If 1 , α 1 , α 2 , α 3 , α 4 are the roots of z 5 - 1 = 0 and ω is a cube root of…

$\therefore z^5-1=(z-1)(z-{\alpha }_1)(z-{\alpha }_2)(z-{\alpha }_3)(z-{\alpha }_4)$

$\therefore {\omega }^5-1=(\omega -1)(\omega -{\alpha }_1)(\omega -{\alpha }_2)(\omega -{\alpha }_3)(\omega -{\alpha }_4)$

$(\omega -1)(\omega -{\alpha }_1)(\omega -{\alpha }_2)(\omega -{\alpha }_3)(\omega -{\alpha }_4) + \omega = {\omega }^5-1+ \omega$

$= {\omega }^3\times {\omega }^2+\omega -1={\omega }^2+\omega -1 (\because {\omega }^3=1)$

$=-1-1=-2 (\because {\omega }^2+\omega +1=0)$