FP1 January 2011 Q1
1. \[z = 5 - 3\mathrm{i}, \qquad w = 2 + 2\mathrm{i}\]
Express in the form \(a + b\mathrm{i}\), where \(a\) and \(b\) are real constants,
(a) \(z^2\), (2)
(b) \(\dfrac{z}{w}\). (3)
| Scheme | Marks |
|---|---|
| \(z = 5 - 3\mathrm{i}, \quad w = 2 + 2\mathrm{i}\) \(z^2 = (5 - 3\mathrm{i})(5 - 3\mathrm{i})\) | |
| \(= 25 - 15\mathrm{i} - 15\mathrm{i} + 9\mathrm{i}^2\) \(= 25 - 15\mathrm{i} - 15\mathrm{i} - 9\) An attempt to multiply out the brackets to give four terms (or four terms implied). \(zw\) is M0 | M1 |
| \(= 16 - 30\mathrm{i}\) \(16 - 30\mathrm{i}\) Answer only 2/2 | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{z}{w} = \dfrac{(5 - 3\mathrm{i})}{(2 + 2\mathrm{i})}\) | |
| \(= \dfrac{(5 - 3\mathrm{i})}{(2 + 2\mathrm{i})} \times \dfrac{(2 - 2\mathrm{i})}{(2 - 2\mathrm{i})}\) Multiplies \(\dfrac{z}{w}\) by \(\dfrac{(2 - 2\mathrm{i})}{(2 - 2\mathrm{i})}\) | M1 |
| \(= \dfrac{10 - 10\mathrm{i} - 6\mathrm{i} - 6}{4 + 4}\) Simplifies realising that a real number is needed on the denominator and applies \(\mathrm{i}^2 = -1\) on their numerator expression and denominator expression. | M1 |
| \(= \dfrac{4 - 16\mathrm{i}}{8}\) | |
| \(= \dfrac{1}{2} - 2\mathrm{i}\) \(\dfrac{1}{2} - 2\mathrm{i}\) or \(a = \dfrac{1}{2}\) and \(b = -2\) or equivalent Answer as a single fraction A0 | A1 |
| (3) | |
| [5] |