A2 June 2019 Paper 1 Q9

OCR ACurrent spec12 marksComplex Numbers De Moivre's Theorem

9 In this question you must show detailed reasoning.

You are given the complex number \(\omega = \cos\frac{2}{5}\pi + \mathrm{i}\sin\frac{2}{5}\pi\) and the equation \(z^5 = 1\).

(a) Show that \(\omega\) is a root of the equation. [2]

(b) Write down the other four roots of the equation. [1]

(d) Hence show that \(\left(\omega + \dfrac{1}{\omega}\right)^2 + \left(\omega + \dfrac{1}{\omega}\right) - 1 = 0\). [3]

(e) Hence determine the value of \(\cos\frac{2}{5}\pi\) in the form \(a + b\sqrt{c}\) where \(a\), \(b\) and \(c\) are rational numbers to be found. [4]

Mark scheme (a)

Scheme	Marks	AO
DR \(\omega = \cos\dfrac{2\pi}{5} + \mathrm{i}\sin\dfrac{2\pi}{5}\) \(\Rightarrow \omega^5 = \left(\cos\dfrac{2\pi}{5} + \mathrm{i}\sin\dfrac{2\pi}{5}\right)^5 = \cos 2\pi + \mathrm{i}\sin 2\pi = 1 + 0\mathrm{i} = 1\)	M1 A1	2.1 1.1
	[2]

Notes

M1: Finding \(\omega^5\)

A1: AG

Use of exponentials is satisfactory. Could be argued backwards

Mark scheme (b)

Scheme	Marks	AO
\(\omega^2, \omega^3, \omega^4, 1\)	B1	1.1
	[1]

Notes

Alternative: Roots are \(\cos\dfrac{2k\pi}{5} + \mathrm{i}\sin\dfrac{2k\pi}{5}\) for \(k = 2, 3, 4\) and 1 (or \(k = 5\))

Exponentials satisfactory

Mark scheme (c)

Scheme	Marks	AO
DR \(\omega^5 - 1 = 0\) \(\Rightarrow (\omega - 1)(\omega^4 + \omega^3 + \omega^2 + \omega + 1) = 0\)	M1	1.1a
\(\Rightarrow \omega^4 + \omega^3 + \omega^2 + \omega = -1\)	A1	2.1
	[2]

Notes

M1: Use equation and \(\omega\)

A1: AG

Alternative method

Scheme	Marks
\(1 + \omega + \omega^2 + \omega^3 + \omega^4 = \dfrac{1 - \omega^5}{1 - \omega} = \dfrac{0}{1 - \omega}\)	M1
or \(\omega + \omega^2 + \omega^3 + \omega^4 = \omega\left(\dfrac{1 - \omega^4}{1 - \omega}\right) = \left(\dfrac{\omega - \omega^5}{1 - \omega}\right) = \left(\dfrac{\omega - 1}{1 - \omega}\right) = -1\)	A1
	[2]

Alternative

sum of roots \(= -\dfrac{b}{a}\) where \(b = 0\) M1 – needs explanation – i.e. coefficient of \(z^4\) term \(= 0\)

Mark scheme (d)

Scheme	Marks	AO
AG \(\left(\omega + \dfrac{1}{\omega}\right)^2 + \left(\omega + \dfrac{1}{\omega}\right) - 1 = \omega^2 + 2 + \dfrac{1}{\omega^2} + \omega + \dfrac{1}{\omega} - 1\)	M1	2.1
\(= \dfrac{1}{\omega^2}\left(\omega^4 + \omega^2 + 1 + \omega^3 + \omega\right) = 0\)	A1	1.1
Since \(\dfrac{1}{\omega^2} \neq 0,\ \omega^4 + \omega^2 + 1 + \omega^3 + \omega = 0\) or from part (c)	A1	2.2a
	[3]

Notes

M1: Multiply out

Alternative method

Scheme	Marks
\(\omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0\) \(\Rightarrow \omega^2\left(\omega^2 + \omega + 1 + \dfrac{1}{\omega} + \dfrac{1}{\omega^2}\right) = 0\)	M1
\(\Rightarrow \omega^2\left(\left(\omega^2 + 2 + \dfrac{1}{\omega^2}\right) + \left(\omega + \dfrac{1}{\omega}\right) + 1 - 2\right) = 0\)	A1
\(\Rightarrow \omega^2\left(\left(\omega + \dfrac{1}{\omega}\right)^2 + \left(\omega + \dfrac{1}{\omega}\right) - 1\right) = 0\) Since \(\omega^2 \neq 0,\ \left(\omega + \dfrac{1}{\omega}\right)^2 + \left(\omega + \dfrac{1}{\omega}\right) - 1 = 0\)	A1
	[3]

M1: For extraction of \(\omega^2\)

A1: For dealing with the 2

Mark scheme (e)

Scheme	Marks	AO
\(\dfrac{1}{\omega} = \cos\dfrac{2\pi}{5} - \mathrm{i}\sin\dfrac{2\pi}{5}\) \(\Rightarrow \left(\omega + \dfrac{1}{\omega}\right) = 2\cos\dfrac{2\pi}{5}\)	B1	3.1a
From (d) solving quadratic: \(\left(\omega + \dfrac{1}{\omega}\right) = \dfrac{-1 \pm \sqrt{5}}{2}\)	B1	3.1a
\(\Rightarrow 2\cos\dfrac{2\pi}{5} = \dfrac{\sqrt{5} - 1}{2} \Rightarrow \cos\dfrac{2\pi}{5} = \dfrac{\sqrt{5} - 1}{4}\)	M1	2.2a
\(= -\dfrac{1}{4} + \dfrac{\sqrt{5}}{4}\) or \(-\dfrac{1}{4} + \dfrac{1}{4}\sqrt{5}\) or \(-0.25 + 0.25\sqrt{5}\)	A1	2.3
	[4]

Notes

B1: \(\omega + \dfrac{1}{\omega}\) may be seen in (d)

B1: BC

M1: Equating

A1: For taking the valid value and presenting in correct form oe. No other forms acceptable

(corrected from the printed mark scheme: the printed scheme says “From (iii)”; the quadratic is the one in part (d).)