A2 June 2024 Paper 1 Q2

OCR MEICurrent spec7 marksComplex Numbers

2 Two complex numbers are given by \(u = -1 + \mathrm{i}\) and \(v = -2 - \mathrm{i}\).

(a)

(i) Find \(u - v\) in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real. [1]

(ii) In this question you must show detailed reasoning.
Find \(\dfrac{u}{v}\) in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real. [3]

(b) Express \(u\) in exact modulus-argument form. [3]

Mark scheme (a)

Scheme	Marks	AO
(i) \(1 + 2\mathrm{i}\)	B1	1.1
	[1]
(ii) DR \(\dfrac{u}{v} = \dfrac{(-1 + \mathrm{i})(-2 + \mathrm{i})}{(-2 - \mathrm{i})(-2 + \mathrm{i})}\)	M1*	1.1a
\(= \dfrac{2 - 2\mathrm{i} - \mathrm{i} + \mathrm{i}^2}{5}\)	M1dep	2.1
\(= \frac{1}{5} - \frac{3}{5}\mathrm{i}\)	A1	1.1
	[3]

Notes

(a)(i)

B1: cao

(a)(ii)

M1*: \(\times\) numerator and denominator by \((-2 + \mathrm{i})\) or \((2 - \mathrm{i})\)

M1dep: Numerator expanded to include at least three terms with correct denominator seen (must be expanded but need not be simplified). Allow one sign slip in numerator only.

A1: Or simplified equivalent, e.g. \(\frac{1 - 3\mathrm{i}}{5}\)

Alternative method

Scheme	Marks
\(-1 + \mathrm{i} = (-2 - \mathrm{i})(a + b\mathrm{i})\) \(= -2a - 2b\mathrm{i} - a\mathrm{i} + b\)	M1*
\(-1 = -2a + b\) and \(1 = -2b - a\)	M1dep
\(a = \frac{1}{5}, b = -\frac{3}{5} \Rightarrow \frac{u}{v} = \frac{1}{5} - \frac{3}{5}\mathrm{i}\)	A1
	[3]

M1*: Expanding \((-2 - \mathrm{i})(a + b\mathrm{i})\). Allow one slip only.

M1dep: Equating both real and imaginary coefficients

A1: Or simplified equivalent, e.g. \(\frac{1 - 3\mathrm{i}}{5}\)

Mark scheme (b)

Scheme	Marks	AO
\(\sqrt{2}\)	B1	1.1
\(\frac{3}{4}\pi\)	B1	1.1
\(\sqrt{2}\left(\cos\frac{3\pi}{4} + \mathrm{i}\sin\frac{3\pi}{4}\right)\)	B1	1.1
	[3]

Notes

B1: Correct modulus seen

B1: Correct argument seen. Condone \(135^\circ\)

B1: Must be exact; condone \(135^\circ\)