IB DP Mathematics: Analysis and Approaches – Complex numbers (HL) Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the complex numbers unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 1.12–1.14, and every question 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.

HL Paper 1 allows no technology; Papers 2 and 3 require it. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give calculator-allowed answers exactly or to 3 significant figures, and arguments in the range −π < θ ≤ π unless stated.

Learn the methods first in the complex numbers study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) Let z = 2 − 3i and w = 4 + i.

(a) Find zw in the form a + bi. [2] (b) Find z/w in the form a + bi, where a, b ∈ ℚ. [3]

2. (calculator-free) Let z = −√3 − i. Find |z| and arg z, and hence write z in polar form and in Euler form. [4]

3. (calculator-free) Find the complex number z that satisfies z + 2z* = 6 − 3i. [3]

4. (calculator-free) Find the two square roots of −21 − 20i, giving your answers in the form a + bi. [5]

5. (calculator-free) Find (1 + i)⁸ / (√3 − i)⁴ in the form a + bi. [5]

6. (calculator-free) On an Argand diagram, the point P represents z = 4 + 2i and the point Q represents w = iz.

(a) Find w. [1] (b) Describe the geometric transformation that maps P to Q. [1] (c) The point R represents z + w. Show that OPRQ is a square and find its area. [3]

7. (calculator allowed) Let z₁ = 3e^(iπ/5) and z₂ = 2e^(−iπ/4).

(a) Find z₁z₂ in the form re^(iθ). [2] (b) Find z₁ + z₂ in the form a + bi. [2] (c) Hence find |z₁ + z₂| and arg(z₁ + z₂). [2]

8. (calculator allowed) Let w = 5 − 12i.

(a) Write w in the form re^(iθ). [2] (b) Find the four solutions of z⁴ = w in the form a + bi. [4]

9. (calculator-free) Prove by mathematical induction that (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for all n ∈ ℤ⁺. [6]

10. (calculator-free) The equation z³ + az² + bz + 10 = 0, where a, b ∈ ℝ, has a root 2 + i.

(a) Find the other two roots and the values of a and b. [6] (b) The three roots are plotted on an Argand diagram. Find the area of the triangle they form. [2]

11. (calculator-free)

(a) Write −4 + 4√3 i in the form r cis θ. [2] (b) Hence solve z³ = −4 + 4√3 i, giving your answers in the form r cis θ, where −π < θ ≤ π. [4] (c) Find the exact area of the triangle whose vertices are the three solutions. [3]

12. (calculator-free)

(a) Use De Moivre’s theorem to show that cos 4θ = 8cos⁴θ − 8cos²θ + 1. [5] (b) Hence find the exact value of cos²(π/8). [3]

Answers

1. (a) zw = 8 + 2i − 12i − 3i² [1] = 11 − 10i [1] (b) Multiply by w*/w*: (2 − 3i)(4 − i)/((4 + i)(4 − i)) [1]; numerator 8 − 2i − 12i + 3i² = 5 − 14i, denominator 17 [1]; z/w = 5/17 − (14/17)i [1] Examiner insight: The question asks for the form a + bi; an answer with i still in the denominator is not in that form and loses the accuracy mark.

2. |z| = √(3 + 1) = 2 [1]. The point is in the third quadrant with reference angle π/6 [1], so arg z = −5π/6 [1]. z = 2 cis(−5π/6) = 2e^(−5πi/6) [1] Examiner insight: arctan(1/√3) = π/6 gives the wrong quadrant; without a sketch or quadrant statement the argument mark is lost even if the modulus is right.

3. Let z = a + bi, so z* = a − bi [1]. Then a + bi + 2a − 2bi = 3a − bi = 6 − 3i [1]. So a = 2, b = 3 and z = 2 + 3i [1] Examiner insight: The method mark is for substituting a + bi and equating real and imaginary parts; guessing and checking earns nothing if the guess is wrong.

4. Let (a + bi)² = −21 − 20i with a, b real [1]. Then a² − b² = −21 and 2ab = −20 [1]. Substituting b = −10/a: a⁴ + 21a² − 100 = 0 [1], so (a² + 25)(a² − 4) = 0 and a² = 4 [1]. The roots are 2 − 5i and −2 + 5i [1] Examiner insight: You must reject a² = −25 because a is real; stating why keeps the reasoning complete.

5. 1 + i = √2 cis(π/4) [1] and √3 − i = 2 cis(−π/6) [1]. By De Moivre, (1 + i)⁸ = 16 cis 2π = 16 and (√3 − i)⁴ = 16 cis(−2π/3) [1]. The quotient is cis(2π/3) [1] = −1/2 + (√3/2)i [1] Examiner insight: Expanding with the binomial theorem is valid but slow and error-prone; the method marks reward the polar conversion.

6. (a) w = i(4 + 2i) = −2 + 4i [1] (b) Rotation of π/2 anticlockwise about the origin [1] (c) |OP| = |OQ| = √20 and OP is perpendicular to OQ, because Q is P rotated by π/2 [1]. R = z + w, so OPRQ is a parallelogram by vector addition, and a parallelogram with equal perpendicular adjacent sides is a square [1]. Area = |z|² = 20 [1] Examiner insight: “Rotation” alone scores nothing; the transformation mark needs the angle, the direction and the centre.

7. (a) Multiply moduli and add arguments: 6e^(i(π/5 − π/4)) [1] = 6e^(−iπ/20) [1] (b) z₁ = 2.427 + 1.763i and z₂ = 1.414 − 1.414i [1], so z₁ + z₂ = 3.84 + 0.349i [1] (c) |z₁ + z₂| = 3.86 [1]; arg(z₁ + z₂) = 0.0906 [1] Examiner insight: Carry unrounded values from (b) into (c); using the 3 s.f. values can change the third figure, and early rounding costs the accuracy mark.

8. (a) |w| = 13 [1]; arg w = −arctan(12/5) = −1.176, so w = 13e^(−1.18i) [1] (b) Each root has modulus 13^(1/4) = 1.899 [1]. Arguments are (−1.176 + 2kπ)/4 for k = 0, 1, 2, 3 [1]. For k = 0: z = 1.82 − 0.550i [1]. The roots are spaced by π/2, so each is i times the previous: 0.550 + 1.82i, −1.82 + 0.550i, −0.550 − 1.82i [1] Examiner insight: A GDC list of four roots with no modulus or argument working risks losing the method marks if any value is wrong; show the general argument.

9. Let P(n) be the statement. For n = 1, both sides equal cos θ + i sin θ, so P(1) is true [1]. Assume P(k) is true for some k ∈ ℤ⁺ [1]. Then (cos θ + i sin θ)^(k+1) = (cos kθ + i sin kθ)(cos θ + i sin θ) [1] = (cos kθ cos θ − sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ) [1] = cos(k + 1)θ + i sin(k + 1)θ by the compound angle identities, so P(k + 1) is true [1]. P(1) is true and P(k) true implies P(k + 1) true, so P(n) is true for all n ∈ ℤ⁺ by mathematical induction [1] Examiner insight: The concluding sentence must refer to both the n = 1 case and the step from k to k + 1; a bare “hence true” does not earn the reasoning mark.

10. (a) The coefficients are real, so 2 − i is also a root [1]. The pair has sum 4 and product 5, giving the factor z² − 4z + 5 [1]. Write the cubic as (z² − 4z + 5)(z − c); comparing constants, −5c = 10 [1], so the third root is −2 [1]. Expanding (z² − 4z + 5)(z + 2) = z³ − 2z² − 3z + 10 [1], so a = −2, b = −3 [1] (b) The base joins 2 + i and 2 − i, length 2; the height is the distance from x = 2 to x = −2, which is 4 [1]. Area = ½ × 2 × 4 = 4 [1] Examiner insight: The conjugate root 2 − i earns its own mark; write it down with its reason (real coefficients) rather than using it silently inside the quadratic factor.

11. (a) r = √(16 + 48) = 8 [1]; θ = 2π/3, so 8 cis(2π/3) [1] (b) z = 8^(1/3) cis((2π/3 + 2kπ)/3) [1], modulus 2 [1]. Arguments 2π/9 and 8π/9 [1], and 14π/9 − 2π = −4π/9, so z = 2 cis(2π/9), 2 cis(8π/9), 2 cis(−4π/9) [1] (c) The roots are equally spaced, so each pair subtends 2π/3 at O [1]. Area = 3 × ½ × 2 × 2 × sin(2π/3) [1] = 3√3 [1] Examiner insight: Leaving 14π/9 loses the final accuracy mark in (b) because the question fixed the range −π < θ ≤ π.

12. (a) By De Moivre, cos 4θ + i sin 4θ = (c + is)⁴, where c = cos θ, s = sin θ [1]. The real part of the binomial expansion is c⁴ − 6c²s² + s⁴ [1]. Substitute s² = 1 − c² [1]: c⁴ − 6c²(1 − c²) + (1 − c²)² [1] = c⁴ − 6c² + 6c⁴ + 1 − 2c² + c⁴ = 8cos⁴θ − 8cos²θ + 1 [1] (b) Put θ = π/8: cos(π/2) = 0, so 8c⁴ − 8c² + 1 = 0 [1]. Then c² = (8 ± √32)/16 = (2 ± √2)/4 [1]. cos²(π/8) > cos²(π/4) = 1/2, so take the positive sign: cos²(π/8) = (2 + √2)/4 [1] Examiner insight: In a “show that”, each line of the expansion must be written; jumping from (c + is)⁴ to the final identity earns no marks after the first.

Where marks are usually lost

• Arguments found with arctan(b/a) and no quadrant check, as in questions 2 and 11.
• Final arguments left outside the stated range after using the general formula for roots.
• n-th roots where the modulus is left as R rather than R^(1/n).
• Decimal answers on calculator-free questions where exact surds are needed.
• Conjugate roots used without stating that the coefficients are real.
• Induction proofs that omit “assume true for n = k” or the concluding sentence.
• “Show that” identities with algebra steps missing between the binomial expansion and the result.
• Rounding intermediate values to 3 s.f. before a later calculation.
• Transformations described without the angle, direction or centre.

Next steps

• Recap the key facts in the complex numbers revision notes.
• Re-learn any weak topic in the complex numbers study guide.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Plan your final weeks with the AA exam preparation guide.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

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