Lesson 21: Complex Number Extensions (HL) HL ONLY

Lesson 8 gave complex numbers their arithmetic; this HL extension (AA Topic 1) turns the polar form into an engine. The key that unlocks everything is De Moivre's theorem: (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ) for any integer n — powers of a complex number raise the modulus to the power and multiply the argument. Its consequences cascade: painless high powers, roots of unity and all solutions of zⁿ = a, polynomial equations with complex roots, loci (geometric conditions in the Argand plane), and the exponential form e^(iθ) = cos θ + i sin θ that fuses algebra with the calculus of Lessons 10 and 13.

De Moivre as a computational engine

Worked Example 1 — a power that algebra cannot touch. Express (1 + i)⁸ and (√3 − i)⁶ in Cartesian form.

Solution. 1 + i has modulus √2 and argument π/4, so (1 + i)⁸ has modulus (√2)⁸ = 16 and argument 8 × π/4 = 2π: the value is 16(cos 2π + i sin 2π) = 16. (Squaring repeatedly confirms: (1 + i)² = 2i, (2i)² = −4, (−4)² = 16 ✓.) For √3 − i: modulus 2, argument −π/6; the sixth power has modulus 64 and argument −π, giving 64(cos(−π) + i sin(−π)) = −64. Direct expansion of either would take a page; De Moivre takes three lines. The method marks are in the conversion — modulus and argument *first*, powers second.

Roots of unity and solving zⁿ = a

The equation zⁿ = a (a ≠ 0) has exactly n distinct roots: if a = R(cos Θ + i sin Θ), the roots are z_k = R^(1/n) [cos((Θ + 2kπ)/n) + i sin((Θ + 2kπ)/n)] for k = 0, 1, …, n − 1. Geometrically they sit at the vertices of a regular n-gon on the circle of radius R^(1/n). The special case a = 1 gives the roots of unity, 1, ω, ω², …, with ω = e^(2πi/n); they satisfy 1 + ω + … + ωⁿ⁻¹ = 0 (a geometric series with ratio ω ≠ 1, sum (ωⁿ − 1)/(ω − 1) = 0) — IB adores this result.

Worked Example 2 — solving z³ = 8i (past-paper style). Find all roots in modulus–argument form and plot them.

Solution. 8i = 8(cos(π/2) + i sin(π/2)). Cube roots: modulus 8^(1/3) = 2; arguments (π/2 + 2kπ)/3 for k = 0, 1, 2: π/6, 5π/6, 3π/2. Roots: 2∠(π/6) = √3 + i; 2∠(5π/6) = −√3 + i; 2∠(3π/2) = −2i. They form an equilateral triangle on the circle radius 2 ✓, and each cubes to 8i — spot-check the last: (−2i)³ = −8i³ = 8i ✓ (since i³ = −i). Reporting only the principal root (k = 0) is the classic half-answer: n-th roots come in sets of n, symmetric by construction.

Polynomial equations in ℂ

Over ℂ, every degree-n polynomial factorises completely (fundamental theorem of algebra), and for real coefficients, non-real roots arrive in conjugate pairs. Strategy: use given roots, build quadratic factors from conjugate pairs ((z − (a + bi))(z − (a − bi)) = z² − 2az + (a² + b²)), divide, repeat.

Worked Example 3 — building from one root. The equation z³ − 3z² + 4z − 2 = 0 has root z = 1 + i. Solve it completely.

Solution. Real coefficients ⟹ 1 − i is also a root. The pair gives the quadratic factor z² − 2z + 2 (sum of roots 2, product |1 + i|² = 2). Divide: z³ − 3z² + 4z − 2 = (z² − 2z + 2)(z − 1) (expand to verify: z³ − z² − 2z² + 2z + 2z − 2 ✓). Remaining root z = 1. Solutions: 1, 1 + i, 1 − i. Sum of roots 3 = −(coefficient of z²): Vieta's relations offer a free check on every such problem.

Loci in the complex plane

A locus is the set of points z satisfying a geometric condition: |z − a| = r is the circle centre a radius r (points at fixed distance from a); |z − a| = |z − b| is the perpendicular bisector of the segment ab (points equidistant from a and b); arg(z − a) = θ is a ray from a at angle θ (excluding a itself); |z − a| ≤ r is a disc. Translate the condition into its geometry first, then into an equation only if asked.

Sketching example. |z − 2 − 3i| = 4: circle, centre (2, 3), radius 4 — it crosses the real axis where (x − 2)² + 9 = 16, x = 2 ± √7. And |z − 1| = |z + 3i|: the perpendicular bisector of (1, 0) and (0, −3) — midpoint (1/2, −3/2), slope of the segment is (0 + 3)/(1 − 0) = 3, so the bisector has slope −1/3: equation y + 3/2 = −(1/3)(x − 1/2), or in complex form found by expanding |z − 1|² = |z + 3i|² with z = x + iy: (x − 1)² + y² = x² + (y + 3)² → −2x + 1 = 6y + 9 → x + 3y + 4 = 0 ✓ consistent.

The exponential form and Euler's identity

e^(iθ) = cos θ + i sin θ makes De Moivre a triviality (e^(iθ))ⁿ = e^(inθ) and turns multiplication into exponent addition. At θ = π: e^(iπ) = −1, i.e. e^(iπ) + 1 = 0 — Euler's identity, linking e, i, π, 1, 0 in one line. Practically: write products, quotients and powers in e-form whenever angles accumulate — and know that z = re^(iθ) with z̄ = re^(−iθ) gives zz̄ = r² in one glance.

Real-world modelling: phasors and signal roots

Roots of unity are the skeleton of the discrete Fourier transform (signal processing, audio compression): sampling a signal at n points literally solves with ω = e^(2πi/n). Phasor multiplication (Lesson 8) is De Moivre at n = 1. When an IA or Paper 3 asks "where is this used?", the honest answer is: every digital recording you own.

Common mistakes

• Roots reported with modulus R instead of R^(1/n), or arguments not divided by n.
• Only the principal root given for zⁿ = a — the full set has n members, equally spaced by 2π/n.
• Conjugate-pair assumption applied to non-real coefficients — the theorem needs real coefficients; check before using.
• Locus |z − a| = |z − b| drawn as a circle — it is a straight line (perpendicular bisector).
• arg(z − a) drawn as a full line — it is a ray (one direction only), and the point a is excluded.
• Degrees in De Moivre working, or arguments left outside a sensible range without comment.

IB Exam Experience

Question 1 [2 marks]. Evaluate (1 + i√3)³ using De Moivre's theorem. *Marking notes:* M1: modulus 2, argument π/3; A1: 2³∠(3 × π/3) = 8∠π = −8. (Binomial expansion confirms: (1 + i√3)³ = 1 + 3i√3 − 9 − 3i√3 = −8 ✓ — the imaginary parts cancel by design.)

Question 2 [3 marks]. Solve z² = −5 + 12i, giving the roots in the form a + bi. *Marking notes:* M1 for setting (a + bi)² = −5 + 12i and equating parts: a² − b² = −5, 2ab = 12; M1 for solving (b = 6/a → a² − 36/a² = −5 → a⁴ + 5a² − 36 = 0 → a² = 4); A1: roots 2 + 3i and −2 − 3i. (Polar method equally acceptable: modulus 13, roots modulus √13.)

Question 3 [4 marks]. On an Argand diagram, sketch the locus of points satisfying |z + 2i| ≤ 3 and arg(z) = π/4 simultaneously, and state the resulting set precisely. *Marking notes:* M1: |z + 2i| ≤ 3 is the closed disc centre (0, −2) radius 3; M1: arg(z) = π/4 is the ray from the origin at 45° (origin excluded); A1: the intersection is the segment of that ray inside the disc — the ray enters the disc since the origin's distance to the centre is 2 < 3; A1: the intersection is the portion of the ray lying inside the disc: points z = s·e^(iπ/4) with 0 < s ≤ s_max, where the ray meets the circle |z + 2i| = 3 at s satisfying s² + 2√2 s − 5 = 0, giving s_max = √7 − √2 ≈ 1.23. Examiners reward the translation discipline: each condition drawn, then the overlap described.

Extended problem (Paper 3 style, HL) [10 marks]. Let ω = e^(2πi/5), and consider the five roots of z⁵ = 1. (a) Write all five roots in exponential form. [2] (b) Show that 1 + ω + ω² + ω³ + ω⁴ = 0. [3] (c) The roots are plotted; state the sum of all five roots and the product of all five roots, justifying each by symmetry or by Vieta. [3] (d) Hence evaluate cos(2π/5) + cos(4π/5) exactly, given that ω + ω⁴ = 2cos(2π/5) and ω² + ω³ = 2cos(4π/5). [2] *Marking notes:* (a) A2: 1, e^(2πi/5), e^(4πi/5), e^(6πi/5), e^(8πi/5). (b) M1: geometric series with ratio ω: sum = (ω⁵ − 1)/(ω − 1); A1: ω⁵ = 1 so the numerator is 0; A1: and ω ≠ 1 so the division is legal — sum 0. (c) M1: by (b) the sum is 0 (or by the pentagon's symmetry — vector sum cancels); A1: product: by Vieta on the monic polynomial z⁵ − 1, the product of the roots equals −(constant term) = −(−1) = 1; A1: check via arguments: e^(i(0 + 2π/5 + 4π/5 + 6π/5 + 8π/5)) = e^(i·4π) = 1 ✓. (d) M1: taking real parts of (b): 1 + 2cos(2π/5) + 2cos(4π/5) = 0; A1: cos(2π/5) + cos(4π/5) = −1/2. (The individual values, (√5 − 1)/4 and −(√5 + 1)/4, follow by also using cos(2π/5) − cos(4π/5) = √5/2 — a beautiful extension if time allows.)

Exploration (IA) idea: build the escape-time picture for z → z² + c at several c values and analyse the geometry you find; or use roots of unity to model equally-spaced sensors/microphones and the interference pattern — De Moivre with data attached.