AQA A-Level Further Maths: Complex Numbers — In-Depth Revision Guide (7367)
AQA A-Level Further Maths: Complex Numbers — In-Depth Revision Guide (7367)
Complex numbers are the gateway topic of AQA A-Level Further Mathematics (7367). They are usually the first piece of genuinely new mathematics a Further Maths student meets, and they sit at the heart of the compulsory Pure content examined on Papers 1 and 2. More than that, complex numbers are a beautiful example of how extending a number system unlocks results that are invisible from inside the old one: once you accept a square root of −1, every polynomial factorises completely, rotations become multiplications, and trigonometric identities fall out of algebra almost for free.
This guide is the in-depth companion for AQA's complex numbers course. It runs from the imaginary unit and the arithmetic of complex numbers, through conjugates and their role in finding polynomial roots, into the geometry of the Argand diagram, modulus-argument (polar) form, multiplication and division in polar form, loci, De Moivre's theorem, and the elegant structure of the nth roots of unity. Each section gives the core skills, a worked example, the common pitfalls, and a link to the matching lesson.
The aim is not to replace working through problems — fluency with complex numbers comes only from doing many of them until the algebra and the geometry feel like one thing. The aim of this guide is to give you a clear map of what AQA can examine, in roughly the order they teach it, so your revision is targeted rather than scattered.
What the AQA 7367 Specification Covers
AQA A-Level Further Mathematics (7367) is a linear qualification assessed through three written papers, each 2 hours long, each worth 100 marks, and each carrying 33⅓% of the marks. Papers 1 and 2 are compulsory and cover the Pure content — which is where complex numbers, matrices and further calculus all live. Paper 3 is the optional applications paper, where you choose two of Mechanics, Statistics and Discrete Mathematics. Because complex numbers are compulsory Pure, they can appear on either Paper 1 or Paper 2, often woven into larger questions on polynomials, series or geometry.
Further Mathematics is one of the most highly valued A-Levels for any STEM degree, and is effectively expected for mathematics and many physics and engineering courses at Oxbridge and other competitive universities. The complex-numbers content also overlaps heavily with the kind of structured reasoning rewarded by the STEP and MAT admissions tests.
The table below shows the complex-numbers sub-topics, what each involves, and why it matters across the course.
| Topic | What it involves | Why it matters |
|---|---|---|
| The imaginary unit and complex numbers | Defining i, real and imaginary parts, equality | Foundation for everything that follows |
| Complex number arithmetic | Adding, subtracting, multiplying, dividing | The core algebraic toolkit |
| Complex conjugates and polynomial roots | Conjugate pairs, the conjugate root theorem | Factorising polynomials completely |
| Argand diagrams | Plotting complex numbers as points and vectors | Turns algebra into geometry |
| Modulus and argument (polar form) | r(cosθ+isinθ), converting between forms | Unlocks powers, roots and rotations |
| Multiplication and division in polar form | Multiply moduli, add arguments | The geometric meaning of complex multiplication |
| Loci in the Argand diagram | Circles, perpendicular bisectors, half-lines | Combines geometry, modulus and argument |
| De Moivre's theorem | (cosθ+isinθ)n=cosnθ+isinnθ | Powers, roots and trig identities |
| nth roots of unity | The n solutions of zn=1 | Elegant structure; regular polygons |
These topics build on one another tightly. The arithmetic underpins the conjugate work; the Argand diagram makes modulus-argument form natural; polar form makes De Moivre's theorem almost obvious; and De Moivre gives you the roots of unity. Revising them in order pays off.
The Imaginary Unit and Complex Numbers
The whole subject begins with a single definition: the imaginary unit i satisfies i2=−1. From this, a complex number is any number of the form z=a+bi, where a and b are real. We call a the real part, written Re(z), and b the imaginary part, written Im(z). Note carefully that the imaginary part is the real number b, not bi.
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This "comparing real and imaginary parts" step is one of the most powerful moves in the whole topic: a single complex equation is secretly two real equations, which is exactly what you need to solve for two unknowns.
The powers of i cycle with period four: i1=i, i2=−1, i3=−i, i4=1, and then the pattern repeats. To simplify a high power such as i23, divide the exponent by 4 and use the remainder: 23=4×5+3, so i23=i3=−i.
A short worked example. Find real numbers x and y such that (x+yi)+3(2−i)=7+i. Expand: x+6+(y−3)i=7+i. Comparing real parts gives x+6=7, so x=1. Comparing imaginary parts gives y−3=1, so y=4.
A common pitfall is treating the imaginary part as including the i, which causes sign and bookkeeping errors later. Another is forgetting that i2=−1 turns a multiplication into a sign change. For full coverage, see the Imaginary Unit and Complex Numbers lesson.
Complex Number Arithmetic
Arithmetic with complex numbers follows the ordinary rules of algebra, with the single extra fact that i2=−1. Addition and subtraction act componentwise: (a+bi)±(c+di)=(a±c)+(b±d)i. Multiplication uses the distributive law exactly as you would expand any pair of brackets, after which you replace every i2 by −1 and collect.
Division is the step that needs a technique. To divide by a complex number, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of c+di is c−di, and the product (c+di)(c−di)=c2+d2 is always a real number. This clears the imaginary part out of the denominator, leaving an answer in standard a+bi form.
A short worked example. Simplify 1−i3+2i. Multiply top and bottom by the conjugate 1+i:
1−i3+2i×1+i1+i=(1−i)(1+i)(3+2i)(1+i)=1−i23+3i+2i+2i2=21+5i=21+25i.
A common pitfall is forgetting to apply i2=−1 during expansion, which leaves an answer that is wrong by a sign. Another is multiplying only the numerator by the conjugate, not the denominator — you must multiply by c−dic−di, which equals 1, to keep the value unchanged. For practice in every operation, see the Complex Number Arithmetic lesson.
Complex Conjugates and Polynomial Roots
The complex conjugate of z=a+bi is zˉ=a−bi. Conjugates have three properties AQA expects you to use fluently: z+zˉ=2a (twice the real part, so always real), zzˉ=a2+b2 (the square of the modulus, always real and non-negative), and the conjugate of a sum, product or power is the sum, product or power of the conjugates.
The single most important application is the conjugate root theorem: if a polynomial has real coefficients and z=a+bi is a root, then its conjugate zˉ=a−bi is also a root. Complex roots of real polynomials therefore come in conjugate pairs. This is what lets you factorise completely: a cubic with real coefficients and one complex root must have its conjugate as a second root, leaving a single real root.
A worked example. Given that z=2+i is a root of z3−7z2+17z−15=0, find the other roots. By the conjugate root theorem, 2−i is also a root. The quadratic factor with these roots is (z−(2+i))(z−(2−i))=z2−4z+5. Dividing the cubic by this quadratic gives the linear factor z−3, so the third root is z=3. The full root set is 2+i, 2−i, 3.
A common pitfall is applying the conjugate root theorem to a polynomial whose coefficients are not all real — the theorem only holds for real coefficients. Another is making sign errors when expanding the conjugate-pair quadratic; remember the constant term is a2+b2 and the z-coefficient is −2a. For worked factorisations, see the Complex Conjugates and Polynomial Roots lesson.
Argand Diagrams
The Argand diagram is the plane in which we picture complex numbers. The complex number z=a+bi is plotted at the point with coordinates (a,b): the horizontal axis is the real axis and the vertical axis is the imaginary axis. This single idea — that a complex number is a point in a plane — turns the whole subject into geometry.
A complex number can also be read as a vector from the origin to the point (a,b). This vector picture makes addition visual: adding two complex numbers corresponds to adding their vectors head-to-tail, exactly the parallelogram rule. Subtraction z1−z2 corresponds to the vector that points from z2 to z1, which is the key to interpreting distances on the diagram.
The modulus ∣z∣ is the length of this vector, the distance from the origin to the point — by Pythagoras, ∣z∣=a2+b2. The distance between two points z1 and z2 is ∣z1−z2∣. Recognising distances as moduli of differences is exactly what unlocks loci later.
A short worked example. Plot z1=3+2i and z2=−1+4i, and find the distance between them. The distance is ∣z1−z2∣=∣(3−(−1))+(2−4)i∣=∣4−2i∣=16+4=20=25.
A common pitfall is confusing the roles of the axes, plotting the imaginary part horizontally. Another is forgetting that the modulus is always non-negative — it is a length. For diagrams and vector interpretations, see the Argand Diagrams lesson.
Modulus and Argument (Polar Form)
Instead of describing z by its real and imaginary parts, we can describe it by how far it is from the origin and in which direction it points. The distance is the modulus r=∣z∣=a2+b2, and the direction is the argument θ=arg(z), the angle the vector makes with the positive real axis, measured anticlockwise. This gives the modulus-argument (polar) form:
z=r(cosθ+isinθ).
By convention, AQA uses the principal argument in the range −π<θ≤π. To find the argument you must take account of the quadrant the point lies in: the bare value of tan−1(b/a) only gives the correct angle in the first and fourth quadrants. Always sketch the point first, then adjust the calculator's principal value by π (or 180∘) when the point is in the second or third quadrant.
A worked example. Express z=−1+3i in modulus-argument form. The modulus is r=(−1)2+(3)2=1+3=2. The point lies in the second quadrant. The reference angle is tan−1(3/1)=3π, so the principal argument is θ=π−3π=32π. Therefore z=2(cos32π+isin32π).
A common pitfall is quoting the calculator value of tan−1(b/a) without adjusting for the quadrant, which gives a wrong argument for second- and third-quadrant points. Another is leaving the argument outside the principal range. For conversion practice both ways, see the Modulus and Argument lesson.
Multiplication and Division in Polar Form
Polar form reveals the geometric meaning of complex multiplication, and this is one of the genuinely beautiful results of the course. When you multiply two complex numbers, you multiply the moduli and add the arguments:
z1z2=r1r2(cos(θ1+θ2)+isin(θ1+θ2)).
When you divide, you divide the moduli and subtract the arguments:
z2z1=r2r1(cos(θ1−θ2)+isin(θ1−θ2)).
Geometrically, multiplying by a complex number of modulus r and argument θ scales the plane by r and rotates it by θ. In particular, multiplying by i (modulus 1, argument 2π) is a rotation through a right angle anticlockwise, with no change of size. This rotation-and-scaling picture is the bridge to De Moivre's theorem.
A worked example. Let z1=2(cos6π+isin6π) and z2=3(cos4π+isin4π). Then z1z2=6(cos125π+isin125π), because 2×3=6 and 6π+4π=125π.
A common pitfall is adding the moduli or multiplying the arguments — the rule is the other way round. Another is failing to bring a resulting argument back into the principal range when the sum exceeds π. For worked products and quotients, see the Multiplication and Division in Polar Form lesson.
Loci in the Argand Diagram
A locus is the set of all points satisfying a given condition. Because distances and angles on the Argand diagram are exactly moduli and arguments, geometric conditions on z translate into algebraic equations and back. AQA examines three standard loci.
| Condition | Geometric meaning |
|---|---|
| ∣z−a∣=r | Circle of radius r, centre the point a |
| ∣z−a∣=∣z−b∣ | Perpendicular bisector of the segment joining a and b |
| arg(z−a)=θ | Half-line from a at angle θ to the real axis |
The reasoning behind each is worth holding onto. The first says "the distance from z to the fixed point a is constant," which is the definition of a circle. The second says "z is equidistant from a and b," which is exactly the perpendicular bisector. The third fixes the direction from a, giving a ray (a half-line, not a full line, because the argument fixes both direction and sense).
A worked example. Describe the locus ∣z−(2+i)∣=3 and find where it meets the real axis. This is a circle of radius 3 centred at (2,1). On the real axis, z=x (so y=0): ∣x−2−i∣=3 gives (x−2)2+1=9, so (x−2)2=8 and x=2±22.
A common pitfall is reading arg(z−a)=θ as a full line rather than a half-line, and forgetting that the endpoint a itself is excluded (the argument is undefined there). Another is mis-centring the circle by dropping the sign inside the modulus. For sketched loci and intersection work, see the Loci in the Argand Diagram lesson.
De Moivre's Theorem
De Moivre's theorem states that for any integer n,
(cosθ+isinθ)n=cosnθ+isinnθ.
It follows directly from the multiplication rule in polar form: raising to the nth power multiplies the modulus to the nth power and multiplies the argument by n. With a general modulus, zn=rn(cosnθ+isinnθ). The theorem is the workhorse for computing high powers of complex numbers, which would be hopeless by repeated expansion.
The most elegant application is deriving trigonometric identities. Expanding (cosθ+isinθ)n with the binomial theorem and then comparing real and imaginary parts with cosnθ+isinnθ produces multiple-angle formulae for cosnθ and sinnθ in one stroke. The reverse direction — writing powers such as cosnθ in terms of multiple angles — uses the companion results z+z1=2cosθ and z−z1=2isinθ, where z=cosθ+isinθ.
A worked example. Use De Moivre to express cos3θ in terms of cosθ. Expand (cosθ+isinθ)3=cos3θ+3cos2θ(isinθ)+3cosθ(isinθ)2+(isinθ)3. The real part is cos3θ−3cosθsin2θ. Setting this equal to cos3θ and using sin2θ=1−cos2θ gives cos3θ=4cos3θ−3cosθ.
A common pitfall is forgetting that the multiple-angle expression appears as the real or imaginary part of the expansion, not the whole expansion. Another is dropping the rn factor when the modulus is not 1. For both directions of identity work, see the De Moivre's Theorem lesson.
The nth Roots of Unity
The nth roots of unity are the n solutions of the equation zn=1. Writing 1 in polar form as cos0+isin0 — but remembering that the argument is only fixed up to multiples of 2π — and applying De Moivre in reverse gives the n distinct roots:
zk=cos(n2πk)+isin(n2πk),k=0,1,2,…,n−1.
These roots have a striking structure. They all have modulus 1, so they lie on the unit circle. Their arguments are equally spaced by n2π, so they sit at the vertices of a regular n-gon inscribed in that circle, with one vertex at z=1. Two more facts AQA likes: the roots sum to zero for n≥2 (by symmetry, or as the sum of a geometric series), and every root is a power of the single root ω=cosn2π+isinn2π, so the roots are 1,ω,ω2,…,ωn−1.
The same method finds the nth roots of any complex number: write the number in polar form r(cosθ+isinθ), take the real nth root r1/n of the modulus, and set the arguments to nθ+2πk for k=0,…,n−1.
A worked example. Find the cube roots of unity. Here n=3, so the roots are at arguments 0,32π,34π on the unit circle: z0=1, z1=−21+23i, and z2=−21−23i. As a check, z0+z1+z2=0, and z1,z2 are a conjugate pair, as they must be since z3−1 has real coefficients.
A common pitfall is finding only the obvious real root and missing the complex ones, which happens when you forget that the argument of 1 includes 2πk. Another is using degrees and radians inconsistently. For regular-polygon sketches and roots of general numbers, see the nth Roots of Unity lesson.
Putting It Together: Applications and Problem Solving
The hardest exam questions combine several of these ideas at once. A typical Paper 1 or Paper 2 question might give a complex root of a quartic and ask you to use the conjugate root theorem to factorise it, then plot the roots on an Argand diagram, then describe the locus passing through two of them. Or it might ask you to use De Moivre to derive an identity and then use that identity to solve a trigonometric equation. The skill being tested is moving fluently between the algebraic view (real and imaginary parts), the geometric view (the Argand diagram), and the polar view (modulus and argument).
A short synoptic worked example. The points representing 2+i and 4+3i are joined by a segment; find the equation of the locus of points equidistant from them, and verify it passes through the midpoint. Equidistance gives ∣z−(2+i)∣=∣z−(4+3i)∣, the perpendicular bisector. Squaring and writing z=x+yi: (x−2)2+(y−1)2=(x−4)2+(y−3)2. Expanding and cancelling the quadratic terms gives −4x+4−2y+1=−8x+16−6y+9, which simplifies to 4x+4y=20, i.e. x+y=5. The midpoint (3,2) satisfies 3+2=5, as required.
A common pitfall in these multi-part questions is treating each part in isolation and re-deriving facts already established. Read the whole question first; the earlier parts are usually scaffolding for the later ones. For mixed problem sets and exam-style questions, see the Applications and Problem Solving lesson.
Common Mark-Loss Patterns Across Complex Numbers
Across the whole topic, a small set of habits accounts for a disproportionate share of lost marks. Most are not about content you do not know — they are about content you do know, applied carelessly under exam pressure.
- Forgetting that i2=−1 during expansion, leaving an answer that is wrong by a sign.
- Quoting the calculator value of tan−1(b/a) as the argument without adjusting for the quadrant. Always sketch the point first.
- Leaving an argument outside the principal range −π<θ≤π.
- Applying the conjugate root theorem when the coefficients are not all real — the theorem only holds for real coefficients.
- Adding moduli or multiplying arguments in polar-form multiplication. Multiply the moduli, add the arguments.
- Reading arg(z−a)=θ as a full line rather than a half-line, and forgetting the endpoint is excluded.
- Taking the whole binomial expansion as the multiple-angle formula in De Moivre work, rather than its real or imaginary part.
- Finding only the real root of zn=1 and missing the complex ones spaced around the unit circle.
- Mixing degrees and radians within a single question.
- Not showing enough working. AQA mark schemes award method marks generously when the working is clear; a bare final answer can score fewer marks than a flawed answer with clean method.
A revision plan that explicitly drills these habits — not just the content — will move your grade more than another pass through the textbook.
How LearningBro's AQA Further Maths Complex Numbers Course Helps
LearningBro's AQA A-Level Further Maths: Complex Numbers course is built around the structure of this guide. Each lesson covers one focus area of the compulsory Pure content — the imaginary unit and arithmetic, conjugates and roots, Argand diagrams, polar form, multiplication and division, loci, De Moivre's theorem, the nth roots of unity, and applications — with worked examples, mark-scheme-style solutions, and practice questions designed for spaced revisits.
The course is designed to be used in two ways. As a first pass, you can work through the lessons in order, building the geometry on top of the algebra exactly as the subject is best learned. As a revision tool, you can drop into any single lesson — drilling loci and De Moivre together in the week before a mock, say — and work the practice independently. The AI tutor is available throughout to give targeted hints when you get stuck, without giving away full solutions, and to mark your written working with structured feedback.
If you want one place to revise complex numbers well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the AQA A-Level Further Maths: Complex Numbers course.