Problem Statement
(a) Solve the equation $z^5 = 1$, giving all roots in exponential form $e^{i\theta}$ where $-\pi < \theta \le \pi$. [2 marks]
(b) By considering the sum of roots of $z^5 - 1 = 0$, show that $\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}$. [3 marks]
(c) Hence, deduce that $\cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} - 1}{4}$. [2 marks]
Verified Solution & Marking Scheme
Part (a): Find fifth roots of unity
The equation $z^5 = e^{i 2k\pi}$ has roots $z_k = e^{i 2k\pi / 5}$ for $k \in \{0, 1, 2, -1, -2\}$:
$$z \in \left\{ 1, \; e^{i 2\pi/5}, \; e^{i 4\pi/5}, \; e^{-i 2\pi/5}, \; e^{-i 4\pi/5} \right\}$$
Part (b): Sum of roots identity
Since the sum of the roots of $z^5 - 1 = 0$ is the coefficient of $z^4$ (which is 0):
$$1 + e^{i 2\pi/5} + e^{-i 2\pi/5} + e^{i 4\pi/5} + e^{-i 4\pi/5} = 0$$
Using Euler identity $e^{i\theta} + e^{-i\theta} = 2\cos\theta$:
$$1 + 2\cos\left(\frac{2\pi}{5}\right) + 2\cos\left(\frac{4\pi}{5}\right) = 0 \implies \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}$$ (AG)
Part (c): Express as quadratic in cos(2pi/5)
Using double-angle identity $\cos(4\pi/5) = 2\cos^2(2\pi/5) - 1$. Let $u = \cos(2\pi/5)$:
$$u + 2u^2 - 1 = -\frac{1}{2} \implies 4u^2 + 2u - 1 = 0$$
Solving using quadratic formula:
$$u = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4}$$
Since $2\pi/5 = 72^\circ$ is in the first quadrant, $\cos(2\pi/5) > 0$. Therefore, we choose the positive root: $\cos(2\pi/5) = \frac{\sqrt{5} - 1}{4}$.