A2 June 2022 Paper 2 Q4

Notes

M1: A complete method to find both \(z_1\) and \(z_2\) in the form \(a + b\mathrm{i}\) and adds them together.

dM1: Dependent on previous method mark, finds the modulus and argument of \(z_1 + z_2\). They must show their method, just stating modulus = 12 and argument \(= \dfrac{2\pi}{3}\) is not sufficient as this is a show question.

Alternative 1: Factorises out 12 and find the argument

Alternative 2: uses \(12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}} = 12\left(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\right) = \ldots\)

A1*: Achieves the correct answer following no errors or omissions.
Alternatively shows that \(12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}} = -6 + 6\sqrt{3}\mathrm{i}\) and concludes therefore \(z_1 + z_2 = 12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\) *

Alternative 1 and Alternative 2 (for the dM1 and A1*)

Alternative 3

M1: Factorises out 12 and writes in the form \(12\left[\ldots\cos\left(\dfrac{\pi}{3}\right) + \ldots\mathrm{i}\sin\left(\dfrac{\pi}{3}\right) + \ldots\cos\left(\dfrac{5\pi}{6}\right) + \ldots\mathrm{i}\sin\left(\dfrac{5\pi}{6}\right)\right]\)

dM1: Dependent on previous mark. Writes in the form \(12(a + b\mathrm{i})\) leading to the form \(12(\cos\theta + \mathrm{i}\sin\theta)\)

A1*: Achieves the correct answer following no errors or omissions.

Alternative 4

M1: Factorises out 6 and writes in the form \(6\mathrm{e}^{\frac{\pi}{3}\mathrm{i}}\left(1 + \sqrt{3}\,\mathrm{e}^{\frac{\pi}{2}\mathrm{i}}\right) = 6\mathrm{e}^{\frac{\pi}{3}\mathrm{i}}(1 + a\mathrm{i})\)

dM1: Dependent on previous method mark, finds the modulus and argument of \((1 + a\mathrm{i})\) or \(12(a + b\mathrm{i})\) leading to the form \(12(\cos\theta + \mathrm{i}\sin\theta)\)

A1*: Achieves the correct answer following no errors or omissions.

Alternative 5

Scheme	Marks	AO
Uses geometry to show that \(z_1\), \(z_2\) and \(z_1 + z_2\) form a right-angled triangle	M1	3.1a
\(\arg(z_1 + z_2) = \dfrac{\pi}{3} + \tan^{-1}\left(\dfrac{6\sqrt{3}}{6}\right) = \ldots\left\{\dfrac{2\pi}{3}\right\}\) \(|z_1 + z_2| = \sqrt{(6)^2 + \left(6\sqrt{3}\right)^2} = \ldots\{12\}\)	dM1	1.1b
\(z_1 + z_2 = 12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\) *	A1*	1.1b
	(3)

(In the printed diagram the axis labels Re and Im are the wrong way round: the real axis is the horizontal one.)

M1: Draws a diagram to show that \(z_1\), \(z_2\) and \(z_1 + z_2\) form a right-angled triangle.

dM1: Dependent on previous method mark, finds the modulus and argument of \(z_1 + z_2\)

A1*: Achieves the correct answer following no errors or omissions.

Note: Writing \(\arg(z_1 + z_2) = \arctan\left(\dfrac{6\sqrt{3}}{-6}\right) = -\dfrac{\pi}{3}\) therefore \(\arg(z_1 + z_2) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}\) with no diagram or finding \(z_1 + z_2\) is M0dM0A0

Notes

(In the printed diagram the axis labels Re and Im are the wrong way round: the real axis is the horizontal one.)

M1: Draws a diagram and recognises that the shortest distance will form a right-angled triangle.

M1: Uses trigonometry to find the shortest length.

A1: Correct exact value.

Alternative 1

M1: Finds the equation of the half-line by attempting \(m = -\tan\left(\dfrac{\pi}{3}\right)\) \(c = 5\tan\left(\dfrac{\pi}{3}\right)\). Finds \(x^2 + y^2\) in terms of \(x\), differentiates, sets \(= 0\) and finds the value of \(x\).

M1: Uses their value of \(x\) to find the minimum value of \(\sqrt{x^2 + y^2}\)

A1: Correct exact value.

Alternative 2

M1: Finds the equation of the half-line by attempting \(m = -\tan\left(\dfrac{\pi}{3}\right)\) \(c = 5\tan\left(\dfrac{\pi}{3}\right)\). Finds the equation of the line perpendicular which passes through the origin. Finds the point of intersection of the lines

M1: Finds the distance from the origin to their point of intersection

A1: Correct exact value.