A2 June 2025 Paper 1 Q9

OCR MEICurrent spec10 marksComplex Numbers De Moivre's Theorem

9 The figure below shows an Argand diagram with a regular pentagon ABCDE. The point A represents the real number 1. The point B represents the complex number \(w\).

Argand diagram: regular pentagon ABCDE centred at O, with A at 1 on the positive real axis, B in the first quadrant, C in the second, D in the third and E in the fourth quadrant

(a)

(i) Write down, in terms of \(w\), the complex numbers represented by the points C, D and E. [1]

(ii) Write down an equation whose roots are the complex numbers represented by the points A, B, C, D and E. [1]

(iii) Show that the sum of these roots is zero. [2]

(b)

(i) Find \(w\). Give your answer in the form \(r(\cos\theta + \mathrm{i}\sin\theta)\), where \(r \gt 0\) and \(\theta = k\pi\), where \(k\) is a positive constant to be found. [1]

(ii) By considering the line segment AB, show that the length of each side of the pentagon is \(2\sin\dfrac{\pi}{5}\). [5]

Mark scheme (a)

Scheme	Marks	AO
(i) \(w^2\), \(w^3\) and \(w^4\)	B1	1.2
	[1]
(ii) \(z^5 = 1\) oe	B1	1.2
	[1]
(iii) \(\left[1 + w + w^2 + w^3 + w^4 =\right] \frac{1 - w^5}{1 - w}\)	M1	2.4
\(= 0\) as \(w^5 = 1\)	A1	2.1
	[2]

Notes

(a)(i)

B1: or \(w\mathrm{e}^{\frac{2\pi}{5}\mathrm{i}}\), \(w\mathrm{e}^{\frac{4\pi}{5}\mathrm{i}}\) and \(w\mathrm{e}^{\frac{-4\pi}{5}\mathrm{i}}\) (or \(w\mathrm{e}^{\frac{6\pi}{5}\mathrm{i}}\) or \(w^*\)) oe. Allow misattribution between C, D, E.

(a)(ii)

B1: allow any variable for \(z\)
or \(z^5 = \cos 2\pi + \mathrm{i}\sin 2\pi\)
or \(z^5 = \cos 2k\pi + \mathrm{i}\sin 2k\pi\) (or \(\mathrm{e}^{2k\pi\mathrm{i}}\)) provided \(k \in \mathbb{Z}\) seen

(a)(iii)

M1: correct use of geometric series formula in terms of \(w\) or exponentials. Cannot be implied.

A1: \(w^5 = 1\) or \(1 - \mathrm{e}^{2\pi\mathrm{i}}\) seen before completion.

Alternative method

Scheme	Marks
Sum of roots = coefficient of \(z^4\) [in \(z^5 - 1 = 0\)]	M1
This coefficient is zero so sum of roots is zero	A1

“sum of roots \(= -\frac{b}{a} = 0\)” alone is insufficient

Mark scheme (b)

Scheme	Marks	AO
(i) \([w =] \cos\frac{2\pi}{5} + \mathrm{i}\sin\frac{2\pi}{5}\)	B1	1.1
	[1]
(ii) \([w - 1 =] \cos\frac{2\pi}{5} - 1 + \mathrm{i}\sin\frac{2\pi}{5}\)	M1	3.1a
\(\sqrt{\left(\cos\frac{2\pi}{5} - 1\right)^2 + \sin^2\frac{2\pi}{5}}\)	M1	1.1
\(= \sqrt{\cos^2\dfrac{2\pi}{5} - 2\cos\dfrac{2\pi}{5} + 1 + \sin^2\dfrac{2\pi}{5}}\)	A1	1.1
\(= \sqrt{2 - 2\left(1 - 2\sin^2\dfrac{\pi}{5}\right)}\)	M1	3.1a
\(= 2\sin\dfrac{\pi}{5}\)	A1	2.1
	[5]

Notes

(b)(ii)

M1: or \(\begin{pmatrix} \cos\frac{2\pi}{5} - 1 \\ \sin\frac{2\pi}{5} \end{pmatrix}\); or equivalent for \(1 - w\). Soi by correct modulus calculation.

M1: finding the modulus or modulus² of their \(w - 1\) (or \(1 - w\))

A1: or \(\sqrt{2 - 2\cos\frac{2\pi}{5}}\) or correct expression for AB²

M1: correct use of double angle formula, must be seen. Accept \(2\left(1 - \cos\frac{2\pi}{5}\right) = 2\left(2\sin^2\frac{\pi}{5}\right)\) without intermediate step but not \(2 - 2\cos\frac{2\pi}{5} = 2\left(2\sin^2\frac{\pi}{5}\right)\).

A1: AG

Alternative method 1

Scheme	Marks
\(\mathrm{AB}^2 = 1^2 + 1^2 - 2(1)(1)\cos\frac{2\pi}{5}\)	M2
\(= 2 - 2\cos\frac{2\pi}{5}\)	A1
\(= 2 - 2\left(1 - 2\sin^2\frac{\pi}{5}\right)\)	M1
\(\mathrm{AB} = 2\sin\frac{\pi}{5}\)	A1

M2: correct use of cosine rule; or correct expression for AB

A1: simplified expression for AB² or AB

A1: AG

Alternative method 2

Scheme	Marks
Considering triangle AOB	M1
Considering right angled triangle OAM or OBM where M is the midpoint of AB	M1
\(\frac{1}{2}\mathrm{AB} = \sin\frac{\pi}{5}\) or \(\mathrm{AM} = \sin\frac{\pi}{5}\)	A1
\(\mathrm{AB} = 2\sin\frac{\pi}{5}\)	A2

M1: may be stated or seen in diagram; condone missing labels if the triangle is clearly isosceles

M1: may be stated or seen in diagram; condone missing labels

A1: must be seen

A2: AG

Alternative method 3

Scheme	Marks
\(\dfrac{\mathrm{AB}}{\sin\frac{2\pi}{5}} = \dfrac{1}{\sin\frac{3\pi}{10}}\)	M1*
\(\mathrm{AB} = \dfrac{\sin\frac{2\pi}{5}}{\sin\frac{3\pi}{10}}\)	A1
\(\sin\dfrac{2\pi}{5} = 2\sin\dfrac{\pi}{5}\cos\dfrac{\pi}{5}\)	M1dep
\(\sin\dfrac{3\pi}{10} = \cos\dfrac{\pi}{5}\)	M1dep
\(\mathrm{AB} = 2\sin\dfrac{\pi}{5}\)	A1

M1*: using sine rule correctly

A1: correct expression for AB

M1dep: correct use of double angle formula

M1dep: using \(\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)\)

A1: AG