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EnglandFurther MathsSyllabus dot point

How do complex numbers extend the real numbers, and how do you add, multiply, divide and represent them geometrically?

Solving quadratic, cubic and quartic equations with complex roots, arithmetic of complex numbers, the Argand diagram, modulus-argument form, de Moivre's theorem and loci.

A focused answer to the AQA A-Level Further Mathematics complex numbers content, covering the arithmetic of complex numbers, the Argand diagram, modulus-argument and exponential form, de Moivre's theorem, complex roots of polynomials, roots of unity and loci.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. Arithmetic and the conjugate
  3. The Argand diagram and modulus-argument form
  4. De Moivre's theorem
  5. Roots of polynomials
  6. Roots of unity and the nth roots of a complex number
  7. Loci on the Argand diagram

What this dot point is asking

AQA wants you to work fluently with complex numbers in Cartesian, modulus-argument and exponential form, plot them on an Argand diagram, use de Moivre's theorem to find powers and roots, solve polynomial equations with complex roots, and sketch loci defined by modulus and argument conditions.

Arithmetic and the conjugate

For z=x+yiz = x + yi the complex conjugate is z=xyiz^* = x - yi. Multiplying gives zz=x2+y2=z2zz^* = x^2 + y^2 = |z|^2, which is always real and non-negative. To divide, multiply numerator and denominator by the conjugate of the denominator.

The Argand diagram and modulus-argument form

The modulus is r=z=x2+y2r = |z| = \sqrt{x^2 + y^2} and the argument θ=argz\theta = \arg z is the angle from the positive real axis, measured in (π,π](-\pi, \pi]. Always sketch the point first so you pick the correct quadrant for θ\theta.

De Moivre's theorem

For any integer nn, (cosθ+isinθ)n=cosnθ+isinnθ(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta. This is the engine for powers, for the nnth roots of a complex number, and for deriving multiple-angle identities such as cos3θ=4cos3θ3cosθ\cos 3\theta = 4\cos^3\theta - 3\cos\theta.

Roots of polynomials

For a polynomial with real coefficients, complex roots occur in conjugate pairs. If 2+3i2 + 3i is a root then so is 23i2 - 3i, and their sum and product give a real quadratic factor.

Roots of unity and the nth roots of a complex number

To find all nnth roots of a complex number w=R(cosϕ+isinϕ)w = R(\cos\phi + i\sin\phi), write ww in modulus-argument form, take the real nnth root of the modulus, and divide the argument (with all its 2πk2\pi k equivalents) by nn. The nn roots are

zk=R1/n(cosϕ+2πkn+isinϕ+2πkn),k=0,1,,n1.z_k = R^{1/n}\left(\cos\frac{\phi + 2\pi k}{n} + i\sin\frac{\phi + 2\pi k}{n}\right), \qquad k = 0, 1, \ldots, n-1.

They all share the modulus R1/nR^{1/n} and are equally spaced by 2πn\frac{2\pi}{n} around a circle, so on an Argand diagram they form the vertices of a regular nn-gon. The nnth roots of unity in particular are powers of ω=e2πi/n\omega = e^{2\pi i/n}, and they sum to zero because they are the roots of zn1=0z^n - 1 = 0, whose zn1z^{n-1} coefficient is zero. This geometric picture is examined directly: AQA often asks you to mark the roots on a diagram and comment on their symmetry.

Loci on the Argand diagram

Conditions on zz describe curves, and the marks come from translating the algebra into a labelled sketch. za=r|z - a| = r is a circle of radius rr centred at the point aa, because za|z - a| is the distance from zz to aa. za=zb|z - a| = |z - b| is the perpendicular bisector of the segment joining aa and bb, the set of points equidistant from the two. arg(za)=θ\arg(z - a) = \theta is a half-line (ray) starting at aa and making angle θ\theta with the positive real direction, with the start point aa itself excluded since arg0\arg 0 is undefined. Inequalities such as zar|z - a| \leq r shade a region (here the closed disc), and intersections of two loci pick out the points or arcs satisfying both conditions at once.

These tools combine on every paper: arithmetic and the conjugate for division, de Moivre for powers and roots, the conjugate-pair fact for polynomials, and the Argand diagram for loci and the geometry of roots of unity.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20195 marksThe equation z35z2+17z13=0z^3 - 5z^2 + 17z - 13 = 0 has z=1z = 1 as one root. Find the other two roots, giving your answers in the form a+bia + bi.
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Method markers reward: factor out the known real root, then solve the remaining quadratic.

Since z=1z = 1 is a root, (z1)(z - 1) is a factor. Dividing gives z35z2+17z13=(z1)(z24z+13)z^3 - 5z^2 + 17z - 13 = (z - 1)(z^2 - 4z + 13).

Solve z24z+13=0z^2 - 4z + 13 = 0 by the quadratic formula: z=4±16522=4±362=4±6i2=2±3iz = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i.

So the other two roots are 2+3i2 + 3i and 23i2 - 3i, a conjugate pair as expected for a real polynomial.

AO1 method marks: 1 for the factor, 1 for the quadratic factor, 1 for using the discriminant, 2 for both roots correct in a+bia + bi form.

AQA 20216 marksUse de Moivre's theorem to express cos4θ\cos 4\theta in terms of powers of cosθ\cos\theta.
Show worked answer →

By de Moivre, cos4θ+isin4θ=(cosθ+isinθ)4\cos 4\theta + i\sin 4\theta = (\cos\theta + i\sin\theta)^4.

Expand with the binomial theorem, writing c=cosθc = \cos\theta, s=sinθs = \sin\theta:
(c+is)4=c4+4c3(is)+6c2(is)2+4c(is)3+(is)4=c4+4ic3s6c2s24ics3+s4(c + is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4 = c^4 + 4ic^3 s - 6c^2 s^2 - 4ic s^3 + s^4.

Take the real part: cos4θ=c46c2s2+s4\cos 4\theta = c^4 - 6c^2 s^2 + s^4.

Replace s2=1c2s^2 = 1 - c^2: cos4θ=c46c2(1c2)+(1c2)2=c46c2+6c4+12c2+c4\cos 4\theta = c^4 - 6c^2(1 - c^2) + (1 - c^2)^2 = c^4 - 6c^2 + 6c^4 + 1 - 2c^2 + c^4.

Collecting, cos4θ=8cos4θ8cos2θ+1\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1.

Markers reward de Moivre stated, a correct binomial expansion, taking the real part, and the Pythagorean substitution to eliminate sinθ\sin\theta.

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