FP1 June 2015 Q4

4. \[z_1 = 3\mathrm{i} \text{ and } z_2 = \frac{6}{1 + \mathrm{i}\sqrt{3}}\]

(a) Express \(z_2\) in the form \(a + \mathrm{i}b\), where \(a\) and \(b\) are real numbers. (2)

(b) Find the modulus and the argument of \(z_2\), giving the argument in radians in terms of \(\pi\). (4)

(c) Show the three points representing \(z_1\), \(z_2\) and \((z_1 + z_2)\) respectively, on a single Argand diagram. (2)

Mark scheme (a)

Scheme	Marks
\(z_2 = \dfrac{6(1 - \mathrm{i}\sqrt{3})}{(1 + \mathrm{i}\sqrt{3})(1 - \mathrm{i}\sqrt{3})} = \dfrac{6(1 - \mathrm{i}\sqrt{3})}{4}\)	M1
\(z_2 = \dfrac{6(1 - \mathrm{i}\sqrt{3})}{4}\ \left(= \dfrac{3}{2} - \mathrm{i}\dfrac{3}{2}\sqrt{3}\right)\)	A1
	(2)

Notes

M1: Multiplies numerator and denominator by \(1 - \mathrm{i}\sqrt{3}\)
A1: any correct equivalent with real denominator.

Mark scheme (b)

Scheme	Marks
\(\|z_2\| = \sqrt{\left(\dfrac{3}{2}\right)^2 + \left(\dfrac{3\sqrt{3}}{2}\right)^2} = \sqrt{\dfrac{9}{4} + \dfrac{27}{4}}\)	M1
The modulus of \(z_2\) is 3	A1
\(\tan\theta = (\pm)\sqrt{3}\) and attempts to find \(\theta\)	M1
and the argument is \(-\dfrac{\pi}{3}\)	A1
	(4)

Notes

M1: Uses correct method for modulus for their \(z_2\) in part (a)
A1: for 3 only
M1: Uses tan or inverse tan
A1: \(-\dfrac{\pi}{3}\) accept \(\dfrac{5\pi}{3}\)
NB Answers only then award 4/4 but arg must be in terms of \(\pi\)