AS June 2019 Paper 1 Q8
8 In this question you must show detailed reasoning.
You are given that i is a root of the equation \(z^4 - 2z^3 + 3z^2 + az + b = 0\), where \(a\) and \(b\) are real constants.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\mathrm{i}^4 - 2\mathrm{i}^3 + 3\mathrm{i}^2 + a\mathrm{i} + b = 0\) | M1 | 2.1 |
| \(\Rightarrow 1 + 2\mathrm{i} - 3 + a\mathrm{i} + b = 0\) | A1 | 1.1 |
| \(\Rightarrow 2 + a = 0,\ -2 + b = 0\) | M1 | 2.1 |
| \(\Rightarrow a = -2,\ b = 2\) | E1 | 2.2a |
| [4] |
Notes
M1: (1st) substituting \(z = \mathrm{i}\) into eqn
M1: (2nd) equating real and im parts
condone verification
E1: AG
Alternative solution
| Scheme | Marks |
|---|---|
| roots \(\mathrm{i}, -\mathrm{i}, \alpha, \beta\) | |
| \(\Sigma\alpha = \alpha + \beta = 2\) | B1 |
| \(\Sigma\alpha\beta = 1 + \alpha\beta = 3 \Rightarrow \alpha\beta = 2\) | B1 |
| \(\Sigma\alpha\beta\gamma = \alpha + \beta = -a \Rightarrow a = -2\) | B1 |
| \(\alpha\beta\gamma\delta = \alpha\beta = b \Rightarrow b = 2\) | B1 |
or \(c + \mathrm{i}d\), \(c - \mathrm{i}d\)
B1: (1st) \(2c = 2\)
B1: (2nd) \(\Rightarrow 1 + c^2 + d^2 = 3 \Rightarrow c = d\)
B1: (3rd) or using \(1 + \mathrm{i}\), \(1 - \mathrm{i}\) from \(\alpha^2 - 2\alpha + 2 = 0\) AG (corrected from the printed mark scheme, which has \(\alpha^2 - 2a + 2 = 0\))
if \(1 + \mathrm{i}\), \(1 - \mathrm{i}\) roots used but not established allow B0 B0 B1 B1
B1: (4th) or using \(1 + \mathrm{i}\), \(1 - \mathrm{i}\) AG
Alternative solution
| Scheme | Marks |
|---|---|
| \((z + \mathrm{i})(z - \mathrm{i}) = z^2 + 1\) | B1 |
| \(z^4 - 2z^3 + 3z^2 + az + b = (z^2 + cz + d)(z^2 + 1)\) | M1 |
| \(z^3\): \(c = -2\), \(z^2\): \(d + 1 = 3\), so \(d = 2\) | A1 |
| \(z\): \(a = c = -2\), constants: \(b = d = 2\) | A1 |
B1: used (see below)
M1: comparing coeffs
may be inferred from multiplication or long division
A1: (1st) finding \(c\), \(d\)
A1: (2nd) verifying \(a\), \(b\)
| Scheme | Marks | AO |
|---|---|---|
| DR Another root is \(-\mathrm{i}\) | B1 | 1.1a |
| \((z + \mathrm{i})(z - \mathrm{i}) = z^2 + 1\) | B1 | 3.1a |
| \(z^4 - 2z^3 + 3z^2 - 2z + 2 = (z^2 + 1)(z^2 + cz + d)\) | M1 | 1.1 |
| \(z^3\) terms: \(-2 = c\) | A1 | 1.1 |
| constants: \(2 = 1 \times d \Rightarrow d = 2\) | A1 | 1.1 |
| \(z^2 - 2z + 2 = 0 \Rightarrow z = \dfrac{2 \pm \sqrt{-4}}{2}\) | M1 | 1.1 |
| roots are \(\mathrm{i}, -\mathrm{i}, 1 + \mathrm{i}, 1 - \mathrm{i}\) | B1 | 1.1 |
| [7] |
Notes
B1: (2nd) working backwards:
\((z - 1 - \mathrm{i})(z - 1 + \mathrm{i}) = z^2 - 2z + 2\) B1
\((z^2 + 1)(z^2 - 2z + 2) = \ldots\)
\(= z^4 - 2z^3 + 3z^2 - 2z + 2\) B1
max 4 marks
M1: (1st) equating coeffs or long divn
M1: (2nd) solving their quad factor = 0 or \((z - 1 - \mathrm{i})(z - 1 + \mathrm{i})\)
Alternative solution
| Scheme | Marks |
|---|---|
| Another root is \(-\mathrm{i}\) | B1 |
| Other roots \(\alpha, \beta\) | M1 |
| \(\mathrm{i} + (-\mathrm{i}) + \alpha + \beta = 2 \Rightarrow \alpha + \beta = 2\) | M1 |
| \(\mathrm{i}(-\mathrm{i})\alpha\beta = 2 \Rightarrow \alpha\beta = 2\) | M1 |
| \(\Rightarrow \alpha^2 - 2\alpha + 2 = 0\) | A1 |
| \(\Rightarrow \alpha = \dfrac{2 \pm \sqrt{-4}}{2} = 1 \pm \mathrm{i}\) | M1 |
| roots are \(1 + \mathrm{i}, 1 - \mathrm{i}, \mathrm{i}, -\mathrm{i}\) | B1 |
M1: (1st) or \(c + \mathrm{i}d\), \(c - \mathrm{i}d\)
M1: (2nd) or \(c + \mathrm{i}d + c - \mathrm{i}d = 2\)
M1: (3rd) or \((c + \mathrm{i}d)(c - \mathrm{i}d) = 2\)
A1: or \(c^2 + d^2 = 2\)
M1: (4th) \(c = 1\), \(d = 1\)