Question

Solve and write in standard form: z = (-√3 + i )^5

Answer 1

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