FP1 June 2014 Q1

1. The complex numbers \(z_1\) and \(z_2\) are given by \[z_1 = p + 2\mathrm{i} \text{ and } z_2 = 1 - 2\mathrm{i}\] where \(p\) is an integer.

(a) Find \(\dfrac{z_1}{z_2}\) in the form \(a + b\mathrm{i}\) where \(a\) and \(b\) are real. Give your answer in its simplest form in terms of \(p\). (4)

Given that \(\left|\dfrac{z_1}{z_2}\right| = 13\),

(b) find the possible values of \(p\). (4)

Mark scheme (a)

Scheme	Marks
\(\dfrac{z_1}{z_2} = \dfrac{p + 2i}{1 - 2i} \cdot \dfrac{1 + 2i}{1 + 2i}\) Multiplying top and bottom by conjugate	M1
\(= \dfrac{p + 2pi + 2i - 4}{5}\) At least 3 correct terms in the numerator, evidence that \(\mathrm{i}^2 = -1\) and denominator real.	M1
\(= \dfrac{p - 4}{5}, \qquad + \dfrac{2p + 2}{5}\mathrm{i}\) Real + imaginary with i factored out. Accept single denominator with numerator in correct form. Accept ‘a=’ and ‘b=’.	A1, A1
	(4)

Mark scheme (b)

Scheme	Marks
\(\left\|\dfrac{z_1}{z_2}\right\|^2 = \left(\dfrac{p - 4}{5}\right)^2 + \left(\dfrac{2p + 2}{5}\right)^2\) Accept their answers to part (a). Any erroneous i or \(\mathrm{i}^2\) award M0	M1
\(\left(\dfrac{p - 4}{5}\right)^2 + \left(\dfrac{2p + 2}{5}\right)^2 = 13^2\) or \(\sqrt{\left(\dfrac{p - 4}{5}\right)^2 + \left(\dfrac{2p + 2}{5}\right)^2} = 13\) \(\left\|\dfrac{z_1}{z_2}\right\|^2 = 13^2\) or \(\left\|\dfrac{z_1}{z_2}\right\| = 13\)	dM1
\(\dfrac{p^2 - 8p + 16}{25} + \dfrac{4p^2 + 8p + 4}{25} = 169 \text{ or } 13^2\)
\(5p^2 + 20 = 4225\)
\(p^2 = 841 \Rightarrow p = \pm 29\) dM1: Attempt to solve their quadratic in \(p\), dependent on both previous Ms. A1: both 29 and \(-29\)	dM1A1
OR
\(\dfrac{\|z_1\|}{\|z_2\|} = \dfrac{\sqrt{p^2 + 4}}{\sqrt{5}}\) Finding moduli Any erroneous i or \(\mathrm{i}^2\) award M0	M1
\(\dfrac{\sqrt{p^2 + 4}}{\sqrt{5}} = 13\) oe Equating to 13	dM1
\(\dfrac{p^2 + 4}{5} = 169 \text{ or } 13^2\) oe
\(p^2 = 841 \Rightarrow p = \pm 29\) dM1: Attempt to solve their quadratic in \(p\), dependent on both previous Ms A1: both 29 and \(-29\)	dM1A1
	(4)
	(8 marks)