a regular polygon to its corners balance. For the cube roots of unity, 1 + ω + ω² = 0 is used constantly: it lets you simplify any expression in ω.

On Paper 2 the argument will often not be a special angle. The method is identical; give the roots in polar form with the modulus and arguments to 3 s.f., and check the arguments are in the range the question asks for.

7Multiple-angle identities

The guide links De Moivre's theorem to the compound angle identities of AHL 3.10. Expand (cos θ + i sin θ)ⁿ two ways, by De Moivre and by the binomial theorem, and equate real and imaginary parts. Write c = cos θ and s = sin θ.

Worked example 6. Show that cos 3θ = 4cos³θ − 3cos θ.

Equating the imaginary parts gives sin 3θ = 3c²s − s³, which becomes 3 sin θ − 4 sin³θ using c² = 1 − s²; that is Q4 of Try it.

A related Paper 3 idea: if z = cis θ then zⁿ + z⁻ⁿ = cis nθ + cis(−nθ) = 2cos nθ. Expanding (z + 1/z)³ = (2cos θ)³ then writes cos³θ in terms of cos 3θ and cos θ: cos³θ = ¼(cos 3θ + 3cos θ).