A2 June 2025 Paper 2 Q9
9
(a) It is given that, for the complex number \(z\),\[\left|\frac{z}{z + 1}\right| = 1\]
Find \(\mathrm{Re}(z)\) [3 marks]
(b) Show that the only solutions of the equation\[\left(\frac{w}{w + 1}\right)^3 = 1\]
are \(w = \dfrac{\mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}\) and \(w = \dfrac{\mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}\) [4 marks]
(c) Use the results of part (a) and part (b) to find \(\mathrm{Re}\left(\dfrac{\mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}\right)\)
Fully justify your answer. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Deduces that \(|z| = |z + 1|\) or \(\dfrac{|z|}{|z + 1|} = 1\) | B1 | 2.2a |
| Obtains and solves an equation in \(\mathrm{Re}(z)\) | M1 | 1.1a |
| Obtains \(-\dfrac{1}{2}\) | A1 | 1.1b |
| (3) |
Typical solution
Let \(z = a + b\mathrm{i}\)
\[|z| = |z + 1|\]\[a^2 + b^2 = (a + 1)^2 + b^2\]\[a^2 + b^2 = a^2 + 2a + 1 + b^2\]\[2a + 1 = 0\]\[a = -\frac{1}{2}\]\[\mathrm{Re}(z) = -\frac{1}{2}\]| Scheme | Marks | AO |
|---|---|---|
| Uses the complex cube roots of unity Or Expands and obtains a quadratic equation. | M1 | 3.1a |
| Solves an equation in \(w\) to obtain at least one correct root. | M1 | 1.1a |
| Explains why one root is impossible Or Converts at least one of the given solutions into the form \(a + \mathrm{i}b\) | E1 | 2.4 |
| Completes a reasoned argument to show that the (only) solutions of the equation are \(w = \dfrac{\mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}\) and \(w = \dfrac{\mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}\) May use \(\pm\) notation. | R1 | 2.1 |
| (4) |
Typical solution
\[\left(\frac{w}{w + 1}\right)^3 = 1\]\[\frac{w}{w + 1} = 1,\ \frac{w}{w + 1} = \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}},\ \frac{w}{w + 1} = \mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}\]\[\frac{w}{w + 1} = 1 \Rightarrow w = w + 1 \quad \text{(impossible)}\]\[\frac{w}{w + 1} = \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}} \Rightarrow w = w\left(\mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}\right) + \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}\]\[w\left(1 - \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}\right) = \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}\]\[w = \dfrac{\mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}\]Also
\[\frac{w}{w + 1} = \mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}} \Rightarrow w = \dfrac{\mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}\]So the only solutions of the equation are
\[w = \dfrac{\mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{\frac{2\pi\mathrm{i}}{3}}} \quad \text{and} \quad w = \dfrac{\mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}{1 - \mathrm{e}^{-\frac{2\pi\mathrm{i}}{3}}}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(-\dfrac{1}{2}\) | M1 | 2.2a |
| Obtains \(-\dfrac{1}{2}\) from correct reasoning using part (a) and/or part (b). | R1 | 2.1 |
| (2) | ||
| (9 marks) |