AS June 2024 Paper 1 Q8
8
(a) The complex number \(z\) is given by \(z = x + \mathrm{i}y\) where \(x, y \in \mathbb{R}\)
(i) Write down the complex conjugate \(z^*\) in terms of \(x\) and \(y\) [1 mark]
(ii) Hence prove that \(zz^*\) is real for all \(z \in \mathbb{C}\) [2 marks]
(b) The complex number \(w\) satisfies the equation\[3w + 10\mathrm{i} = 2w^* + 5\]
(i) Find \(w\) [3 marks]
(ii) Calculate the value of \(w^2(w^*)^2\) [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| (i) States \(x - y\mathrm{i}\) | B1 | 1.2 |
| (1) | ||
| (ii) Obtains a correct expansion and replaces \(\mathrm{i}^2\) with \(-1\) PI | M1 | 1.1a |
| Simplifies to \(x^2 + y^2\) and explains \(zz^*\) (for all \(z \in \mathbf{C}\)) is real with a reference to \(x\) and \(y\) being real. Condone \(x^2\) and \(y^2\) for \(x\) and \(y\) in their explanation. | R1 | 2.1 |
| (2) |
Typical solution
(i)
\[z^* = x - \mathrm{i}y\](ii)
\[\begin{aligned} zz^* &= (x + \mathrm{i}y)(x - \mathrm{i}y) \\ &= x^2 - \mathrm{i}xy + \mathrm{i}xy - \mathrm{i}^2y^2 \\ &= x^2 + y^2\end{aligned}\]As \(x\) and \(y\) are both real then \(zz^*\) is real for all \(z \in \mathbf{C}\)
| Scheme | Marks | AO |
|---|---|---|
| (i) Substitutes \(x + \mathrm{i}y\) for \(w\) and \(x - \mathrm{i}y\) for \(w^*\) | M1 | 3.1a |
| Obtains \(\mathrm{Re}(w) = 5\) or \(\mathrm{Im}(w) = -2\) | A1 | 1.1b |
| Obtains \(5 - 2\mathrm{i}\) | A1 | 1.1b |
| (3) | ||
| (ii) Obtains 841 FT their \(w\) of the form \(a + \mathrm{i}b\) where \(a\) and \(b\) are non-zero | B1F | 1.1b |
| (1) | ||
| (7 marks) |
Typical solution
(i)
Let \(w = x + \mathrm{i}y\) where \(x, y \in \mathbf{R}\)
\[3(x + \mathrm{i}y) + 10\mathrm{i} = 2(x - \mathrm{i}y) + 5\]\[3x + 3\mathrm{i}y + 10\mathrm{i} = 2x - 2\mathrm{i}y + 5\]Comparing real and imaginary parts:
\[3x = 2x + 5 \quad \text{and} \quad 3y + 10 = -2y\]\[x = 5 \quad \text{and} \quad 5y = -10\]\[y = -2\]\[w = 5 - 2\mathrm{i}\](ii)
\[w^2(w^*)^2 = 841\]