Complex numbers.

HookThe 'impossible' number that solved a real cubic

ModelArithmetic in x + iy — one new symbol, all the old rules

A complex number \(z=x+iy\) has real part \(\operatorname{Re}(z)=x\) and imaginary part \(\operatorname{Im}(z)=y\) — note that the imaginary part is the real number \(y\), not \(iy\). Addition and subtraction work componentwise, exactly like vectors. Multiplication is ordinary expansion of brackets followed by one substitution, \(i^2=-1\): so \((2+3i)(1-i)=2-2i+3i-3i^2=5+i\).

Division is the move students meet last and need most. You cannot 'divide by \(i\)' directly, so you make the denominator real: multiply top and bottom by the complex conjugate of the denominator, \(z^*=x-iy\). The engine behind this is the identity \(zz^*=x^2+y^2\) — a complex number times its conjugate is always real and non-negative. The conjugate also behaves beautifully under algebra: \((z+w)^*=z^*+w^*\) and \((zw)^*=z^*w^*\), which is the fact that powers the conjugate-pairs theorem in the next block.

The other workhorse skill is equating real and imaginary parts: if two complex numbers are equal, their real parts match and their imaginary parts match — one complex equation is secretly two real ones. AQA uses this constantly: finding square roots of a complex number, solving \(z+2z^*=6-4i\), or fixing unknown real constants in a polynomial.

ModelPolynomials — non-real roots hunt in pairs

A quadratic with real coefficients and negative discriminant no longer 'has no roots' — it has two, and they are conjugates of each other: \(z^2-4z+13=0\) gives \(z=2\pm 3i\). This is a special case of the theorem that runs the whole leaf: if a polynomial has real coefficients, its non-real roots occur in conjugate pairs. The proof is one line of conjugate algebra — conjugating the equation \(p(z)=0\) conjugates each root while leaving the real coefficients untouched, so \(p(z^*)=0\) too.

The exam payoff is 'finishing off' cubics and quartics. Hand a cubic with real coefficients one non-real root and you instantly own two: the root and its conjugate. Multiply them into a quadratic factor using \(\big(z-\alpha\big)\big(z-\alpha^*\big)=z^2-2\operatorname{Re}(\alpha)\,z+|\alpha|^2\) — sum and product, no messy expansion — then divide or compare coefficients to find the remaining linear factor. A quartic with two given non-real roots (not conjugates of each other) yields two quadratic factors the same way. Count degrees before you start: a cubic has exactly three roots in \(\mathbb{C}\), a quartic four, multiplicities included. If a cubic with real coefficients has one non-real root, the third root must be real — a conjugate pair plus one more non-real root would need a fourth partner.

ModelThe Argand diagram and modulus-argument form

Plot \(z=x+iy\) as the point \((x,y)\) — real axis horizontal, imaginary axis vertical — and complex numbers become geometry. Addition is vector addition; conjugation is reflection in the real axis. Every non-zero \(z\) then has a second address: its distance from the origin, the modulus \(|z|=\sqrt{x^2+y^2}\), and the angle it makes with the positive real axis, the argument, with the principal value taken in \(-\pi<\theta\le\pi\). Together they give the modulus-argument form \(z=r(\cos\theta+i\sin\theta)\).

Converting is where marks leak. \(x=r\cos\theta\), \(y=r\sin\theta\) one way; \(r=\sqrt{x^2+y^2}\) and \(\tan\theta=y/x\) the other — but \(\arctan(y/x)\) alone cannot tell which quadrant you are in. Always sketch the point first. For \(z=-1+i\), blind calculator work gives \(\arctan(-1)=-\frac{\pi}{4}\); the sketch shows the point in the second quadrant, so \(\arg z=\frac{3\pi}{4}\).

Why bother with a second form? Because multiplication becomes trivial: moduli multiply, arguments add. \(|zw|=|z||w|\) and \(\arg(zw)=\arg z+\arg w\) (adjusting by \(2\pi\) to stay principal); for division, divide the moduli and subtract the arguments. Geometrically, multiplying by \(w\) rotates the whole plane by \(\arg w\) and enlarges it by \(|w|\) — multiplication by \(i\) is exactly a quarter-turn. Cartesian form is built for adding; modulus-argument form is built for multiplying. Fluency means picking the right form before you start.

MechanismLoci — geometry written in modulus and argument

Read \(|z-a|\) as 'the distance from \(z\) to the fixed point \(a\)' and every standard locus becomes a sentence. \(|z-a|=r\): all points at distance \(r\) from \(a\) — a circle, centre \(a\), radius \(r\). \(|z-a|>r\): the region outside that circle (dashed boundary, since the inequality is strict). \(|z-a|=|z-b|\): points equidistant from \(a\) and \(b\) — the perpendicular bisector of the segment joining them. \(\arg(z-a)=\theta\): the half-line from \(a\) at angle \(\theta\) to the positive real direction, excluding the endpoint \(a\) itself, where the argument is undefined.

AQA's favourite escalation is intersecting these: shade the region where \(|z-3-4i|\le 2\) and \(0\le\arg(z-3-4i)\le\frac{\pi}{2}\), say. Translate each condition into geometry separately, draw accurately, then take the overlap. The second favourite is optimisation: the least and greatest values of \(|z|\) (or \(|z-b|\)) on a circle locus. No calculus needed — the extremes always lie on the straight line through the two centres, at distance (centre distance) \(\mp\) radius.

Algebra backstops the picture when you are unsure: substitute \(z=x+iy\) and grind. \(|z-a|=r\) becomes \((x-a_1)^2+(y-a_2)^2=r^2\), the Cartesian circle. But the geometric reading is faster and is what the mark scheme is built around — centre and radius stated, boundary type correct, region shaded.

MechanismDe Moivre's theorem and the exponential form

Multiplying complex numbers adds arguments, so raising to a power multiplies the argument: de Moivre's theorem states \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\) for integer \(n\) — provable for positive integers by induction, the same engine as section A. Euler's relation compresses all of this into one line: \(e^{i\theta}=\cos\theta+i\sin\theta\), so every complex number is \(z=re^{i\theta}\), and de Moivre is just an index law: \(\big(e^{i\theta}\big)^n=e^{in\theta}\). Multiplication, division and powers all become exponent arithmetic.

Two exam applications dominate. Multiple-angle identities: expand \((\cos\theta+i\sin\theta)^n\) binomially, then equate real and imaginary parts with \(\cos n\theta+i\sin n\theta\) — out fall formulae like \(\cos 3\theta\) in powers of \(\cos\theta\). Running the machine the other way, with \(z=e^{i\theta}\), the identities \(z+\frac{1}{z}=2\cos\theta\) and \(z^n+\frac{1}{z^n}=2\cos n\theta\) convert powers like \(\cos^4\theta\) into multiple angles — which is exactly what you need before integrating them.

Series summation: a trigonometric series such as \(C=\sum\cos(k\theta)\) pairs with its sibling \(S=\sum\sin(k\theta)\) to give \(C+iS=\sum e^{ik\theta}\) — a geometric series with common ratio \(e^{i\theta}\). Sum it with the GP formula, tidy, then take real and imaginary parts to read off \(C\) and \(S\) separately. One geometric series, two closed forms.

Casenth roots — the regular polygon inside every equation

The equation \(z^n=re^{i\theta}\) has exactly \(n\) distinct solutions, and the exponential form hands them over: the trap is forgetting that the right-hand side's argument is only defined up to whole turns. Write \(z^n=re^{i(\theta+2k\pi)}\) for \(k=0,1,\dots,n-1\), then take the \(n\)th root: \(z_k=r^{1/n}e^{i(\theta+2k\pi)/n}\). All \(n\) roots share the same modulus \(r^{1/n}\), and consecutive arguments differ by \(\frac{2\pi}{n}\) — so they sit at the vertices of a regular \(n\)-gon centred on the origin. Find one root, rotate by \(\frac{2\pi}{n}\) repeatedly, and you have them all.

The special case \(z^n=1\) gives the roots of unity: \(1,\ \omega,\ \omega^2,\dots,\omega^{n-1}\) where \(\omega=e^{2\pi i/n}\). Two facts do nearly all the geometric work AQA asks for. First, \(1+\omega+\omega^2+\cdots+\omega^{n-1}=0\) — a geometric series with sum \(\frac{\omega^n-1}{\omega-1}=0\); physically, \(n\) equally spaced unit vectors balance perfectly. Second, the roots of \(z^n=a\) are (any one root) \(\times\) (each root of unity), which is why every root diagram is a rotated, scaled copy of the unity picture.

Geometric problems then reverse the flow: given that three points form an equilateral triangle centred at the origin, they must be \(z,\ \omega z,\ \omega^2 z\) for a cube root of unity \(\omega\); side lengths come from \(|z_1-z_0|\); areas from trigonometry on the isosceles triangles at the centre. The polygon is not decoration — it is the answer.

VocabularyKey terms the mark scheme pays for

TrapsMisconceptions that cost marks

ExamWhat examiners want