FP1 June 2012 Q7

EdexcelOld spec11 marksComplex Numbers

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

(a) Calculate \(\arg z\), giving your answer in radians to 2 decimal places. (2)

Use algebra to express

(b) \(z + z^2\) in the form \(a + b\mathrm{i}\sqrt{3}\), where \(a\) and \(b\) are integers, (3)

Given that \[w = \lambda - 3\mathrm{i}\] where \(\lambda\) is a real constant, and \(\arg(4 - 5\mathrm{i} + 3w) = -\dfrac{\pi}{2}\),

(d) find the value of \(\lambda\). (2)

Mark scheme (a)

Scheme	Marks
\(\arg z = -\tan^{-1}\left(\tfrac{\sqrt{3}}{2}\right)\) \(\tan^{-1}\left(\pm\tfrac{\sqrt{3}}{2}\right)\) or \(\tan^{-1}\left(\pm\tfrac{2}{\sqrt{3}}\right)\) seen or evaluated	M1
\(= -0.7137243789\ldots = -0.71\ (2\text{ dp})\) awrt \(-0.71\) or awrt 5.57	A1
	[2]

Notes

Awrt \(\pm 0.71\) or awrt \(\pm 0.86\) can be taken as evidence for the method mark.
Or \(\pm 40.89\) or \(\pm 49.10\) if working in degrees

NB \(\tan\left(\dfrac{\sqrt{3}}{2}\right) = 1.18\) and \(\tan\left(\dfrac{2}{\sqrt{3}}\right) = 2.26\) and both score M0

Mark scheme (b)

Scheme	Marks
\(z^2 = \left(2 - \mathrm{i}\sqrt{3}\right)\left(2 - \mathrm{i}\sqrt{3}\right)\) \(= 4 - 2\mathrm{i}\sqrt{3} - 2\mathrm{i}\sqrt{3} + 3\mathrm{i}^2\) An attempt to multiply out the brackets to give four terms (or four terms implied).	M1
\(= 2 - \mathrm{i}\sqrt{3} + \left(4 - 4\mathrm{i}\sqrt{3} - 3\right)\) \(= 2 - \mathrm{i}\sqrt{3} + \left(1 - 4\mathrm{i}\sqrt{3}\right)\) \(= 3 - 5\mathrm{i}\sqrt{3}\)\(\quad\)(Note: \(a = 3, b = -5\).) M1: An understanding that \(\mathrm{i}^2 = -1\) and an attempt to add \(z\) and put in the form \(a + b\mathrm{i}\sqrt{3}\) A1: \(3 - 5\mathrm{i}\sqrt{3}\)	M1A1
	[3]

Notes

\(z + z^2 = 2 - \mathrm{i}\sqrt{3} + \left(4 - 4\mathrm{i}\sqrt{3} + 3\right) = 9 - 5\mathrm{i}\sqrt{3}\) scores M1M0A0 (No evidence of \(\mathrm{i}^2 = -1\))

Alternative (Way 2)

Scheme	Marks
\(z + z^2 = z(1 + z)\)
\(= \left(2 - \mathrm{i}\sqrt{3}\right)\left(1 + \left(2 - \mathrm{i}\sqrt{3}\right)\right)\) \(= \left(2 - \mathrm{i}\sqrt{3}\right)\left(3 - \mathrm{i}\sqrt{3}\right)\) \(= 6 - 2\mathrm{i}\sqrt{3} - 3\mathrm{i}\sqrt{3} + 3\mathrm{i}^2\) An attempt to multiply out the brackets to give four terms (or four terms implied).	M1
\(= 6 - 2\mathrm{i}\sqrt{3} - 3\mathrm{i}\sqrt{3} - 3\) M1: An understanding that \(\mathrm{i}^2 = -1\) and an attempt to put in the form \(a + b\mathrm{i}\sqrt{3}\)	M1
\(= 3 - 5\mathrm{i}\sqrt{3}\)\(\quad\)(Note: \(a = 3, b = -5\).) \(3 - 5\mathrm{i}\sqrt{3}\)	A1
	[3]

Mark scheme (c)

Scheme	Marks
\(\dfrac{z + 7}{z - 1} = \dfrac{2 - \mathrm{i}\sqrt{3} + 7}{2 - \mathrm{i}\sqrt{3} - 1}\) Substitutes \(z = 2 - \mathrm{i}\sqrt{3}\) into both numerator and denominator.	M1
\(= \dfrac{\left(9 - \mathrm{i}\sqrt{3}\right)}{\left(1 - \mathrm{i}\sqrt{3}\right)} \times \dfrac{\left(1 + \mathrm{i}\sqrt{3}\right)}{\left(1 + \mathrm{i}\sqrt{3}\right)}\) Simplifies \(\dfrac{z + 7}{z - 1}\) and* multiplies by \(\dfrac{\textbf{their } \left(1 + \mathrm{i}\sqrt{3}\right)}{\textbf{their } \left(1 + \mathrm{i}\sqrt{3}\right)}\)*	dM1
\(= \dfrac{9 + 9\mathrm{i}\sqrt{3} - \mathrm{i}\sqrt{3} + 3}{1 + 3}\) \(= \dfrac{12 + 8\mathrm{i}\sqrt{3}}{4}\) Simplifies realising that a real number is needed in the denominator and applies \(\mathrm{i}^2 = -1\) in their numerator expression and denominator expression.	M1
\(= 3 + 2\mathrm{i}\sqrt{3}\)\(\quad\)(Note: \(c = 3,\ d = 2\).) \(3 + 2\mathrm{i}\sqrt{3}\)	A1
	[4]

Mark scheme (d)

Scheme	Marks
\(w = \lambda - 3\mathrm{i}\), and \(\arg(4 - 5\mathrm{i} + 3w) = -\dfrac{\pi}{2}\)
\((4 - 5\mathrm{i} + 3w = 4 + 3\lambda - 14\mathrm{i})\)
So real part of \((4 - 5\mathrm{i} + 3w) = 0\) or \(4 + 3\lambda = 0\) States real part of \((4 - 5\mathrm{i} + 3w) = 0\) or \(4 + 3\lambda = 0\)	M1
So, \(\lambda = -\tfrac{4}{3}\) \(-\tfrac{4}{3}\)	A1
	[2]
	11 marks

Notes

Allow \(\pm\left(\dfrac{14}{3\lambda + 4}\right) = \pm\infty \Rightarrow 3\lambda + 4 = 0\) M1 \(\Rightarrow \lambda = -\dfrac{4}{3}\) A1