Problem Statement
Consider the complex polynomial equation $z^5 - 1 = 0$, where $z \in \mathbb{C}$. Let $\omega = e^{i \frac{2\pi}{5}}$.
**(a)** [3 marks] Show that $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$.
**(b)** [4 marks] Hence, by considering the real part of $\omega + \omega^4$ and $\omega^2 + \omega^3$, prove that $\cos\left(\frac{2\pi}{5}\right) = \frac{-1 + \sqrt{5}}{4}$.
**(c)** [2 marks] Deduce the exact area of the regular pentagon inscribed in the unit circle $|z| = 1$.
Verified Solution & Marking Scheme
Part (a) [M1]
Recognizing sum of finite geometric series $\sum_{r=0}^4 \omega^r = \frac{1 - \omega^5}{1 - \omega}$ with ratio $\omega \neq 1$
Part (a) [A1]
Using $\omega^5 = (e^{i 2\pi/5})^5 = e^{i 2\pi} = 1$
Part (a) [A1]
Conclude $\frac{1 - 1}{1 - \omega} = 0$, hence $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ (AG)
Part (b) [M1]
Let $u = \omega + \omega^4 = 2\cos(2\pi/5)$ and $v = \omega^2 + \omega^3 = 2\cos(4\pi/5)$
Part (b) [A1]
Sum $u + v = -1$, and product $uv = (\omega + \omega^4)(\omega^2 + \omega^3) = \omega^3 + \omega^4 + \omega^6 + \omega^7 = \omega^3 + \omega^4 + \omega + \omega^2 = -1$
Part (b) [M1]
Form quadratic equation $X^2 + X - 1 = 0$ with roots $X = \frac{-1 \pm \sqrt{5}}{2}$
Part (b) [R1]
Since $2\pi/5$ is in quadrant 1, $u > 0 \implies 2\cos(2\pi/5) = \frac{-1 + \sqrt{5}}{2} \implies \cos(2\pi/5) = \frac{-1 + \sqrt{5}}{4}$ (AG)
Part (c) [M1]
Area of pentagon = $5 \times \left(\frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(2\pi/5)\right) = \frac{5}{2}\sin(2\pi/5)$
Part (c) [A1]
Exact value $\frac{5}{8}\sqrt{10 + 2\sqrt{5}}$