A2 June 2025 Paper 1 Q3
3. The complex number \(z = a + b\mathrm{i}\) where \(a\) and \(b\) are real constants.
Given that \(\dfrac{z}{z^*}\) is purely imaginary,
Given also that \(zz^* = 50\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{z}{z^*} = \dfrac{a + b\mathrm{i}}{a - b\mathrm{i}} = \dfrac{(a + b\mathrm{i})(a + b\mathrm{i})}{(a - b\mathrm{i})(a + b\mathrm{i})} = \dfrac{\left(a^2 - b^2\right) + 2ab\mathrm{i}}{a^2 + b^2}\) | M1 | 2.1 |
| \(a^2 - b^2 = 0 \Rightarrow a^2 = b^2\) * | A1* | 1.1b |
| (2) |
Notes
M1: Finds an expression for \(\dfrac{z}{z^*}\) by rationalising the denominator and collects real parts, eliminating \(\mathrm{i}^2\) terms.
e.g. Accept \(\dfrac{a^2 - b^2 + 2ab\mathrm{i}}{a^2 + b^2}\) or \(\dfrac{a^2 - b^2}{a^2 + b^2} + \dfrac{2ab\mathrm{i}}{a^2 + b^2}\)
but \(\dfrac{a^2 + 2ab\mathrm{i} - b^2}{a^2 + b^2}\) is insufficient unless they then go on to identify the real parts
Allow sign slips but must be a correct use of rationalising the denominator.
A1*: Sets real part equal to zero and then achieves the correct equation, with no errors seen.
\(\dfrac{z}{z^*} = \dfrac{a^2 - b^2 + 2ab\mathrm{i}}{a^2 + b^2} \Rightarrow a^2 = b^2\) is insufficient, as they have not set or identified their real part as zero.
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{z}{z^*} = \dfrac{a + b\mathrm{i}}{a - b\mathrm{i}} = k\mathrm{i} \Rightarrow a + b\mathrm{i} = k\mathrm{i}(a - b\mathrm{i})\) \(\Rightarrow a + b\mathrm{i} = ka\mathrm{i} + kb\) | M1 | 2.1 |
| \(\Rightarrow a = kb,\ b = ka\) e.g. \(a = \dfrac{b}{a} \times b\) or \((k =)\dfrac{a}{b} = \dfrac{b}{a}\) \(\Rightarrow a^2 = b^2\) | A1* | 1.1b |
| (2) |
M1: Substitutes \(z = a + b\mathrm{i}\) and \(z^* = a - b\mathrm{i}\), sets equal to an imaginary number such as \(k\mathrm{i}\) but not \(\mathrm{i}\). Then expands brackets, eliminating \(\mathrm{i}^2\) terms, achieving an equation equivalent to \(a + b\mathrm{i} = ka\mathrm{i} + kb\)
A1*: Equates real parts and imaginary parts and uses their pair of equations to eliminate \(k\) and then achieve the correct equation, with no errors seen.
If you see a method that involves using exponentials, or modulus argument form, or working backwards, or any other method which may be worthy of credit, please send to review.
| Scheme | Marks | AO |
|---|---|---|
| \(zz^* = (a + b\mathrm{i})(a - b\mathrm{i}) = a^2 + b^2 = 50\) Solves simultaneously with \(a^2 = b^2\) to find a value for \(a\) or \(b\) \(2a^2 = 50 \Rightarrow a = \ldots\) or \(2b^2 = 50 \Rightarrow b = \ldots\) | M1 | 3.1a |
| \((z) = 5 - 5\mathrm{i},\ 5 + 5\mathrm{i},\ -5 - 5\mathrm{i},\ -5 + 5\mathrm{i}\) | A1 A1 | 1.1b 2.2a |
| (3) | ||
| (5 marks) |
Notes
M1: Uses the information \(zz^* = 50\) to form another equation for \(a\) and \(b\), with no imaginary terms present. Then solves simultaneously to find a value for \(a\) or \(b\)
If they obtain \(a^2 - b^2 = 50\) this will be M0
A1: At least two correct complex numbers. Accept e.g. \(\pm(5 + 5\mathrm{i})\)
A1: Deduces all 4 correct complex numbers. Accept e.g. \(\pm 5 \pm 5\mathrm{i}\) or \(\pm(5 \pm 5\mathrm{i})\)
Note: Correct answers with no workings can achieve M1A1A1