IB · Mathematics: Analysis & Approaches SL & HL · Number & algebra

Complex numbers: IB Mathematics: Analysis & Approaches questions with answers

Free Complex numbers questions for IB Mathematics: Analysis & Approaches (SL & HL), HL only, part of Number & algebra. Each question has the answer and the marking points an examiner would look for.

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Complex numbers practice questions

Question 1Structured · 5 marks

Find the complex number z that satisfies z + 4z* = 30 + 6i.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
z = 6 − 2i
let z = a + bi, so z* = a − bi M1
a + bi + 4(a − bi) = 30 + 6i (A1)
real: 5a = 30; imaginary: −3b = 6 M1
a = 6 A1
b = −2 A1
5
Question 2Structured · 6 marks

Find the two square roots of 15 − 8i, giving your answers in the form a + bi.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
±(4 − i)
(u + vi)2 = 15 − 8i M1
u2 − v2 = 15, 2uv = −8 A1A1
attempt to solve, e.g. substituting v = −4u M1
u4 − 15u2 − 16 = 0 ⇒ u2 = 16 A1
±(4 − i) A1
6
Question 3Structured · 6 marksA* challenge
(a)

Express z = 1 − i in the form reiθ, where r > 0 and −π < θ ≤ π.

[3]
(b)

Find the smallest positive integer n such that zn is real.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)√2e−πi/4
r = √2 A1
attempt to find the argument in the correct quadrant (M1)
θ = −π4 A1
3
(b)n = 4
zn = (√2)nen·−πi/4 (M1)
n·−π4 = kπ (M1)
n = 4 A1
3
Question 4Structured · 5 marks

Find the complex number z that satisfies z − 2z* = 5 − 9i.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
z = −5 − 3i
let z = a + bi, so z* = a − bi M1
a + bi − 2(a − bi) = 5 − 9i (A1)
real: −a = 5; imaginary: 3b = −9 M1
a = −5 A1
b = −3 A1
5
Question 5Structured · 6 marks

Find the two square roots of 21 + 20i, giving your answers in the form a + bi.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
±(5 + 2i)
(u + vi)2 = 21 + 20i M1
u2 − v2 = 21, 2uv = 20 A1A1
attempt to solve, e.g. substituting v = 10u M1
u4 − 21u2 − 100 = 0 ⇒ u2 = 25 A1
±(5 + 2i) A1
6
Question 6Structured · 6 marksA* challenge
(a)

Express z = −√3 + i in the form reiθ, where r > 0 and −π < θ ≤ π.

[3]
(b)

Find the smallest positive integer n such that zn is real.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)2e5πi/6
r = 2 A1
attempt to find the argument in the correct quadrant (M1)
θ = 5π6 A1
3
(b)n = 6
zn = (2)nen·5πi/6 (M1)
n·5π6 = kπ (M1)
n = 6 A1
3

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in