e

Let ${\rm exp}(i\pi/5) = z$ . Then $z$ is a primitive $10$ th root of $1$ . So $z^5+1 = 0$ but $z+1 \ne 0$ , so after dividing we get $z^4-z^3+z^2-z+1 = 0$ (this is the 10th cyclotomic polynomial).

Dividing through by $z^2$ and rearranging, we get $z^2 + 1/z^2 + 1 = z + 1/z$ . Since the left side is $a^2-1$ , this gives $a^2-1 = a$ , so $a^2-a = \fbox{1}$ .

euler"s representation of any complex number

$cosa + isina = e^{ia}$

thus the above becomes

$2cos(\pi/5) = a$

then by substituting values we get 1

$\begin{aligned} e^{\frac{\pi}{5}i} + e^{-\frac{\pi}{5}i} & = \cos{\left(\frac{ \pi}{5}\right)} + i \sin{\left(\frac{\pi}{5}\right)} + \cos{\left(- \frac{ \pi}{5}\right)} + i \sin{\left(- \frac{\pi}{5}\right)} \\ & = \cos{\left(\frac{ \pi}{5}\right)} + i \sin{\left(\frac{\pi}{5}\right)} + \cos{\left(\frac{ \pi}{5}\right)} - i \sin{\left(\frac{\pi}{5}\right)} \\ & = 2\cos{\left(\frac{\pi}{5}\right)} = a \end{aligned}$

$\begin{aligned} \Rightarrow a^2 - a & = 4\cos^2{\left(\frac{\pi}{5}\right)} - 2\cos{\left(\frac{\pi}{5}\right)} \\ & = \color{#3D99F6} {2\cos{\left( \frac{2\pi}{5}\right)}} + 2 - 2\cos{\left(\frac{\pi}{5}\right)} \\ & = \color{#3D99F6} {-2\cos{\left( \frac{3\pi}{5}\right)}} - 2\cos{\left(\frac{\pi}{5}\right)} + 2 \\ & = -2\left(\color{#D61F06} {\cos{\left( \frac{3\pi}{5}\right)} + \cos{\left(\frac{\pi}{5}\right)}}\right) + 2 \quad \quad \color{#D61F06} {\text{[See Note]}} \\ & = -2 \left(\color{#D61F06}{\frac{1}{2}}\right) + 2 = \boxed{1} \end{aligned}$

$\color{#D61F06}{\text{Note: }} z^5 = 1 \Longrightarrow e^{\frac{2k\pi}{5}i}$ are the fifth roots of unity. From Argand's diagram, we have $\color{#D61F06}{\cos{\frac{\pi}{5}} + \cos{\frac{3\pi}{5}} = \frac{1}{2}}$ .