FP1 January 2010 Q1
1. The complex numbers \(z_1\) and \(z_2\) are given by \[z_1 = 2 + 8\mathrm{i} \quad \text{and} \quad z_2 = 1 - \mathrm{i}\]
Find, showing your working,
(a) \(\dfrac{z_1}{z_2}\) in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real, (3)
(b) the value of \(\left|\dfrac{z_1}{z_2}\right|\), (2)
(c) the value of \(\arg\dfrac{z_1}{z_2}\), giving your answer in radians to 2 decimal places. (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{z_1}{z_2} = \dfrac{2 + 8\mathrm{i}}{1 - \mathrm{i}} \times \dfrac{1 + \mathrm{i}}{1 + \mathrm{i}}\) | M1 |
| \(= \dfrac{2 + 2\mathrm{i} + 8\mathrm{i} - 8}{2} = -3 + 5\mathrm{i}\) | A1 A1 |
| (3) |
Notes
(a) \(\times\dfrac{1 + \mathrm{i}}{1 + \mathrm{i}}\) and attempt to multiply out for M1
−3 for first A1, +5i for second A1
| Scheme | Marks |
|---|---|
| \(\left|\dfrac{z_1}{z_2}\right| = \sqrt{(-3)^2 + 5^2} = \sqrt{34}\) (or awrt 5.83) | M1 A1ft |
| (2) |
Notes
(b) Square root required without i for M1
\(\dfrac{|z_1|}{|z_2|}\) award M1 for attempt at Pythagoras for both numerator and denominator
| Scheme | Marks |
|---|---|
| \(\tan\alpha = -\dfrac{5}{3}\) or \(\dfrac{5}{3}\) | M1 |
| \(\arg\dfrac{z_1}{z_2} = \pi - 1.03\ldots = 2.11\) | A1 |
| (2) | |
| [7] |
Notes
(c) tan or \(\tan^{-1}\), \(\pm\dfrac{5}{3}\) or \(\pm\dfrac{3}{5}\) seen with their 3 and 5 award M1
2.11 correct answer only award A1