Question

How do you find the two square roots of `-2+2sqrt3i`?

Answer 1

`+-sqrt(-2+2sqrt(3)i) = +-2bar(omega) = +-(1+sqrt(3)i)`

Explanation:

Note that:

`-2+2sqrt(3)i = 4omega`

where:

`omega = cos((2pi)/3)+i sin((2pi)/3) = -1/2+sqrt(3)/2i`

is the primitive complex cube root of `1`

So `omega^3 = 1` and `(omega^2)^2 = omega^3 omega = omega`

That is:

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

`+-sqrt(-2+2sqrt(3)i) = +-sqrt(4)bar(omega) = +-(1+sqrt(3)i)`