Lesson 12: Complex Numbers (Intro)

Complex numbers begin with a simple impossibility: no real number squares to −1. Rather than stopping there, mathematics invents i with i² = −1 — and gains the Argand diagram, elegant solutions to every polynomial, and the language of AC circuits and quantum mechanics. This introductory lesson covers complex arithmetic, the Argand diagram, modulus and argument, and the first applications. Further Maths goes deeper (de Moivre, roots of unity); here we build the foundations every STEM student needs.

Inventing i — arithmetic of complex numbers

Define i by i² = −1. A complex number is z = x + iy (x, y real): x the real part Re(z), y the imaginary part Im(z). Arithmetic follows from i² = −1: (3 + 2i) + (1 − 5i) = 4 − 3i; (3 + 2i)(1 − 5i) = 3 − 15i + 2i − 10i² = 3 − 13i + 10 = 13 − 13i. Multiplication uses FOIL plus the i² replacement — the only new rule in the whole subject.

Division uses the complex conjugate: z̄ = x − iy. Then z·z̄ = (x + iy)(x − iy) = x² + y² (real!). To divide, multiply top and bottom by the conjugate: (3 + 2i)/(1 − i) = (3 + 2i)(1 + i)/((1 − i)(1 + i)) = (3 + 3i + 2i + 2i²)/2 = (1 + 5i)/2 = 0.5 + 2.5i. The conjugate is the workhorse: it realises denominators, and z + z̄ = 2Re(z), z − z̄ = 2i Im(z) extract parts. Note z̄z = |z|² — preview of modulus.

Why is this legitimate, not wordplay? Because the system is consistent: all field axioms hold, and every polynomial equation gains solutions — x² + 1 = 0 gives x = ±i. The fundamental theorem of algebra (every degree-n polynomial has n complex roots) makes ℂ algebraically complete where ℝ was not. Cardano's 16th-century cubic formula forced mathematicians to compute with √(−1) centuries before they understood it — utility preceded rigour, as often in mathematics.

The Argand diagram — geometry of ℂ

Plot z = x + iy as the point (x, y): the Argand diagram turns complex numbers into plane geometry. Addition is the parallelogram law (vectors!); the conjugate z̄ is the reflection in the real axis. Multiplication by i rotates by 90° anticlockwise: i(1 + i) = i + i² = −1 + i — check on the diagram: (1, 1) → (−1, 1), indeed a quarter-turn about the origin. This rotation property is why complex numbers model rotations, waves, and alternating current.

The modulus |z| = √(x² + y²) is the distance from the origin — Pythagoras again. |z − w| is the distance between points z and w: so |z − (3 + 4i)| = 5 is the circle centre (3, 4) radius 5, and |z − a| = |z − b| is the perpendicular bisector of a and b (locus language from Lesson 3, now in ℂ). Worked: |(2 + 3i) − (5 − i)| = |(−3 + 4i)| = 5.

Triangle inequality: |z + w| ≤ |z| + |w| — the shortest path is direct, visible on the Argand diagram as the triangle inequality for vectors. Equality when z and w are positive real multiples (same direction). These geometric readings turn algebra into pictures: 'find the minimum of |z − (1 + i)| given |z| = 2' is a nearest-point-to-circle problem, solved by drawing, not computing.

Modulus–argument (polar) form

Write z = r(cos θ + i sin θ), where r = |z| and θ = arg z, the angle from the positive real axis (−π < θ ≤ π, the principal argument). Example: z = 1 + √3 i: r = 2, and tan θ = √3 with x > 0 → θ = π/3, so z = 2(cos(π/3) + i sin(π/3)). For z = −1 + i: r = √2, θ = 3π/4 (second quadrant — check the signs, not just arctan!). The argument's quadrant care matches the R-formula care in Lesson 7 — same underlying issue.

Multiplication in polar form is beautiful: r₁(cos θ₁ + i sin θ₁) × r₂(cos θ₂ + i sin θ₂) = r₁r₂(cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)) — proved via compound-angle formulae. Moduli multiply, arguments add: multiplication is scaling plus rotation. So multiplying by (cos(π/6) + i sin(π/6)) rotates by 30°; powers become trivial: zⁿ = rⁿ(cos nθ + i sin nθ) — de Moivre's theorem (proved by induction), the engine of Further Maths.

Worked: (1 + i)¹⁰. Polar: √2(cos(π/4) + i sin(π/4)). Power: (√2)¹⁰ (cos(10π/4) + i sin(10π/4)) = 32(cos(5π/2) + i sin(5π/2)) = 32(0 + i) = 32i. Ten lines of binomial expansion avoided. Division: moduli divide, arguments subtract. Roots: z^{1/n} has n values with arguments (θ + 2πk)/n — the n-th roots of unity sit on the unit circle as a regular n-gon, one vertex at angle θ/n.

Worked example (de Moivre in Cartesian clothing). Find (1 − i√3)³. Polar form: r = √(1 + 3) = 2, arg = −π/3 (fourth quadrant: x > 0, y < 0). Cube: 2³ (cos(−π) + i sin(−π)) = 8(−1 + 0i) = −8. A purely real answer from a thoroughly complex calculation — expand (1 − i√3)³ by binomial if you doubt it: 1 − 3i√3 + 3(−3) ... the cross-terms cancel to leave −8. Polar form turns three pages of expansion into three lines.

Loci and regions in ℂ

Locus problems in ℂ mirror coordinate geometry. |z − (2 + i)| = 3: circle centre (2, 1), radius 3. |z − 1| = |z + 1|: perpendicular bisector of (1, 0) and (−1, 0) — the imaginary axis (x = 0): verify: |x + iy − 1| = |x + iy + 1| → (x − 1)² = (x + 1)² → x = 0 ✓. arg(z − (1 + i)) = π/4: the ray from (1, 1) at 45° (excluding the starting point — arg is undefined at zero).

Inequalities give regions: |z| < 2 is the open disc of radius 2; |z − 1| ≤ 2 and Re(z) > 0 is a half-disc — sketch both conditions and take the overlap. Regions in ℂ are drawn exactly like regions in coordinate geometry, because they are coordinate geometry.

Worked: find the locus of z satisfying |z − 3| = 2|z + 1|. Square: |z − 3|² = 4|z + 1|² → (x − 3)² + y² = 4((x + 1)² + y²) → x² − 6x + 9 + y² = 4x² + 8x + 4 + 4y² → 3x² + 14x + 3y² − 5 = 0 → x² + (14/3)x + y² = 5/3 → (x + 7/3)² + y² = 49/9 + 15/9 = 64/9. Circle centre (−7/3, 0), radius 8/3 — the Apollonius circle again (Lesson 3's locus, now in ℂ).

Applications and exam technique

Applications: AC circuits — impedance Z = R + iX combines resistance and reactance; voltage and current as complex phasors turn differential equations into algebra (this is why electrical engineers learn ℂ first). Quantum mechanics: wavefunctions are complex-valued; probabilities come from |ψ|². Fractals: the Mandelbrot set is zₙ₊₁ = zₙ² + c iterated in ℂ. Even GCSE quadratics with negative discriminants gain solutions: x² + 4x + 13 = 0 → x = −2 ± 3i — the 'no real roots' case now has two complex ones.

Exam technique: (1) for division, always multiply by the conjugate and show the real denominator; (2) give complex answers as x + iy with real x, y (simplify i² immediately); (3) for arg, determine the quadrant from signs of x, y — never blind arctan; (4) principal argument in (−π, π]; (5) for loci, square moduli to get Cartesian equations, then complete the square to identify the shape; (6) in 'show that' questions with conjugates, use z + z̄ = 2x and zz̄ = x² + y² freely.

Complex roots of polynomials

The fundamental theorem of algebra: every degree-n polynomial has exactly n complex roots (counting multiplicity). Quadratics with negative discriminants give conjugate pairs: x² + 4x + 13 = 0 gives x = −2 ± 3i — note the roots are conjugates, always, for real-coefficient polynomials (proved: if P(z) = 0 with real coefficients, conjugating gives P(z̄) = 0). So non-real roots of real polynomials come in pairs — a cubic must have at least one real root.

Worked: factorise x³ − 1 fully over ℂ. One real root x = 1; divide: x³ − 1 = (x − 1)(x² + x + 1). The quadratic gives x = (−1 ± i√3)/2 = cos(±2π/3) + i sin(±2π/3) — the complex cube roots of unity, sitting at 120° intervals on the unit circle. In general the n-th roots of unity are the vertices of a regular n-gon — algebra drawing geometry.

Exam technique: given one complex root of a real polynomial, write down its conjugate free; divide or use sums/products of roots (Vieta: for ax² + bx + c, sum = −b/a, product = c/a) to find the rest. Example: if 1 + 2i is a root of a real cubic, so is 1 − 2i; their sum 2 and product 5 give the quadratic factor x² − 2x + 5; polynomial division yields the remaining linear factor and the real root. Conjugates halve the work.

Common mistakes

• Leaving i² unsimplified. Any answer still containing i² (or i³, i⁴) is unfinished: i² = −1, i³ = −i, i⁴ = 1. Reduce every power on sight.
• Blind arctan for the argument. arctan(y/x) returns values in (−π/2, π/2) only. For z = −1 + i it suggests −π/4; the true principal argument is 3π/4. Plot the point and choose the quadrant.
• Dividing without the conjugate. (3 + 2i)/(1 − i) cannot be simplified term-by-term. Multiply top and bottom by the conjugate of the denominator so the denominator becomes the real number |z|².
• Treating Im(z) as including the i. For z = 3 + 4i, Im(z) = 4, not 4i. The imaginary part is a real number by definition.
• Forgetting the conjugate root. For real-coefficient polynomials, non-real roots come in conjugate pairs — if 2 − i is a root, 2 + i is too, and a 'find all roots' answer missing it is incomplete.

Exam Experience

Questions in the style of the A-level pure paper. Attempt them before reading the mark scheme.

1. Express (4 + 2i)/(1 − i) in the form a + bi, where a, b are real. [2]

2. Let z = −3 + 3i. Find |z| and arg z. [2]

3. Show that z = 2 − i is a root of the equation z² − 4z + 5 = 0, and write down the other root. [3]

Mark scheme.

• 1. Multiply top and bottom by the conjugate 1 + i: (4 + 2i)(1 + i)/((1 − i)(1 + i)) (M1) for the conjugate method with real denominator 2. Numerator = 4 + 4i + 2i + 2i² = 2 + 6i; result = 1 + 3i (A1).
• 2. |z| = √(9 + 9) = √18 = 3√2 (A1). z is in the second quadrant with reference angle π/4, so arg z = π − π/4 = 3π/4 (A1) — the quadrant reasoning, not just arctan, earns the mark.
• 3. Substitute: (2 − i)² − 4(2 − i) + 5 = (4 − 4i + i²) − 8 + 4i + 5 (M1) for expanding with i² handled. = (3 − 4i) − 8 + 4i + 5 = 0 (A1) — the target 0 reached, so 2 − i is a root. The other root is the conjugate, 2 + i (A1) (real coefficients ⟹ roots in conjugate pairs).