FP1 June 2009 Q1

1. The complex numbers \(z_1\) and \(z_2\) are given by \[z_1 = 2 - \mathrm{i} \quad \text{and} \quad z_2 = -8 + 9\mathrm{i}\]

(a) Show \(z_1\) and \(z_2\) on a single Argand diagram. (1)

Find, showing your working,

(b) the value of \(|z_1|\), (2)

(d) \(\dfrac{z_2}{z_1}\) in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real. (3)

Scheme	Marks
	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(\|z_1\| = \sqrt{2^2 + (-1)^2} = \sqrt{5}\) (or awrt 2.24)	M1 A1
	(2)

Notes

(b) M1 Attempt at Pythagoras to find modulus of either complex number

A1 condone correct answer even if negative sign not seen in (−1) term

A0 for \(\pm\sqrt{5}\)

Mark scheme (c)

Scheme	Marks
\(\alpha = \arctan\left(\dfrac{1}{2}\right)\) or \(\arctan\left(-\dfrac{1}{2}\right)\)	M1
\(\arg z_1 = -0.46\) or 5.82 (awrt) (answer in degrees is A0 unless followed by correct conversion)	A1
	(2)

Notes

(c) \(\arctan 2\) is M0 unless followed by \(\boxed{\tfrac{3\pi}{2} + \arctan 2}\) or \(\boxed{\tfrac{\pi}{2} - \arctan 2}\) Need to be clear that \(\arg z = -0.46\) or 5.82 for A1

Mark scheme (d)

Scheme	Marks
\(\dfrac{-8 + 9\mathrm{i}}{2 - \mathrm{i}} \times \dfrac{2 + \mathrm{i}}{2 + \mathrm{i}}\)	M1
\(= \dfrac{-16 - 8\mathrm{i} + 18\mathrm{i} - 9}{5} = -5 + 2\mathrm{i}\) i.e. \(a = -5\) and \(b = 2\) or \(-\tfrac{2}{5}a\)	A1 A1ft
	(3)
	[8]

Alternative method to part (d)

Scheme	Marks
\(-8 + 9\mathrm{i} = (2 - i)(a + bi)\), and so \(2a + b = -8\) and \(2b - a = 9\) and attempt to solve as far as equation in one variable	M1
So \(a = -5\) and \(b = 2\)	A1 A1cao

Notes

(d) M1 Multiply numerator and denominator by conjugate of their denominator

A1 for −5 and A1 for 2i (should be simplified)

Alternative scheme for (d) Allow slips in working for first M1