CIE A-Level Mathematics Revision
Explore CIE A-Level Mathematics revision topics. Check the official specification for complete requirements for your exam year and route.
Course details
Assessment The assessment includes Paper 1 and Paper 3; see the official specification for the full structure.
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Reviewed course guidance
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Open a point to read what to know, how to use it in an assessment, a plain-language explanation and a self-check. Answers stay hidden until you choose to reveal them.
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A-level component combination
- Pure Mathematics 1 + Pure Mathematics 3 + Mechanics + Probability & Statistics 1
- Pure Mathematics 1 + Pure Mathematics 3 + Probability & Statistics 1 + Probability & Statistics 2
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Topic group
Reviewed requirements
- Core
Scope: Core · A-level component combination: any of Pure Mathematics 1 + Pure Mathematics 3 + Mechanics + Probability & Statistics 1, Pure Mathematics 1 + Pure Mathematics 3 + Probability & Statistics 1 + Probability & Statistics 2 · Pure Mathematics 1, §1.1 Quadratics — solve by substitution a simultaneous pair with one linear and one quadratic equation
Open revision guidance
What to know
- A simultaneous system with one linear and one quadratic equation can be reduced to a single quadratic. Substitute the expression for one variable into the other equation, solve, then recover the second variable.
What to do in an assessment
- Rearrange the linear equation to make one variable the subject, substitute into the quadratic, factor or solve the resulting quadratic, and substitute every root back into the linear equation.
In plain language
For x+y=4 and y=x²−2, use the first equation to write y=4−x. Equating gives 4−x=x²−2, so x²+x−6=0=(x+3)(x−2). Thus x=−3 or 2; the matching y values are 7 and 2. Both ordered pairs satisfy the original equations, so neither solution should be dropped without a reason.
Self-check
Solve x+y=5 and y=x²−2 by substitution.
Show the answer
5−x=x²−2 gives x²+x−7=0, so x=(−1±√29)/2; use y=5−x for each matching solution.
- Core
Scope: Core · A-level component combination: any of Pure Mathematics 1 + Pure Mathematics 3 + Mechanics + Probability & Statistics 1, Pure Mathematics 1 + Pure Mathematics 3 + Probability & Statistics 1 + Probability & Statistics 2 · Pure Mathematics 3, §3.9 Complex numbers — Cartesian form, real and imaginary parts, conjugates, arithmetic and conjugate roots for real-coefficient polynomials
Open revision guidance
What to know
- Write a complex number as z=a+bi, where i²=−1. Its real and imaginary parts are Re(z)=a and Im(z)=b; its conjugate is z*=a−bi. Equality requires matching real and imaginary parts. Real-coefficient polynomials have non-real roots in conjugate pairs.
What to do in an assessment
- Add and subtract real and imaginary parts separately. Multiply using i²=−1. For division, multiply numerator and denominator by the denominator's conjugate so the denominator becomes real.
In plain language
Complex arithmetic follows ordinary algebra once i² is replaced by −1. For example, (3+2i)(1−i)=3−3i+2i−2i²=5−i. A conjugate changes the sign of the imaginary part; this is useful both for division and for checking roots. If a real-coefficient polynomial has 2−3i as a root, its conjugate 2+3i must also be a root. Keep real and imaginary parts visibly separate to avoid treating i as an ordinary variable.
Self-check
Find (4+i)+(2−3i) and the conjugate of 5−2i.
Show the answer
The sum is 6−2i; the conjugate is 5+2i.
- Core
Scope: Core · A-level component combination: any of Pure Mathematics 1 + Pure Mathematics 3 + Mechanics + Probability & Statistics 1, Pure Mathematics 1 + Pure Mathematics 3 + Probability & Statistics 1 + Probability & Statistics 2 · Pure Mathematics 3, §3.9 Complex numbers — modulus, argument, polar form, multiplication/division and square roots
Open revision guidance
What to know
- Polar form is z=r(cosθ+i sinθ), with r=|z| and θ=arg(z). Multiplication multiplies moduli and adds arguments; division divides moduli and subtracts arguments. The argument is given in the syllabus's stated principal range.
What to do in an assessment
- Convert to polar form when multiplying or dividing, apply the modulus and argument rules, and convert back if requested. To find square roots, halve the argument and take the square root of the modulus, giving two opposite roots for a non-zero number.
In plain language
Polar form records a complex number's distance from the origin and its angle. For example, (2 cis π/6)(3 cis π/4)=6 cis 5π/12: the lengths multiply and the angles add. For 5+12i, the modulus is 13 and one square root is 3+2i, since (3+2i)²=5+12i; the other is −3−2i. When reporting an argument, use the required principal interval rather than adding or subtracting a full turn without explanation.
Self-check
What are the two square roots of 5+12i?
Show the answer
3+2i and −3−2i.
- Core
Scope: Core · A-level component combination: any of Pure Mathematics 1 + Pure Mathematics 3 + Mechanics + Probability & Statistics 1, Pure Mathematics 1 + Pure Mathematics 3 + Probability & Statistics 1 + Probability & Statistics 2 · Pure Mathematics 3, §3.9 Complex numbers — loci from distance and argument conditions, including |z−a|<k, |z−a|=|z−b| and arg(z−a)=α
Open revision guidance
What to know
- A complex number z is a point in the Argand diagram. The modulus |z−a| is the distance from z to the fixed point a, while arg(z−a) is the direction of the vector from a to z.
What to do in an assessment
- Translate each condition into a geometric boundary and region. Check whether the inequality is strict, whether the boundary is included, and whether an argument condition describes a ray rather than a full line.
In plain language
The condition |z−a|<k, for k>0, describes the open disk inside the circle of centre a and radius k; the circumference is excluded. The equation |z−a|=|z−b| describes points equally far from a and b, so its locus is their perpendicular bisector. The condition arg(z−a)=α describes a ray starting at a at angle α; the point a itself is excluded because its argument is undefined. For example, |z−1|<2 is the interior of a circle centred at 1 with radius 2.
Self-check
Describe the locus |z−0|=|z−2| in Cartesian coordinates.
Show the answer
The perpendicular bisector of 0 and 2 on the real axis is the vertical line Re(z)=1.
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Mathematics topic list
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