FP1 June 2010 Q1

1. \[z = 2 - 3\mathrm{i}\]

(a) Show that \(z^2 = -5 - 12\mathrm{i}\). (2)

Find, showing your working,

(b) the value of \(|z^2|\), (2)

(d) Show \(z\) and \(z^2\) on a single Argand diagram. (1)

Mark scheme (a)

Scheme	Marks
\((2 - 3\mathrm{i})(2 - 3\mathrm{i}) = \ldots\ldots\) Expand and use \(\mathrm{i}^2 = -1\), getting completely correct expansion of 3 or 4 terms	M1
Reaches \(-5 - 12\mathrm{i}\) after completely correct work (must see \(4 - 9\)) (*)	A1cso
	(2)

Notes

(a) M1: for \(4 - 9 - 12\mathrm{i}\) or \(4 - 9 - 6\mathrm{i} - 6\mathrm{i}\) or \(4 - 3^2 - 12\mathrm{i}\) but must have correct statement seen and see i^2 replaced by −1 maybe later

A1: Printed answer. Must see \(4 - 9\) in working.

Jump from \(4 - 6\mathrm{i} - 6\mathrm{i} + 9\mathrm{i}^2\) to \(-5 - 12\mathrm{i}\) is M0A0

Mark scheme (b)

Scheme	Marks
\(\|z^2\| = \sqrt{(-5)^2 + (-12)^2} = 13\) or \(\|z^2\| = \sqrt{5^2 + 12^2} = 13\)	M1 A1
	(2)

Alternative methods for part (b)

Scheme	Marks
\(\|z^2\| = \|z\|^2 = 2^2 + (-3)^2 = 13\) Or: \(\|z^2\| = zz^* = 13\)	M1 A1
	(2)

Notes

(b) Method may be implied by correct answer. NB \(|z^2| = 169\) is M0 A0

Mark scheme (c)

Scheme	Marks
\(\tan\alpha = \dfrac{12}{5}\) (allow \(-\dfrac{12}{5}\)) or \(\sin\alpha = \dfrac{12}{13}\) or \(\cos\alpha = \dfrac{5}{13}\)	M1
\(\arg(z^2) = -(\pi - 1.176\ldots) = -1.97\) (or 4.32) allow awrt	A1
	(2)

Alternative method for part (c)

Scheme	Marks
\(\alpha = 2 \times \arctan\left(-\tfrac{3}{2}\right)\) (allow \(\tfrac{3}{2}\)) or use \(\dfrac{\pi}{2} + \arctan\dfrac{5}{12}\)	M1
so \(\arg(z^2) = -(\pi - 1.176\ldots) = -1.97\) (or 4.32) allow awrt	A1

Scheme	Marks
Both in correct quadrants. Approximate relative scale No labels needed Allow two diagrams if some indication of scale Allow points or arrows	B1
	(1)
	7 marks