z=2cis(pi/3) in mod-arg form
z^n = 2^n [cos (n*pi/3) + i sin (n*pi/3)]
for it to be real, imaginary part = 0
sin (n*pi/3) = 0
etc.
for imaginary,
cos(n*pi/3) = 0
n*pi/3 = -3pi/2, -pi/2, pi/2, 3pi/2, etc.
there wont be integral n
for the induction thing... I'll just prove the...