We have $x^5=1=\cos 0+i \sin 0$

$\therefore x^5 = \cos (2k\pi) + i\sin (2k\pi)$

$\therefore x = [\cos (2k\pi) + i\sin (2k\pi)]^{1/5} = \cos \dfrac {2k\pi}{5}+i\sin \dfrac {2k\pi}{5}$

Putting k=0,1,2,3,4 we get the five roots,

$x_0 = \cos 0 + i\sin 0=1$

$x_1 = \cos \dfrac {2\pi}{5}+ i\sin \dfrac {2\pi}{5}$ …