FP1 June 2018 Q9

EdexcelOld spec12 marksComplex Numbers

(i) Given that \[\frac{3w + 7}{5} = \frac{p - 4\mathrm{i}}{3 - \mathrm{i}} \qquad \text{where } p \text{ is a real constant}\]

(a) express \(w\) in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real constants.
Give your answer in its simplest form in terms of \(p\). (5)

Given that \(\arg w = -\dfrac{\pi}{2}\)

(b) find the value of \(p\). (1)

(ii) Given that \[(z + 1 - 2\mathrm{i})^* = 4\mathrm{i}z\] find \(z\), giving your answer in the form \(z = x + \mathrm{i}y\), where \(x\) and \(y\) are real constants. (6)

Mark scheme (i)(a)

Scheme	Marks
\(\dfrac{3w + 7}{5} = \dfrac{(p - 4\mathrm{i})}{(3 - \mathrm{i})} \times \dfrac{(3 + \mathrm{i})}{(3 + \mathrm{i})}\)	M1
\(= \left(\dfrac{3p + 4}{10}\right) + \left(\dfrac{p - 12}{10}\right)\mathrm{i}\)	B1
So, \(w = \left(\dfrac{3p - 10}{6}\right) + \left(\dfrac{p - 12}{6}\right)\mathrm{i}\)	dM1 A1 A1
	(5)

Notes

M1: Multiplies by \(\dfrac{(3 + \mathrm{i})}{(3 + \mathrm{i})}\) or divide by \((9 - 3\mathrm{i})\) then multiply by \(\dfrac{(9 + 3\mathrm{i})}{(9 + 3\mathrm{i})}\)

B1: Evidence of \((3 - \mathrm{i})(3 + \mathrm{i}) = 10\) or \(3^2 + 1^2\) or \(9^2 + 3^2\)

dM1: Rearranges to \(w = \ldots\)

A1: At least one of either the real or imaginary part of \(w\) is correct in any equivalent form.

A1: Correct \(w\) in the form \(a + b\mathrm{i}\). Accept \(a + \mathrm{i}b\).

ALT (i) (a)

Scheme	Marks
\((3 - \mathrm{i})(3w + 7) = 5(p - 4\mathrm{i})\)
\(9w + 21 - 3\mathrm{i}w - 7\mathrm{i} = 5p - 20\mathrm{i}\)
\(w(9 - 3\mathrm{i}) = 5p - 21 - 13\mathrm{i}\)
Let \(w = a + b\mathrm{i}\), so \((a + b\mathrm{i})(9 - 3\mathrm{i}) = 5p - 21 - 13\mathrm{i}\)
\(9a + 3b - 3a\mathrm{i} + 9b\mathrm{i} = 5p - 21 - 13\mathrm{i}\)
Real: \(9a + 3b = 5p - 21\) Imaginary: \(-3a + 9b = -13\)	M1 B1
\(b = \dfrac{p - 12}{6},\ a = \dfrac{3p - 10}{6}\)	dM1 A1
\(w = \left(\dfrac{3p - 10}{6}\right) + \left(\dfrac{p - 12}{6}\right)\mathrm{i}\)	A1
	(5)

M1: Sets \(w = a + b\mathrm{i}\) and equates at least either the real or imaginary part.

B1: \(9a + 3b = 5p - 21\)

dM1: Solves to finds \(a = \ldots\) and \(b = \ldots\)

A1: At least one of \(a\) or \(b\) is correct in any equivalent form.

A1: Correct \(w\) in the form \(a + b\mathrm{i}\). Accept \(a + \mathrm{i}b\).

Mark scheme (i)(b)

Scheme	Marks
\(\left\{\arg w = -\dfrac{\pi}{2} \Rightarrow \left(\dfrac{3p - 10}{6}\right) = 0\right\} \Rightarrow p = \dfrac{10}{3}\)	B1ft
	(1)

Notes

B1ft: \(p = \dfrac{10}{3}\). Follow through provided \(p < 12\)

Mark scheme (ii)

Scheme	Marks
\((x + \mathrm{i}y + 1 - 2\mathrm{i})^* = 4\mathrm{i}(x + \mathrm{i}y)\)	M1
\(x - \mathrm{i}y + 1 + 2\mathrm{i} = 4\mathrm{i}(x + \mathrm{i}y)\) or \(x + \mathrm{i}y + 1 - 2\mathrm{i} = -4\mathrm{i}(x - \mathrm{i}y)\)	M1
\(x - \mathrm{i}y + 1 + 2\mathrm{i} = 4\mathrm{i}x - 4y\)
Real: \(x + 1 = -4y\) Imaginary: \(-y + 2 = 4x\)	A1
\(4x + 16y = -4\) \(4x + y = 2\) \(\Rightarrow 15y = -6 \Rightarrow y = \ldots\)	ddM1
So, \(x = \dfrac{3}{5},\ y = -\dfrac{2}{5}\quad \left\{z = \dfrac{3}{5} - \dfrac{2}{5}\mathrm{i}\right\}\)	A1 A1
	(6)
	(12 marks)

Notes

M1: Replaces \(z\) with \(x + \mathrm{i}y\) on both sides of the equation

M1: Fully correct method for applying the conjugate

A1: \(x + 1 = -4y\) and \(-y + 2 = 4x\)

ddM1: Solves two equations in \(x\) and \(y\) to obtain at least one of \(x\) or \(y\)

A1: At least one of either \(x\) or \(y\) are correct

A1: Both \(x\) and \(y\) are correct