AS June 2023 Paper 1 Q7

OCR MEICurrent spec10 marksComplex Numbers

7 In this question you must show detailed reasoning.

The complex number \(\sqrt{3} + \mathrm{i}\) is denoted by \(z\).

(a) By expanding \(\left(\sqrt{3} + \mathrm{i}\right)^5\), express \(z^5\) in the form \(a + b\mathrm{i}\) where \(a\) and \(b\) are real and exact. [3]

(b)

(i) Express \(z\) in modulus-argument form. [3]

(ii) Hence find \(z^5\) in modulus-argument form. [2]

(iii) Use this result to verify your answers to part (a). [2]

Mark scheme (a)

Scheme	Marks	AO
DR \((\sqrt{3} + \mathrm{i})^5 = (\sqrt{3})^5 + 5(\sqrt{3})^4\mathrm{i} + 10(\sqrt{3})^3\mathrm{i}^2 + 10(\sqrt{3})^2\mathrm{i}^3 + 5\sqrt{3}\mathrm{i}^4 + \mathrm{i}^5\)	M1 A1	1.1a 1.1
\(= -16\sqrt{3} + 16\mathrm{i}\) [so \(a = -16\sqrt{3}\) and \(b = 16\)]	A1	1.1
	[3]

Notes

M1: Binomial theorem showing correct pattern and coeffs (could use \({}^5\mathrm{C}_0\), etc

A1: (1st) correct expression (must evaluate \({}^n\mathrm{C}_r\) s)

Alternative method

Scheme	Marks
\((\sqrt{3} + \mathrm{i})^2 = (2 + 2\sqrt{3}\mathrm{i})\)	B1
\((\sqrt{3} + \mathrm{i})^3 = 8\mathrm{i}\) or \((\sqrt{3} + \mathrm{i})^4 = -8 + 8\sqrt{3}\mathrm{i}\)	B1
\((\sqrt{3} + \mathrm{i})^5 = -16\sqrt{3} + 16\mathrm{i}\) so \(a = -16\sqrt{3}\) and \(b = 16\)	B1

B1: (1st) or \(3 + 2\sqrt{3}\mathrm{i} - 1\)

B1: (2nd) Either seen

B1: (3rd) If without working, award no marks

Mark scheme (b)

Scheme	Marks	AO
(i) \(\|\sqrt{3} + \mathrm{i}\| = 2\)	B1	1.1
\(\arg(\sqrt{3} + \mathrm{i}) = \pi/6\)	B1	1.1
So \(z = 2(\cos \pi/6 + \mathrm{i}\sin \pi/6)\)	B1ft	2.5
	[3]
(ii) \(z^5 = 2^5(\cos 5\pi/6 + \mathrm{i}\sin 5\pi/6) = 32(\cos 5\pi/6 + \mathrm{i}\sin 5\pi/6)\)	B1ft B1ft	1.1 1.1
	[2]
(iii) \(32(\cos 5\pi/6 + \mathrm{i}\sin 5\pi/6)\) [\(= 32(-\sqrt{3}/2 + \tfrac{1}{2}\mathrm{i})\)]	B1
\(= -16\sqrt{3} + 16\mathrm{i}\) as before	B1
	[2]

Notes

(b)(i)

B1: (1st) modulus = 2

B1: (2nd) arg = \(\pi/6\) or \(30^\circ\)

B1ft: ft their modulus and argument

(b)(ii)

B1ft: (1st) (their modulus)\(^5\)

B1ft: (2nd) \(5 \times\) their argument

If fully correct (condone \(150^\circ\)) by converting \(-16\sqrt{3} + 16\mathrm{i}\) then allow SC B2 (but see below)

(b)(iii)

Condone no intermediate working for B2 provided 7(b)(ii) correct and from using de Moivre
Or \(32^2 = (-16\sqrt{3})^2 + 16^2\) oe B1, \(\tan^{-1}(-16/16\sqrt{3}) = 5\pi/6\) as in 2^nd quadrant B1 (but do not allow if (b)(ii) done by converting). In this case, if de Moivre used here:
\(32 = 2^5\) or \(\sqrt[5]{32} = 2\) SCB1,
and \(\pi/6 \times 5 = 5\pi/6\) or \(5\pi/6\ /\ 5 = \pi/6\) SCB1
NB if (b)(ii) is not fully correct, award no marks for part (iii)