FP1 January 2012 Q1
1. Given that \(z_1 = 1 - \mathrm{i}\),
(a) find \(\arg(z_1)\). (2)
Given also that \(z_2 = 3 + 4\mathrm{i}\), find, in the form \(a + \mathrm{i}b\), \(a, b \in \mathbb{R}\),
(b) \(z_1 z_2\), (2)
(c) \(\dfrac{z_2}{z_1}\). (3)
In part (b) and part (c) you must show all your working clearly.
| Scheme | Marks |
|---|---|
| \(\arg z_1 = -\arctan(1)\) \(-\arctan(1)\) or \(\arctan(1)\) or \(\arctan(-1)\) | M1 |
| \(= -\dfrac{\pi}{4}\) or \(-45\) or awrt \(-0.785\) (oe e.g \(\dfrac{7\pi}{4}\)) | A1 |
| (2) |
Notes
Correct answer only 2/2
| Scheme | Marks |
|---|---|
| \(z_1 z_2 = (1 - \mathrm{i})(3 + 4\mathrm{i}) = 3 - 3\mathrm{i} + 4\mathrm{i} - 4\mathrm{i}^2\) At least 3 correct terms (Unsimplified) | M1 |
| \(= 7 + \mathrm{i}\) cao | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{z_2}{z_1} = \dfrac{(3 + 4\mathrm{i})}{(1 - \mathrm{i})} = \dfrac{(3 + 4\mathrm{i}).(1 + \mathrm{i})}{(1 - \mathrm{i}).(1 + \mathrm{i})}\) Multiply top and bottom by \((1 + \mathrm{i})\) | M1 |
| \(= \dfrac{(3 + 4\mathrm{i}).(1 + \mathrm{i})}{2}\) \((1 + \mathrm{i})(1 - \mathrm{i}) = 2\) | A1 |
| \(= -\dfrac{1}{2} + \dfrac{7}{2}\mathrm{i}\) or \(\dfrac{-1 + 7\mathrm{i}}{2}\) | A1 |
| (3) | |
| (7 marks) |
Notes
Special case \(\dfrac{z_1}{z_2} = \dfrac{(1 - \mathrm{i})}{(3 + 4\mathrm{i})} = \dfrac{(1 - \mathrm{i}).(3 - 4\mathrm{i})}{(3 + 4\mathrm{i}).(3 - 4\mathrm{i})}\) Allow M1A0A0
Correct answers only in (b) and (c) scores no marks