Question

By `z=-1/2+sqrt(3)/2i` what is the answer for 1+z+z^2 ?

`z=-1/2+sqrt(3)/2i`

Answer 1

`1+z+z^2=0`

Explanation:

As `z=-1/2+sqrt(3)/2i`

`z^2=(-1/2+sqrt(3)/2i)^2`

= `(-1/2)^2+(sqrt3/2i)^2-2*1/2*sqrt3/2i`

= `1/4-3/4-sqrt3/2i`

= `-1/2-sqrt3/2i`

Therefore`z+z^2=-1`

and `1+z+z^2=0`

Additional information - `1,-1/2+sqrt(3)/2i` and `-1/2-sqrt(3)/2i` are cube roots of `1` and their sum and product both are `1`. They are generally represented as `1,omega,omega^2`.