Hello from MrBillDoesMath!
Answer:
16 ( sqrt(3) + i)
Discussion:
As (-sqrt(3) + i) in polar form is 2 ( cos(150) + isin(150) )
(Angles shown in degrees for simplicity.
z = ( -sqrt(3) + i)
z = 2 ( cos(150) + i sin(150) ) =>
z^5 = 2^5 ( cosi(150) + i sin(150)) ^5 (*)
De Moivre's theorem for complex number gives
(cos x+isin x)^n = cos(nx)+i sin(nx)
Evaluating the rhs of (*) gives
z^5 =
= 2^5 ( cos(150*5) + i sin (150*5)
= 32 ( cos(750) + i sin(750) )
= 32 ( sqrt(3) /2 + i (1/2) )
= 16 sqrt(3) + 16 i
= 16 ( sqrt(3) + i)
Thank you,
MrB