Finding complex numbers using DeMoivre's Theorem

Find the cube roots of 2^1/2cis(pi/4)
2^1/2cis(pi/4+ 2pi k), for every integer k
By DeMoivre:
2^1/21/3cis((pi/4+ 2pi k)/3)³2^1/6cis(pi/12+ 2/3pi *k)
Taking k=0,1,2 gives the three cube roots:z₁= 2^1/6cis(pi/12) (k=0)z₂= 2^1/6cis(3pi/4) (k=1)z₃= 2^1/6cis(17pi/12) (k=2)