Question

The value of $\left(\displaystyle \sum _{k = 1}^{4} \left(s i n \frac{2 \pi k}{5} - i c o s \frac{2 \pi k}{5}\right)\right)^{4}$ is (where $i$ is iota)

Answer 1

$\displaystyle \sum _{k = 1}^{4} \left(s i n \frac{2 \pi k}{5} - i c o s \frac{2 \pi k}{5}\right)$
$\displaystyle \sum _{k = 1}^{4} \left(- i^{2} s i n \frac{2 \pi k}{5} - i c o s \frac{2 \pi k}{5}\right)$
$=-i\displaystyle \sum _{k = 1}^{4} e^{i} \frac{2 \pi k}{5}$
$=-i\left[\left(- 1\right) + \left\{e^{i 0} + e^{\frac{i 2 \pi }{5}} + e^{\frac{i 4 \pi }{5}} + e^{i \frac{6 \pi }{5}} + e^{i \frac{8 \pi }{5}}\right\}\right]$ (Sum of roots of the fifth root of unity is zero)
$=i$
$\Rightarrow \left(\left(\displaystyle \sum _{k = 1}^{4} \left(\right. s i n \frac{2 \pi k}{5} - i c o s \frac{2 \pi k}{5}\right)\right)^{4}=i^{4}=1$