FP1 January 2013 Q2
2. \[z = \frac{50}{3 + 4\mathrm{i}}\]
Find, in the form \(a + \mathrm{i}b\) where \(a, b \in \mathbb{R}\),
(a) \(z\), (2)
(b) \(z^2\). (2)
Find
(c) \(|z|\), (2)
(d) \(\arg z^2\), giving your answer in degrees to 1 decimal place. (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{50}{3 + 4\mathrm{i}} = \dfrac{50(3 - 4\mathrm{i})}{(3 + 4\mathrm{i})(3 - 4\mathrm{i})} = \dfrac{50(3 - 4\mathrm{i})}{25} = 6 - 8\mathrm{i}\) | M1 A1cao |
| (2) |
Notes
(a) M for \(\times\dfrac{3 - 4\mathrm{i}}{3 - 4\mathrm{i}}\) (accept use of \(-3 + 4\mathrm{i}\)) and attempt to expand using \(\mathrm{i}^2 = -1\), A for \(6 - 8\mathrm{i}\) only
| Scheme | Marks |
|---|---|
| \(z^2 = (6 - 8\mathrm{i})^2 = 36 - 64 - 96\mathrm{i} = -28 - 96\mathrm{i}\) | M1 A1 |
| (2) |
Notes
(b) M for attempting to expand their \(z^2\) using \(\mathrm{i}^2 = -1\), A for \(-28 - 96\mathrm{i}\) only. If using original \(z\) then must attempt to multiply top and bottom by conjugate and use \(\mathrm{i}^2 = -1\).
| Scheme | Marks |
|---|---|
| \(|z| = \sqrt{6^2 + (-8)^2} = 10\) | M1 A1ft |
| (2) |
Notes
(c) M for \(\sqrt{a^2 + b^2}\), A for ‘their 10’
Alternative
| Scheme | Marks |
|---|---|
| \(|z| = \dfrac{50}{|3 + 4\mathrm{i}|} = 10\) | M1 A1 |
| Scheme | Marks |
|---|---|
| \(\tan\alpha = \dfrac{-96}{-28}\) | M1 |
| so \(\alpha = -106.3^\circ\) or \(253.7^\circ\) | A1cao |
| (2) | |
| [8] |
Notes
(d) M for use of tan or \(\tan^{-1}\) and values from their \(z^2\) either way up ignoring signs. Radians score A0.
Alternative
| Scheme | Marks |
|---|---|
| \(\arg(3 + 4\mathrm{i}) = 53.13..\) so \(\arg\left(\dfrac{50}{3 + 4\mathrm{i}}\right)^2 = -2 \times 53.13\ldots = -106.3\) | M1 A1 |