A. 18
B. 72β362β
C. 36
D. 72
E. 72+362β Solution:
Note that z=cis(45β)
Also note that zk=zk+8 for all positive integers k because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo 8 .
12,52, and 92 are all 1(mod8)
22,62, and 102 are all 4(mod8)
32,72, and 112 are all 1(mod8)
42,82, and 122 are all 0(mod8)
Therefore,
z12=z52=z92=cis(45β)
z22=z62=z102=cis(180β)=β1
z32=z72=z112=cis(45β)
z42=z82=z122=cis(0β)=1
The term thus (z12+z22+z32+β―+z122) simplifies to 6 cis (45β), while the term (z121β+z221β+z321β+β―+z1221β) simplifies to cis(45β)6β. Upon multiplication, the cis (45β) cancels out and leaves us with (C)36β .