How do you find the two square roots of #-2+2sqrt3i#?
1 Answer
Feb 16, 2018
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
So
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)#