IB · Mathematics: Applications & Interpretation SL & HL · Number & algebra

Complex numbers: IB Mathematics: Applications & Interpretation questions with answers

Free Complex numbers questions for IB Mathematics: Applications & Interpretation (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 · 6 marks

Two voltages are given by V1 = 3sin(2t) and V2 = 4cos(2t). By writing V1 and V2 as the imaginary parts of 3e2ti and 4ie2ti, or otherwise,

express V1 + V2 in the form R sin(2t + θ), where R > 0 and 0 < θ < π2.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
5.00sin(2t + 0.927)
V1 + V2 = Im((3 + 4i)e2ti) M1A1
|3 + 4i| = 5.00 A1
arg(3 + 4i) = 0.927 (M1)A1
5.00sin(2t + 0.927) A1
6
Question 2Structured · 4 marks

z = −1 + 2i.

(a)

Find the modulus and argument of z.

[2]
(b)

Find z3, giving your answer in the form reiθ.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)2.24; 2.03
2.24 A1
2.03 A1
2
(b)11.2e6.10i
r3, 3θ (M1)
correct A1
2
Question 3Structured · 6 marks

Two voltages are given by V1 = 8sin(2t) and V2 = 8cos(2t). By writing V1 and V2 as the imaginary parts of 8e2ti and 8ie2ti, or otherwise,

express V1 + V2 in the form R sin(2t + θ), where R > 0 and 0 < θ < π2.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
11.3sin(2t + 0.785)
V1 + V2 = Im((8 + 8i)e2ti) M1A1
|8 + 8i| = 11.3 A1
arg(8 + 8i) = 0.785 (M1)A1
11.3sin(2t + 0.785) A1
6
Question 4Structured · 4 marks

z = −4 − 4i.

(a)

Find the modulus and argument of z.

[2]
(b)

Find z3, giving your answer in the form reiθ.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)5.66; −2.36
5.66 A1
−2.36 A1
2
(b)181e−7.07i
r3, 3θ (M1)
correct A1
2
Question 5Structured · 6 marks

Two voltages are given by V1 = 3sin(5t) and V2 = 6cos(5t). By writing V1 and V2 as the imaginary parts of 3e5ti and 6ie5ti, or otherwise,

express V1 + V2 in the form R sin(5t + θ), where R > 0 and 0 < θ < π2.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
6.71sin(5t + 1.11)
V1 + V2 = Im((3 + 6i)e5ti) M1A1
|3 + 6i| = 6.71 A1
arg(3 + 6i) = 1.11 (M1)A1
6.71sin(5t + 1.11) A1
6
Question 6Structured · 4 marks

z = 3 + 4i.

(a)

Find the modulus and argument of z.

[2]
(b)

Find z3, giving your answer in the form reiθ.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)5.00; 0.927
5.00 A1
0.927 A1
2
(b)125e2.78i
r3, 3θ (M1)
correct A1
2

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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