FP1 June 2014 (R) Q1
1. The roots of the equation \[2z^3 - 3z^2 + 8z + 5 = 0\] are \(z_1\), \(z_2\) and \(z_3\)
Given that \(z_1 = 1 + 2\mathrm{i}\), find \(z_2\) and \(z_3\) (5)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(z) = 2z^3 - 3z^2 + 8z + 5\) | |
| \(1 - 2\mathrm{i}\) (is also a root) seen | B1 |
| \((z - (1 + 2\mathrm{i}))(z - (1 - 2\mathrm{i})) = z^2 - 2z + 5\) Attempt to expand \((z - (1 + 2\mathrm{i}))(z - (1 - 2\mathrm{i}))\) or any valid method to establish the quadratic factor e.g. \(z = 1 \pm 2\mathrm{i} \Rightarrow z - 1 = \pm 2\mathrm{i} \Rightarrow z^2 - 2z + 1 = -4\) \(z = 1 \pm \sqrt{-4} = \dfrac{2 \pm \sqrt{-16}}{2} \Rightarrow b = -2, c = 5\) Sum of roots 2, product of roots 5 \(\therefore z^2 - 2z + 5\) | M1A1 |
| \(\mathrm{f}(z) = (z^2 - 2z + 5)(2z + 1)\) Attempt at linear factor with their \(cd\) in \((z^2 + az + c)(2z + d) = \pm 5\) Or \((z^2 - 2z + 5)(2z + a) \Rightarrow 5a = 5\) | M1 |
| \((z_3) = -\dfrac{1}{2}\) | A1 |
| (5) | |
| (5 marks) |