A-LevelMathematicsComplex numbers9709/33 · May/June 2023 · Q119 Marks

The complex number z is defined by z = (5a - 2i) / (3 + ai), where a is an integer. It is given that arg z = –π/4.

(a)Find the value of a and hence express z in the form x + iy, where x and y are real.[6]
(b)Express z³ in the form re^(iθ), where r > 0 and −π < θ ≤ π. Give the simplified exact values of r and θ.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[M1] Multiply numerator and denominator by (3-ai); Must perform complete multiplications but need not simplify i². Can have errors but no term duplicated or missing. (5a-2i)(3 – ai)/(9-a²) -> (13a-i(5a²+6))/(9-a²) M0 M1 A0 No working so unsure if denominator multiplied by 3-ai M1 M1 A0
[M1] Use i² = −1 at least once and separate real and imaginary parts
[A1] Obtain (13a - i(5a²+6)) / (9+a²) or (13a - 5a²i - 6i) / (9+a²); OE If 15a-2a = 13a seen later award this A1.
[M1] Use arg z to form equation in a; (5a²+6)/(13a) = tan(π/4) or -(5a²+6)/(13a) = tan(π/4) or -13a/(5a²+6) = tan(π/4) or -13a/(5a²+6) = -tan(π/4) or 5a²+6 = ±13a or use - (x + xi) = (13a – i(5a² + 6))/(9 + a²) and eliminate x so 5a² + 6 = ± 13a M1.
[A1] Obtain a = 2; Need to reject a = 3/5 or ignore it in future work. May not see second root, but if present, must be 3/5
[A1] Obtain z = 2-2i only; Allow z = - 2i + 2. Alternative Method 1 for the first four marks: [M1] arg z = arg (5a – 2i) – arg (3 + ai) [M1] = tan⁻¹(-2/(5a)) - tan⁻¹(a/3); Allow one sign error in second M1. [A1] = tan⁻¹((5a²+6)/(13a)) or tan⁻¹(13a/(5a²+6)) [M1] Equate their tan⁻¹((5a²+6)/(13a)) to -π/4 Then as original scheme for final 2 marks. Alternative Method 2 for the first four marks: [M1 A1] (x + iy)(3+ ai) = 5a – 2i; 3x - ay = 5a and ax + 3y = -2 [M1] x = ±y Find x or y in terms of a, e.g. x = 2/(3-a) or x = 5a/(3+a) [M1] Substitute in other equation, for example 3(2/(3-a)) + a(2/(3-a)) = 5a Then as original scheme for final 2 marks.