Expert: Ahmed Salami - 11/5/2004

Question
Can you show me a way to easily calculate the roots of the equation (Z^6) - 1 = 0

Answer
Hi Michael,
z^6 - 1 = 0
z^6 = 1
z = 1^(1/6) which refers to all the sixth roots of 1
Converting 1 to its polar form
1 = 1(cos 0 + i sin 0)
Using De Moivre's theorem
1^(1/6) = 1^(1/6)[cos (0/6) + i sin (0/6)]
       = 1(cos 0 + i sin 0)  
       = 1
But the six roots are spaced 360/6 = 60 degrees from each other with equal magnitude. The six roots are therefore,
1(cos 0 + i sin 0), 1(cos 60 + i sin 60)
1(cos 120 + i sin 120), 1(cos 180 + i sin 180)
1(cos 240 + i sin 240), 1(cos 300 + i sin 300)
= 1 , 1/2 + i(sqrt3)/2 , -1/2 + i(sqrt3)/2 , -1 ,
-1/2 - i(sqrt3)/2 , 1/2 - i(sqrt3)/2
I hope you understand it.
Regards.