Question

The value of ${\left(\sum _{k=1}^4\left(sin\frac{2\pi k}{5}-icos\frac{2\pi k}{5}\right)\right)}^4$ is (where $i$ is iota)

Answer 1

Answer: 1

$\sum _{k=1}^4\left(sin\frac{2\pi k}{5}-icos\frac{2\pi k}{5}\right)$
$\sum _{k=1}^4\left(-i^{2}sin\frac{2\pi k}{5}-icos\frac{2\pi k}{5}\right)$
$=-i\sum _{k=1}^4{e}^i\frac{2\pi k}{5}$
$=-i\left[\left(-1\right)+\left\{e^{i0}+e^{\frac{i2\pi }{5}}+e^{\frac{i4\pi }{5}}+e^{i\frac{6\pi }{5}}+e^{i\frac{8\pi }{5}}\right\}\right]$ (Sum of roots of the fifth root of unity is zero)
$=i$
$\Rightarrow {\left(\left(\sum _{k=1}^4(sin\frac{2\pi k}{5}-icos\frac{2\pi k}{5}\right)\right)}^4=i^4=1$