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Complex numbers on the Argand diagram, modulus-argument and exponential forms, de Moivre's theorem, roots of unity and loci - Edexcel 9FM0 Core Pure 1 and 2 revision with worked examples and mark-scheme logic.
🎯 Learning Objectives
- Convert confidently between the forms a + bi, r(cos θ + i sin θ) and re^iθ, finding the modulus and principal argument with quadrant care (AO1) - Edexcel 9FM0/01 Core Pure 1 and 9FM0/02 Core Pure 2; OCR MEI H645 Further Pure
- Multiply and divide complex numbers in algebraic form and in modulus-argument form, using |z1 z2| = |z1||z2| and arg(z1/z2) = arg z1 - arg z2 (AO1/AO2)
- Apply de Moivre's theorem to integer powers and use it to derive trigonometric identities such as cos 3θ = 4cos³θ - 3cosθ (AO1/AO2)
- Find all nth roots of a complex number and the nth roots of unity, and describe their geometry on the Argand diagram (AO2)
- Sketch and interpret loci and regions: |z - a| = r, |z - a| = |z - b| and arg(z - a) = θ, including intersections of loci (AO2/AO3)
- Use conjugate properties, including zz̄ = |z|², to simplify expressions and solve equations (AO1/AO2)
📋 Key Definitions and Core Concepts
Imaginary unit: i = √(−1), so i² = −1. Powers cycle: i³ = −i, i⁴ = 1, i⁵ = i.
Complex conjugate: For z = a + bi the conjugate is z̄ = a − bi. Then zz̄ = |z|² = a² + b²; the conjugate of a sum, product or quotient is the sum, product or quotient of the conjugates.
Modulus: |z| = √(a² + b²), the distance of the point (a, b) from the origin on the Argand diagram; |z − w| is the distance between the points representing z and w.
Principal argument: The angle arg z measured anticlockwise from the positive real axis, taken in the range (−π, π]; the general argument is arg z + 2πk for any integer k.
Exponential form: re^iθ = r(cos θ + i sin θ) for r ≥ 0 and θ in radians (Euler's form of a complex number).
de Moivre's theorem: For any integer n, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ).
Roots of unity: The n solutions of zⁿ = 1: zk = e^(2πik/n) for k = 0, 1, …, n − 1; they lie equally spaced on the unit circle and sum to 0.
Locus: The set of points in the Argand plane satisfying a condition on z: |z − a| = r is a circle, |z − a| = |z − b| is a perpendicular bisector, arg(z − a) = θ is a half-line.
📌 Key Points
zz̄ = |z|² = a² + b² - the fastest route to a real denominator when dividing.
Multiplying: moduli multiply, arguments add. Dividing: moduli divide, arguments subtract.
Always convert the argument back into (−π, π] before quoting a final answer.
The n nth roots of z are equally spaced by 2π/n on a circle of radius |z|^(1/n) centred at O; the sum of the nth roots of unity is 0.
Loci: |z − a| = r circle; |z − a| = |z − b| perpendicular bisector; arg(z − a) = θ half-line from a.
Complex Numbers and the Argand Diagram
Definition and equality. A complex number has the form z = a + bi, where a and b are real and i² = −1. Here a = Re z is the real part and b = Im z is the imaginary part. Two complex numbers are equal only when both parts match: a + bi = c + di if and only if a = c and b = d. This is the tool you use whenever a question says equating real and imaginary parts - for example, solving two simultaneous equations produced by expanding a product of complex expressions.
The conjugate. The complex conjugate of z = a + bi is z̄ = a − bi, its mirror image in the real axis on the Argand diagram. The facts you must have at your fingertips: z + z̄ = 2a (real), z − z̄ = 2bi (purely imaginary), and above all zz̄ = |z|² = a² + b². The conjugate of a sum, difference, product or quotient is the sum, difference, product or quotient of the conjugates. A consequence used elsewhere in Core Pure: a polynomial with real coefficients has its non-real roots in conjugate pairs, so knowing one root gives you another free.
Division in algebraic form. To write (a + bi)/(c + di) in the form p + qi, multiply numerator and denominator by the conjugate of the denominator, c − di. The denominator becomes the real number c² + d², and the numerator is an ordinary expansion with i² replaced by −1.
The Argand diagram. Plot z = a + bi as the point (a, b): horizontal axis real, vertical axis imaginary. The modulus |z| = √(a² + b²) is the distance from the origin; the argument is the angle measured anticlockwise from the positive real axis, with the principal argument taken in (−π, π]. Since tan θ = b/a only gives an answer in (−π/2, π/2), always fix the quadrant from the signs of a and b before quoting θ.
| Value | cos | sin |
|---|---|---|
| π/6 | √3/2 | 1/2 |
| π/4 | √2/2 | √2/2 |
| π/3 | 1/2 | √3/2 |
Common trap
arg(−1 + i) is 3π/4, not π/4 and not −π/4. Your calculator's inverse tan returns −45° because it only knows quadrants I and IV - adjust to quadrant II by adding 180° (π) whenever the real part is negative and the imaginary part positive.
Modulus-Argument and Exponential Forms: Multiplying and Dividing
Three equivalent forms. Any non-zero complex number can be written as a + bi, as r(cos θ + i sin θ), or in exponential form re^iθ, where r = |z| = √(a² + b²) and θ = arg z. To convert from algebraic to polar form: find r; find the reference angle from |tan θ| = |b/a|; correct the quadrant using the signs of a and b; then write r(cos θ + i sin θ) or re^iθ. To convert back, expand: r cos θ is the real part and r sin θ the imaginary part, with exact surd answers expected from the standard angles above.
Multiplication and division rules. In modulus-argument form the rules are structural, not computational: moduli multiply and arguments add; moduli divide and arguments subtract. Formally, |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg z₁ + arg z₂; |z₁/z₂| = |z₁|/|z₂| and arg(z₁/z₂) = arg z₁ − arg z₂ (adjusting by 2π into the principal range). For conjugates, |z̄| = |z| and arg z̄ = −arg z; for powers, |zⁿ| = |z|ⁿ and arg zⁿ = n arg z. These statements are examinable in their own right - quote them, do not just use them.
Multiplication in algebraic form is bracket expansion with i² = −1: (3 + 4i)(1 − 2i) = 3 − 6i + 4i + 8 = 11 − 2i. Division uses the conjugate trick from Section 1. But for powers, polar form wins every time:
Worked example: evaluate (1 + i)⁸. Since 1 + i = √2(cos(π/4) + i sin(π/4)), de Moivre (next section) gives (1 + i)⁸ = (√2)⁸(cos(8 × π/4) + i sin(2π)) = 16(1 + 0i) = 16. Doing this by repeated bracket expansion is slower and error-prone - examiners award method marks for spotting the polar route.
Exam technique
When asked for a final answer in the form a + bi, convert fully - leaving re^iθ or an argument outside (−π, π] loses the accuracy mark. When asked for re^iθ, take θ in radians. Marks are also commonly attached to stating |z| and arg z separately before combining.
de Moivre's Theorem, Roots of Unity and Loci
de Moivre's theorem. For any integer n, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ). For negative n it reads 1/zⁿ, so work with the reciprocal's argument. Two uses dominate the exams. First, exact powers, as in (1 + i)⁸ above or (√3 − i)⁷: write √3 − i = 2(cos(−π/6) + i sin(−π/6)), so (√3 − i)⁷ = 2⁷(cos(−7π/6) + i sin(−7π/6)) = 128(cos(5π/6) + i sin(5π/6)) = −64√3 + 64i. Second, deriving trig identities: expand (cos θ + i sin θ)³ binomially, apply de Moivre to say it equals cos 3θ + i sin 3θ, then equate real and imaginary parts to get cos 3θ = 4cos³θ − 3cosθ and sin 3θ = 3sinθ − 4sin³θ.
nth roots. The solutions of zⁿ = w, where w = r(cos θ + i sin θ), are zk = r^(1/n)(cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)) for k = 0, 1, …, n − 1. Geometrically: n roots, equally spaced by angle 2π/n, on a circle of radius r^(1/n) centred at the origin. The roots of unity are the special case zⁿ = 1: e^(2πik/n). Key facts examiners love: they sum to 0; their squares sum to 0; and if ω is one root, the full set is 1, ω, ω², …, ωⁿ⁻¹ - a geometric-style listing you may be asked to connect to the sum of a GP.
Worked example: solve z³ = 8i. Write 8i = 2(cos(π/2) + i sin(π/2)). Then zk = 2(cos((π/2 + 2πk)/3) + i sin((π/2 + 2πk)/3)), k = 0, 1, 2, giving 2cis(π/6), 2cis(5π/6), 2cis(3π/2), i.e. √3 + i, −√3 + i, −2i. Quick check: (−2i)³ = −8i³ = −8(−i) = 8i ✓.
Loci in the Argand plane. Three conditions generate every standard locus question:
| Condition | Locus |
|---|---|
| |z − a| = r | Circle, centre a, radius r (inequality gives a disc region) |
| |z − a| = |z − b| | Perpendicular bisector of the segment joining a and b |
| arg(z − a) = θ | Half-line (ray) from a at angle θ to the positive real direction; the start point a is excluded (draw it dotted) |
Regions combine these with inequalities: |z − a| ≤ r shades inside the circle, arg(z − a) ≥ θ shades one side of the ray. Boundaries are dashed where the inequality is strict. Intersection example: the loci |z − 3| = 2 and arg(z − 3) = π/4 meet where z − 3 = 2cis(π/4), i.e. z = (3 + √2) + √2 i.
Mark-scheme habit
Loci sketching questions award marks for: correct curve type, correct centre/line position, radius or angle marked, and start point dotted for arg loci. State the exact intersection point with surds rather than decimals.
💡 Worked Example
Exam-style: (a) Using de Moivre's theorem, express (√3 − i)⁷ in the form p + qi, where p and q are exact. (4 marks)
(b) Solve the equation z³ = 8i, giving your answers in exact form. (4 marks)
Mark-scheme logic: (a) M1: write √3 − i in modulus-argument form: modulus 2 (from √(3 + 1)) and argument −π/6 (tan = −1/√3 with negative imaginary part). M1: apply de Moivre: (√3 − i)⁷ = 2⁷(cos(−7π/6) + i sin(−7π/6)). M1: convert the angle to the principal range by adding 2π: −7π/6 + 2π = 5π/6. A1: 128(cos(5π/6) + i sin(5π/6)) = 128(−√3/2 + i·1/2) = −64√3 + 64i.
(b) M1: write 8i = 2(cos(π/2) + i sin(π/2)). M1: cube roots have modulus 8^(1/3) = 2 and arguments (π/2 + 2πk)/3, k = 0, 1, 2. A1: the three roots are 2cis(π/6), 2cis(5π/6), 2cis(3π/2). A1: in exact form √3 + i, −√3 + i, −2i (a verification such as (−2i)³ = 8i secures the final mark).
❓ Practice Questions & Mark Schemes
z₁ = 3 + 4i and z₂ = 1 − 2i. Find z₁z₂ and z₁/z₂ in the form p + qi. (5 marks)
Answer: z₁z₂ = (3 + 4i)(1 − 2i) = 3 − 6i + 4i − 8i² = 3 − 2i + 8 = 11 − 2i. For the quotient, multiply numerator and denominator by the conjugate 1 + 2i: z₁/z₂ = ((3 + 4i)(1 + 2i))/((1 − 2i)(1 + 2i)) = (3 + 6i + 4i + 8i²)/(1 + 4) = (−5 + 10i)/5 = −1 + 2i.
Write z = −1 + i in the form r(cos θ + i sin θ), where θ is the principal argument. (3 marks)
Answer: r = √((−1)² + 1²) = √2. The point (−1, 1) is in the second quadrant, so the reference angle π/4 (from tan θ = −1) is adjusted: θ = π − π/4 = 3π/4. Hence z = √2(cos(3π/4) + i sin(3π/4)).
Use de Moivre's theorem to evaluate (1 + i)⁸. (3 marks)
Answer: 1 + i = √2(cos(π/4) + i sin(π/4)), so (1 + i)⁸ = (√2)⁸(cos(8 × π/4) + i sin(2π)) = 16(cos 2π + i sin 2π) = 16.
Solve the equation z² = 3 + 4i. (4 marks)
Answer: Let z = a + bi. Then a² − b² = 3 and 2ab = 4, so ab = 2. Also |3 + 4i| = 5 gives a² + b² = 5 (since |z|² = |z²|). Adding the two equations: 2a² = 8, so a = ±2, and then b² = 1 with the sign of b matching a (since ab = 2 > 0): z = 2 + i or z = −2 − i. Check: (2 + i)² = 4 + 4i − 1 = 3 + 4i ✓.
Sketch the loci |z − 3| = 2 and arg(z − 3) = π/4 on an Argand diagram, and find their point of intersection in exact form. (5 marks)
Answer: |z − 3| = 2 is the circle of radius 2 centred at the point 3 on the real axis; arg(z − 3) = π/4 is the half-line starting at 3 (drawn dotted) making angle π/4 with the positive real direction. The ray meets the circle where z − 3 = 2(cos(π/4) + i sin(π/4)), i.e. z = 3 + √2 + √2 i, so the intersection point is ((3 + √2), √2).