IB DP Mathematics: Analysis and Approaches – Complex numbers (HL) Revision Notes

For full explanations and longer worked examples, start with the complex numbers study guide. These notes are for the final weeks.

They cover the complex numbers unit of IB Diploma Programme Mathematics: Analysis and Approaches, aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 1.12–1.14. Everything here is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

When you are ready, test yourself with the complex numbers practice questions. Tick the unit off on the printable syllabus checklist, and find the other units on the IB DP Maths AA course hub.

Definitions (1.12)

• i: the number with i² = −1. Powers cycle: i, −1, −i, 1.
• Cartesian form: z = a + bi, a, b ∈ ℝ.
• Re z = a, Im z = b. Both are real numbers.
• Conjugate: z* = a − bi. Reflection of z in the real axis.
• Modulus: |z| = √(a² + b²). Distance from the origin.
• Argument: arg z, the angle from the positive real axis. These notes use −π < arg z ≤ π unless a question states another range.
• Complex plane (Argand diagram): z = a + bi is plotted at (a, b).
• Equality: a + bi = c + di means a = c and b = d.

Formulas

Result	Form
Polar form	z = r(cos θ + i sin θ) = r cis θ
Euler form	z = re^(iθ)
Polar to Cartesian	a = r cos θ, b = r sin θ
Modulus times conjugate	z z* = |z|²
Product	r₁ cis θ₁ × r₂ cis θ₂ = r₁r₂ cis(θ₁ + θ₂)
Quotient	r₁ cis θ₁ ÷ r₂ cis θ₂ = (r₁/r₂) cis(θ₁ − θ₂)
De Moivre	(r cis θ)ⁿ = rⁿ cis nθ
n-th roots	r^(1/n) cis((θ + 2kπ)/n), k = 0, 1, …, n − 1
Conjugate-pair factor	(z − (a + bi))(z − (a − bi)) = z² − 2az + a² + b²
Equal-modulus sum	1 + e^(iθ) = 2cos(θ/2) e^(iθ/2)

Method in steps

Dividing in Cartesian form

1. Multiply top and bottom by the conjugate of the denominator.
2. The denominator becomes c² + d², a real number.
3. Split into real and imaginary parts.

Finding the argument

1. Sketch the point on the Argand diagram.
2. Find the reference angle α from tan α = |b|/|a|.
3. First quadrant θ = α; second θ = π − α; third θ = −(π − α); fourth θ = −α.

Square roots of a + bi without polar form

1. Let (x + iy)² = a + bi.
2. Equate parts: x² − y² = a and 2xy = b.
3. Substitute y = b/(2x), solve the quadratic in x², keep x² > 0.

Solving zⁿ = w

1. Write w = R cis(φ + 2kπ).
2. z = R^(1/n) cis((φ + 2kπ)/n).
3. Use n consecutive values of k. Adjust each argument into the required range.
4. Convert to Cartesian only if asked.

Proving De Moivre by induction (n ∈ ℤ⁺)

1. Check n = 1.
2. Assume true for n = k.
3. Multiply (cos kθ + i sin kθ) by (cos θ + i sin θ).
4. Group real and imaginary parts and use the compound angle identities.
5. Write the conclusion: true for n = 1, and true for k implies true for k + 1, so true for all n ∈ ℤ⁺.

Multiple-angle identities

1. Write cos nθ + i sin nθ = (c + is)ⁿ by De Moivre.
2. Expand with the binomial theorem, using i² = −1.
3. Equate real parts (for cos nθ) or imaginary parts (for sin nθ).
4. Replace s² with 1 − c², or c² with 1 − s², as needed.

Small worked reminders

Product in polar form. 4 cis(π/3) × 2 cis(π/12) = 8 cis(5π/12).

Quotient. 4 cis(π/3) ÷ 2 cis(π/12) = 2 cis(π/4) = √2 + √2 i.

Power. (1 + √3 i)⁶ = (2 cis(π/3))⁶ = 64 cis 2π = 64.

Conjugate pair. A real cubic has roots 3 and 1 + 2i. Then 1 − 2i is also a root, the pair gives z² − 2z + 5, and the cubic is (z² − 2z + 5)(z − 3) = z³ − 5z² + 11z − 15.

Roots. z³ = 8i gives z = 2 cis(π/6), 2 cis(5π/6), 2 cis(−π/2), that is √3 + i, −√3 + i, −2i.

Rational exponent. The values of 8^(2/3) are 4 cis(2kπ/3) for k = 0, 1, 2: 4 and −2 ± 2√3 i.

Multiple angle. Real part of (c + is)³ gives cos 3θ = 4cos³θ − 3cos θ.

Equation with a conjugate. To solve z + 2z* = 6 − 3i, let z = a + bi. Then a + bi + 2a − 2bi = 3a − bi. Equate parts: 3a = 6 and −b = −3, so z = 2 + 3i. Always substitute a + bi when z and z* appear together.

Square root by equating parts. (x + iy)² = −5 + 12i gives x² − y² = −5 and xy = 6. Then x⁴ + 5x² − 36 = 0, so x² = 4 and the roots are ±(2 + 3i).

Exact value from a product. (1 + i)(√3 + i) = (√3 − 1) + (√3 + 1)i in Cartesian form, and 2√2 cis(5π/12) in polar form. Equating real parts gives cos(5π/12) = (√6 − √2)/4.

Sum of equal-modulus numbers. 1 + cis(2π/3) = 2cos(π/3) cis(π/3) = cis(π/3) = 1/2 + (√3/2)i. Check in Cartesian form: 1 + (−1/2 + (√3/2)i) gives the same.

Geometry on the Argand diagram

• The roots of zⁿ = w are the vertices of a regular n-gon centred at the origin. Area questions use n × ½r² sin(2π/n).
• Points z and iz, with the origin, form a right-angled isosceles triangle. Adding z + iz completes a square.
• |z − w| is the distance between the points z and w.
• Conjugate pairs are symmetric about the real axis, so the roots of a real polynomial have a mirror-image pattern.

Must-know distinctions

• Modulus vs argument. The modulus is a length and is never negative. The argument is an angle and has a range.
• Im z vs bi. Im(2 − 7i) = −7, not −7i.
• Conjugate vs negative. For z = a + bi, z* = a − bi but −z = −a − bi.
• Product vs sum in polar form. Products and quotients are easy in polar or Euler form. Sums are easy in Cartesian form. Convert to suit the operation.
• Multiply vs divide geometrically. Multiplying by r cis θ scales by r and rotates anticlockwise by θ. Dividing scales by 1/r and rotates clockwise by θ. Multiplying by i is a rotation of π/2 anticlockwise.
• Adding geometrically. z₁ + z₂ is the fourth vertex of the parallelogram on the origin, z₁ and z₂.
• Real coefficients vs complex coefficients. Conjugate pairs are guaranteed only when every coefficient is real.
• n-th roots vs a single root. zⁿ = w has exactly n roots (w ≠ 0), equally spaced by 2π/n on a circle of radius |w|^(1/n).
• Exact vs 3 s.f. Calculator-free answers are exact (surds, multiples of π). Calculator-allowed answers are exact or given to 3 significant figures.

Quick self-test

1. Simplify i²³.
2. Find |3 + 4i| and arg(3 + 4i), giving the argument to 3 s.f.
3. Find (1 + 2i)* × (1 + 2i).
4. Write 6 cis(2π/3) in Cartesian form.
5. Write 1 − i in Euler form.
6. Given z = 4 cis(π/3) and w = 2 cis(π/12), find z/w in Cartesian form.
7. Solve z² − 6z + 13 = 0.
8. Find (1 + i)⁶ in Cartesian form.
9. Find the three cube roots of 1 in Cartesian form.
10. A cubic with real coefficients and leading coefficient 1 has roots 1 and 2i. Write it in expanded form.
11. Find |(2 + i)⁴| without expanding.
12. Describe the geometric effect of multiplying a complex number by i.

Answers

1. i²³ = i²⁰ × i³ = −i.
2. |3 + 4i| = 5, arg = 0.927 (radians).
3. (1 − 2i)(1 + 2i) = 1 + 4 = 5.
4. 6 cos(2π/3) = −3 and 6 sin(2π/3) = 3√3, so −3 + 3√3 i.
5. |1 − i| = √2 and arg(1 − i) = −π/4, so √2 e^(−iπ/4).
6. z/w = 2 cis(π/4) = √2 + √2 i.
7. z = (6 ± √(36 − 52))/2 = 3 ± 2i.
8. (√2 cis(π/4))⁶ = 8 cis(3π/2) = −8i.
9. cis(2kπ/3) for k = 0, 1, 2: 1, −1/2 + (√3/2)i, −1/2 − (√3/2)i.
10. −2i is also a root, so (z − 1)(z² + 4) = z³ − z² + 4z − 4.
11. |2 + i|⁴ = (√5)⁴ = 25.
12. A rotation of π/2 anticlockwise about the origin (the modulus is unchanged).

Where marks are usually lost

• Finding the argument with arctan(b/a) and not checking the quadrant, especially for points in the second and third quadrants.
• Giving an argument outside the range the question asks for, such as 7π/6 when −π < θ ≤ π is required.
• On a calculator-free question, giving a decimal such as 1.73 instead of √3, which loses the accuracy mark.
• When solving zⁿ = w, writing only the principal root, or not dividing 2kπ by n.
• Taking the n-th root of the argument but leaving the modulus as R instead of R^(1/n).
• Using the conjugate-root rule when a coefficient of the polynomial is complex.
• In the induction proof, writing “assume true for n” instead of “for n = k”, or leaving out the concluding sentence. Both cost reasoning marks.
• In a “show that” multiple-angle identity, skipping the binomial expansion or the substitution s² = 1 − c². Every step must be visible.
• Leaving a quotient with i in the denominator, which is not the requested form a + bi.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).