Finding complex numbers using DeMoivre's Theorem
Find the cube roots of 21/2cis(pi/4)
21/2cis(pi/4+ 2pi k), for every integer k
By DeMoivre:
21/21/3cis((pi/4+ 2pi k)/3)321/6cis(pi/12+ 2/3pi *k)
Taking k=0,1,2 gives the three cube roots:z1= 21/6cis(pi/12) (k=0)z2= 21/6cis(3pi/4) (k=1)z3= 21/6cis(17pi/12) (k=2)
LR
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