2011 Paper 1 Question 17

(b)

$z^6+z^3+1=0 \quad (1)$

Let $z^3=w$

$\therefore \; w^2+w+1=0$

This is a quadratic equation

$\Rightarrow w = \frac{-1 \pm \sqrt{-3}}{2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$

Recognise $1/2$ and $\sqrt3/2$ as standard trigonometric ratios. Consider the $+$ root

$w_+ = e^{i 2\pi/3} = z^3$

Compute the cube roots

$z=\exp{\left[ i \left( \frac{2}{9}\pi+\frac{2k}{3}\pi \right) \right] }, \quad k = 0,1,2$

Or in full, placing angles in the range $(-\pi,\pi]$

$z=\exp{ (i \theta)}, \quad \theta \in {\frac{2}{9}\pi, \frac{8}{9}\pi, -\frac{4}{9}\pi}$

As equation $(1)$ has real coefficients, the roots occur in complex-conjugate pairs. The full solution set is therefore

$z=\exp{ (i \theta)}, \quad \theta \in {\pm \frac{2}{9}\pi, \pm \frac{8}{9}\pi, \pm \frac{4}{9}\pi}$

NST First Year Maths