how do you find the fifth root of -16 ( 2) ^ (1/2) ( 1+i)
i found that the mod of z^5 is 4. and the argument of ( 1+i ) = (Pi /2)
from there, i said that the arg of z is ( Pi / 10 + 2k Pi / 5)
thus the answer for the fifth root of z would be of the form,
z = 4 (cos ( Pi / 10 + 2k Pi / 5) + i sin (Pi / 10 + 2k Pi / 5) )
am i right to say that?
also is there a fixed method that i can do to compute questions like ( 1-i) ^ ( -3) and ( -1 + i) ^16?
becos now im just doing them via expansion and im wondering if there is a shorter method that i can do to solve it..