A2 June 2020 Paper 1 Q6

AQACurrent spec9 marksComplex Numbers De Moivre's Theorem

6 Let \(w\) be the root of the equation \(z^7 = 1\) that has the smallest argument \(\alpha\) in the interval \(0 \lt \alpha \lt \pi\)

(a) Prove that \(w^n\) is also a root of the equation \(z^7 = 1\) for any integer \(n\). [1 mark]

(b) Prove that \(1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\) [2 marks]

Blank Argand diagram with axes Re(z) and Im(z)

(d) Prove that\[\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2}\] [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Completes a rigorous argument to show that \(w^n\) satisfies the equation \(z^7 = 1\)	R1	2.1

Typical solution

\[(w^n)^7 = w^{7n} = (w^7)^n = 1^n = 1\]

\(\therefore w^n\) satisfies the equation \(z^7 = 1\)

Mark scheme (b)

Scheme	Marks	AO
Deduces that the LHS is the sum of the roots of \(z^7 = 1\) or factorises \(w^7 - 1\) or uses the sum of a geometric series with values for \(n\) and \(a\).	M1	2.2a
Completes a rigorous argument to show \(1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\)	R1	2.1

Typical solution

The roots of \(z^7 - 1 = 0\) are \(1\), \(w\), \(w^2\), \(w^3\), \(w^4\), \(w^5\), and \(w^6\)

\(z^6\) term \(= 0 \therefore\) sum of roots \(= 0\)

and

\[1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\]

as required.

Argand diagram: dashed unit circle with 1 marked on each positive axis and six points on the circle joined to the origin, equally spaced at multiples of 2π/7 and symmetric about the real axis — Mark scheme (c)

Scheme	Marks	AO
Shows the six required points or vectors with correct arguments, approximately correctly spaced and approximately symmetric in the real axis.	M1	1.1a
Clearly shows that points/vectors have modulus 1. PI by “1” marked on an axis. Labelling of points not required.	A1	1.1b

Mark scheme (d)

Scheme	Marks	AO
States that \(w = \cos\frac{2\pi}{7} + \cdots\)	B1	1.1b
Explains that complex conjugate pairs have the same real part.	E1	2.4
Deduces that a sum of pairs of powers of \(w\) equals twice the cosine of a correct angle.	M1	2.2a
Completes a rigorous argument, using \(1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\) to show \(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2}\)	R1	2.1
	(9 marks)

Typical solution

\[w = \cos\frac{2\pi}{7} + \mathrm{i}\sin\frac{2\pi}{7}\]

and

\[\begin{aligned} w^6 &= \cos\left(-\frac{2\pi}{7}\right) + \mathrm{i}\sin\left(-\frac{2\pi}{7}\right) \\ &= \cos\frac{2\pi}{7} - \mathrm{i}\sin\frac{2\pi}{7} \end{aligned}\]

which is the conjugate of \(w\).

\[\therefore w + w^6 = 2\cos\frac{2\pi}{7}\]

Similarly \(\quad w^2 + w^5 = 2\cos\frac{4\pi}{7}\)

and \(\quad w^3 + w^4 = 2\cos\frac{6\pi}{7}\)

From part (b),

\[1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0\]\[(w + w^6) + (w^2 + w^5) + (w^3 + w^4) = -1\]\[2\cos\frac{2\pi}{7} + 2\cos\frac{4\pi}{7} + 2\cos\frac{6\pi}{7} = -1\]\[\therefore \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2}\]

as required.