AS October 2020 Paper 1 Q1
1 In this question you must show detailed reasoning.
Use an algebraic method to find the square roots of \(-77 - 36\mathrm{i}\). [6]
| Scheme | Marks | AO |
|---|---|---|
| DR \((a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i}\) | B1 | 1.1 |
| \(a^2 - b^2 = -77\) and \(2ab = -36\) (where \(a\) and \(b\) are real) | M1 | 1.1 |
| \(b = -\dfrac{18}{a} \Rightarrow a^2 - \left(-\dfrac{18}{a}\right)^2 = -77\) \(\Rightarrow a^4 + 77a^2 - 324 = 0\) | M1 | 1.1 |
| \((a^2 - 4)(a^2 + 81) = 0\) | B1 | 2.1 |
| \(a^2 = 4\) or \(b^2 = 81\) only | A1ft | 2.3 |
| \(2 - 9\mathrm{i}\) and \(-2 + 9\mathrm{i}\) | A1 | 1.1 |
| [6] |
Notes
B1: (1st) Seen or implied in solution
M1: (1st) Comparing real and imaginary parts (no i unless later recovered) from a 3 term expansion
Allow equating real and imaginary considering \((a + \mathrm{i}b)(c + \mathrm{i}d)\)
M1: (2nd) Eliminating \(b\) or \(a\) to obtain 3 term quadratic in \(a^2\) or \(b^2\). Unknowns must not be in denominator and non-zero terms on same side. \(= 0\) seen or implied by solution.
\((b^4 - 77b^2 - 324 = 0)\)
Factorised forms:
\((b^2 - 81)(b^2 + 4)\)
B1: (2nd) DR requires evidence of solving quadratic in \(a^2\) (or \(b^2\)). Can be implied by sight of all 4 solutions.
A1ft: Rogue solutions; \(a^2 = -81\), \(b^2 = -4\). Any rogue solutions must be discarded before A1 awarded.
For follow through need to discard rogue solutions
A1: Both roots. Can use \(\pm\) but not \(\pm 2 \pm 9\mathrm{i}\) and not \(\pm 2 - 9\mathrm{i}\). \(\pm(2 - 9\mathrm{i})\), \(\pm(-2 + 9\mathrm{i})\) or \(\pm 2 \mp 9\mathrm{i}\) are all acceptable.
\(2 - 9\mathrm{i}\) and \(-2 + 9\mathrm{i}\) without working from quartic could score B1M1M1B0A0A1