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

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Answer 1 · 2018-02-16T21:35:53

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

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)#