Note that sqrt( sqrt(3)^2 + 1^2 )
= sqrt( 3 + 1 )
= 2
The (√3+i) = 2 * (√3/2+i/2)
= 2 * ( cos(pi/6) + i * sin(pi/6))
So that (√3+i)^3 = ( 2 * ( cos(pi/6) + i * sin(pi/6)) ) ^ 3
= 8 * ( cos(pi/6) + i * sin(pi/6)) ^ 3
= 8 * ( cos(3*pi/6) + i * sin(3* pi/6))
= 8 * ( cos(pi/2) + i * sin(pi/2)
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