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

Hello from MrBillDoesMath!

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

i could probably can you me with a question of mine?

step-by-step explanation: