A2 June 2023 Paper 2 Q9
9 The complex number \(z\) is such that
\[z = \frac{1 + \mathrm{i}}{1 - k\mathrm{i}}\]where \(k\) is a real number.
(a) Find the real part of \(z\) and the imaginary part of \(z\), giving your answers in terms of \(k\) [2 marks]
(b) In the case where \(k = \sqrt{3}\), use part (a) to show that\[\cos\frac{7\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}\] [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Multiplies numerator and denominator by conjugate of denominator. | M1 | 1.1a |
| Obtains correct real part and correct imaginary part. Condone \(\dfrac{1 + k}{1 + k^2}\mathrm{i}\) | A1 | 1.1b |
| (2) |
Typical solution
\[z = \frac{1 + \mathrm{i}}{1 - k\mathrm{i}} \times \frac{1 + k\mathrm{i}}{1 + k\mathrm{i}} = \frac{1 - k}{1 + k^2} + \mathrm{i}\left(\frac{1 + k}{1 + k^2}\right)\]\[\text{Real part} = \frac{1 - k}{1 + k^2}\]\[\text{Imaginary part} = \frac{1 + k}{1 + k^2}\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(k = \sqrt{3}\) into \(z\) and finds \(|z|\) | M1 | 1.1a |
| Obtains the correct value for \(|z|\) | A1 | 1.1b |
| Obtains \(\arg z = \dfrac{7\pi}{12}\) by a fully correct method. | B1 | 3.1a |
| Forms an equation of the form \(|z|\cos(\arg(z)) = \mathrm{Re}(z)\) | M1 | 3.1a |
| Completes a reasoned argument to show the required result. | R1 | 2.1 |
| (5) | ||
| (7 marks) |
Typical solution
When \(k = \sqrt{3}\), \(\mathrm{Re}(z) = \dfrac{1 - \sqrt{3}}{4}\)
\[|z| = \frac{|1 + \mathrm{i}|}{\left|1 - \sqrt{3}\mathrm{i}\right|} = \frac{\sqrt{2}}{2}\]\[\arg z = \arg(1 + \mathrm{i}) - \arg\left(1 - \sqrt{3}\mathrm{i}\right) = \frac{\pi}{4} - \left(-\frac{\pi}{3}\right) = \frac{7\pi}{12}\]\[\frac{\sqrt{2}}{2}\left(\cos\frac{7\pi}{12}\right) = \frac{1 - \sqrt{3}}{4}\]\[\cos\frac{7\pi}{12} = \frac{\sqrt{2}\left(1 - \sqrt{3}\right)}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}\]