Complex Number Series If
$$
z  + 1/z  = -1
$$
where $z$ is a complex number then value of the sum from r = 1 to r = 99 of $$
( z^r + 1/z^r)^2
$$
is equal to 
A)  198
B)  3
C) 99
D) 0
I tried by putting z = x + iy but that gives, |z| = 1 

Define $$f_n(z) = z^n + z^{-n}.$$  Then note $$f_1(z) f_n(z) = (z+z^{-1})(z^n + z^{-n}) = z^{n+1} + z^{-(n+1)} + z^{n-1} + z^{-(n-1)} = f_{n+1}(z) + f_{n-1}(z).$$  In the case where $f_1(z) = -1$, we find $$f_{n+1}(z) = -f_n(z) - f_{n-1}(z).$$  Also note that $$f_0(z) = z^0 + z^{-0} = 1 + 1 = 2.$$  Unfolding the first few terms of this recurrence, we obtain $$f_2(z) = -(-1 + 2) = -1 = f_1(z), \\
f_3(z) = -(-1 + (-1)) = 2 = f_0(z).$$
This shows that the sequence is periodic:  we will have $f_4(z) = f_5(z) = -1$, $f_6(z) = 2$, etc., and in general, $$f_{3k+1}(z) = f_{3k+2}(z) = -1, \quad f_{3k}(z) = 2.$$  It follows that $$\sum_{r=1}^{99} f_r^2(z) = \sum_{k=1}^{33} f_{3k-2}^2(z) + f_{3k-1}^2(z) + f_{3k}^2(z) = \sum_{k=1}^{33} (-1)^2 + (-1)^2 + 2^2 = 33(1 + 1 + 4) = 198.$$
