Pure Mathematics Problem-Solving Strategies

Papers 1 and 2 are the compulsory Pure core of AQA 7367 — two 2-hour, 100-mark papers drawing from the same pool: complex numbers, matrices, further algebra and functions, further calculus, and polar/hyperbolic work, all underpinned by proof. Recall the per-paper AO balance: AO1 55% / AO2 25% / AO3 20%, so the majority of marks reward fluent, accurate technique, with a substantial quarter for reasoning and proof. This lesson is about how to attack these questions: how to read a "show that", how to structure a proof, how to keep method marks alive in a long multi-step problem, and where candidates routinely leak marks.

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A general method for any Pure question

The Pure papers reward a small number of habits applied relentlessly. Most candidates who underperform do not lack knowledge of complex numbers or matrices; they lose marks in predictable ways — by skipping the method line that earns the M mark, by misreading a command word, by giving a decimal where an exact form was wanted, or by stopping a proof one line before its required conclusion. The strategies below are therefore as much about how you write as about what you know. Before the topic-specific advice, here is a routine that works across the whole paper:

1. Decode the command word. "Show that" and "prove" mean the marks are in the working; "hence" forces the previous result; "find" wants the answer with method.
2. Write the relevant formula or definition first, then substitute. The formula line is usually the method (M) mark, and it is safe even if your arithmetic later wobbles.
3. Work in exact form unless told otherwise — surds, $\pi$π, $e$e, $\ln$ln, fractions.
4. Signpost the logic with $\implies$⟹, $\therefore$∴ and short words ("equating real parts", "since $z$z is real"). AO2 marks are for communicated reasoning.
5. Conclude in words where a question asks you to show, deduce or interpret — a bare final line of algebra is often not enough.

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Complex numbers

Loci in the Argand diagram

The standard loci are worth memorising as a set, because the whole question usually turns on identifying which one you have:

Locus	Geometric meaning
$\lvert z - z_1 \rvert = r$∣z−z1​∣=r	Circle, centre $z_1$z1​, radius $r$r
$\lvert z - z_1 \rvert = \lvert z - z_2 \rvert$∣z−z1​∣=∣z−z2​∣	Perpendicular bisector of the segment joining $z_1$z1​ and $z_2$z2​
$\arg(z - z_1) = \theta$arg(z−z1​)=θ	Half-line from $z_1$z1​ at angle $\theta$θ to the positive real direction
$\lvert z - z_1 \rvert \le r$∣z−z1​∣≤r	Closed disc (filled circle)

Strategy. Sketch first, label centre/radius/key points, then compute. For a combined locus (e.g. a circle intersected with a half-line) solve the two conditions simultaneously and check the intersection actually lies on the half-line (the algebra can produce a point on the full line but on the wrong side).

De Moivre's theorem and roots

$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.$(cosθ+isinθ)n=cosnθ+isinnθ.

For roots of $z^n = w$zn=w: write $w = r\,e^{i\alpha}$w=reiα, then

$z_k = r^{1/n}\,e^{\,i(\alpha + 2k\pi)/n}, \qquad k = 0,1,\dots,n-1,$zk​=r1/nei(α+2kπ)/n,k=0,1,…,n−1,

giving $n$n roots equally spaced by $2\pi/n$2π/n on a circle of radius $r^{1/n}$r1/n.

Exam Tip: A question asking for the $n$n-th roots wants all $n$n of them. Finding only the principal root is the single most common complex-number error. State every argument and put them on a labelled Argand diagram.

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Worked example 1 — complex numbers ("show that", with mark scheme)

Specimen-style. Use De Moivre's theorem to show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$cos3θ=4cos3θ−3cosθ. (5 marks)

Solution.

By De Moivre, $\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3$cos3θ+isin3θ=(cosθ+isinθ)3. Expand the right-hand side with the binomial theorem:

$(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3\cos^2\theta\,(i\sin\theta) + 3\cos\theta\,(i\sin\theta)^2 + (i\sin\theta)^3.$(cosθ+isinθ)3=cos3θ+3cos2θ(isinθ)+3cosθ(isinθ)2+(isinθ)3.

Using $i^2 = -1$i2=−1 and $i^3 = -i$i3=−i:

$= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta.$=cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ.

Equate real parts:

$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta.$cos3θ=cos3θ−3cosθsin2θ.

Replace $\sin^2\theta = 1 - \cos^2\theta$sin2θ=1−cos2θ:

$\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = \cos^3\theta - 3\cos\theta + 3\cos^3\theta = 4\cos^3\theta - 3\cos\theta. \quad \blacksquare$cos3θ=cos3θ−3cosθ(1−cos2θ)=cos3θ−3cosθ+3cos3θ=4cos3θ−3cosθ.■

How the 5 marks are earned:

• M1 — writing $\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3$cos3θ+isin3θ=(cosθ+isinθ)3 (correct use of De Moivre).
• M1 — expanding the cube by the binomial theorem.
• A1 — correct expansion with $i^2,\ i^3$i2, i3 simplified.
• M1 — equating real parts to isolate $\cos 3\theta$cos3θ.
• A1 — using $\sin^2\theta = 1-\cos^2\theta$sin2θ=1−cos2θ to reach the printed result, with a concluding statement.

Where candidates lose marks: equating imaginary parts by mistake (that gives $\sin 3\theta$sin3θ); forgetting that $i^3 = -i$i3=−i; and — on a "show that" — stopping at $\cos^3\theta - 3\cos\theta\sin^2\theta$cos3θ−3cosθsin2θ without converting to the required form. The instruction was to reach the given expression, so the last line is not optional.

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Matrices

Eigenvalues and eigenvectors

Eigenvalues solve $\det(A - \lambda I) = 0$det(A−λI)=0. For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$A=(ac​bd​) this is fastest as

$\lambda^2 - (a+d)\lambda + (ad - bc) = 0, \qquad\text{i.e.}\qquad \lambda^2 - \operatorname{tr}(A)\,\lambda + \det(A) = 0.$λ2−(a+d)λ+(ad−bc)=0,i.e.λ2−tr(A)λ+det(A)=0.

For each $\lambda$λ, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$(A−λI)v=0 for a (non-zero) eigenvector. Any non-zero scalar multiple is a valid eigenvector — quote the simplest integer form.

Transformations and their order

Matrix	Transformation
$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$(cosθsinθ​−sinθcosθ​)	Rotation $\theta$θ anticlockwise about $O$O
$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$(10​0−1​)	Reflection in the $x$x-axis
$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$(01​10​)	Reflection in $y = x$y=x
$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$(k0​0k​)	Enlargement, scale factor $k$k, centre $O$O

Order is the classic trap. If transformation $P$P is applied first, then $Q$Q, the combined matrix is $QP$QP — the second transformation sits on the left, because it multiplies the result of the first. Writing $PQ$PQ is a guaranteed lost mark.

The Cayley–Hamilton theorem

Every square matrix satisfies its own characteristic equation: if $\lambda^2 - p\lambda + q = 0$λ2−pλ+q=0 is the characteristic equation of $A$A, then

$A^2 - pA + qI = 0.$A2−pA+qI=0.

A favourite use is rewriting the inverse without computing it directly: multiplying through by $A^{-1}$A−1 gives $A - pI + qA^{-1} = 0$A−pI+qA−1=0, so

$A^{-1} = \frac{1}{q}\,(pI - A), \qquad q = \det(A) \neq 0.$A−1=q1​(pI−A),q=det(A)=0.

Exam Tip: Cayley–Hamilton turns "find $A^{-1}$A−1" or "express $A^4$A4 in terms of $A$A and $I$I" into pure algebra — far less error-prone than a $3\times 3$3×3 cofactor expansion. State the characteristic equation, then manipulate.

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Proof by induction: the template that scores

Induction is the highest-frequency AO2 task on Papers 1 and 2, and the marks are as much for structure as for algebra:

1. Base case — verify for the first value (state both sides).
2. Inductive hypothesis — "assume true for $n=k$n=k" (label it).
3. Inductive step — state the target for $n=k+1$n=k+1, then derive it using the hypothesis.
4. Conclusion — the formal sentence naming the base case, the implication, and the principle of induction.

Type	Example
Summation	$\sum_{r=1}^n r^2 = \tfrac{1}{6}n(n+1)(2n+1)$∑r=1n​r2=61​n(n+1)(2n+1)
Divisibility	$3^{2n} - 1$32n−1 is divisible by $8$8
Matrix powers	$A^n = \begin{pmatrix} 2^n & 0 \\ 0 & 3^n \end{pmatrix}$An=(2n0​03n​)
Inequality	$2^n > n^2$2n>n2 for $n \ge 5$n≥5

Exam Tip: The conclusion is a marked line. "Since the statement holds for $n=1$n=1, and truth for $n=k$n=k implies truth for $n=k+1$n=k+1, it holds for all $n\ge 1$n≥1 by induction" should be muscle memory. Omitting it is the most common reason a perfect-looking induction loses its final mark.

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Further algebra and functions

Roots of polynomials and series

For a polynomial with roots $\alpha,\beta,\gamma,\dots$α,β,γ,…, the relationships between the roots and the coefficients (the symmetric functions $\sum\alpha$∑α, $\sum\alpha\beta$∑αβ, $\alpha\beta\gamma$αβγ, …) let you answer "find a polynomial whose roots are …" or "find $\sum\alpha^2$∑α2" without finding the roots themselves. The key identity worth carrying is

$\sum \alpha^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta,$∑α2=(∑α)2−2∑αβ,

which converts the (unknown) sum of squares of roots into the (known) coefficients. For series, the standard results

$\sum_{r=1}^{n} r = \tfrac{1}{2}n(n+1), \qquad \sum_{r=1}^{n} r^2 = \tfrac{1}{6}n(n+1)(2n+1), \qquad \sum_{r=1}^{n} r^3 = \tfrac{1}{4}n^2(n+1)^2$∑r=1n​r=21​n(n+1),∑r=1n​r2=61​n(n+1)(2n+1),∑r=1n​r3=41​n2(n+1)2

are the building blocks, and the method of differences handles the rest: if you can write the general term as $f(r) - f(r+1)$f(r)−f(r+1), the sum telescopes and almost everything cancels.

Exam Tip: When a summation looks awkward, try partial fractions on the general term first — it is the usual gateway to a telescoping (method-of-differences) sum.

Further calculus

The Pure papers draw heavily on calculus that extends A-Level:

Task	Key idea
Improper integral	Replace the infinite limit (or singular point) by a variable, integrate, then take the limit; state convergence/divergence
Volume of revolution	$V = \pi\displaystyle\int_a^b y^2\,dx$V=π∫ab​y2dx (about the $x$x-axis) or $\pi\displaystyle\int_c^d x^2\,dy$π∫cd​x2dy (about the $y$y-axis)
Reduction formula	Integrate by parts to express $I_n$In​ in terms of $I_{n-1}$In−1​, then recurse to a base case
Differential equations	Identify the type (separable, integrating factor, second-order with auxiliary equation) before solving

The discipline that earns method marks here is the same throughout: write the standard formula, then substitute, and never drop the constant of integration in an indefinite integral.

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Polar coordinates and hyperbolic functions

Polar and hyperbolic work are firmly part of the compulsory Pure pool. For polar curves, the area swept by a radius vector is

$A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta,$A=21​∫αβ​r2dθ,

so most polar questions reduce to setting up the correct limits $\alpha,\beta$α,β (often where $r = 0$r=0) and integrating $r^2$r2. For hyperbolic functions, the definitions

$\cosh x = \frac{e^x + e^{-x}}{2}, \qquad \sinh x = \frac{e^x - e^{-x}}{2}$coshx=2ex+e−x​,sinhx=2ex−e−x​

generate every identity you need. The most important is the analogue of the Pythagorean identity,

$\cosh^2 x - \sinh^2 x = 1,$cosh2x−sinh2x=1,

and the derivatives $\frac{d}{dx}\sinh x = \cosh x$dxd​sinhx=coshx, $\frac{d}{dx}\cosh x = \sinh x$dxd​coshx=sinhx (note: no minus sign, unlike the circular functions).

Exam Tip: A "show that" on a hyperbolic identity is almost always fastest by substituting the exponential definitions and simplifying — far more reliable than trying to remember a long list of identities under pressure.

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Worked example 2 — multi-step matrix question (M/A scheme + grade-band)