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 
 A: HINT: observe that
$$
\left(z^r+\frac1{z^r}\right)^2=\left(z^{2}\right)^r+\left(\frac1{z^2}\right)^r+2
$$
then split the sum and using the geometric series
$$
\sum_{n=1}^Nq^n=\frac{q-q^{N+1}}{1-q}\;,\;\;\;\;\;q\in\Bbb C
$$
we can evaluate the first two sums:
\begin{align*}
\sum_{r=1}^{99}\left(z^r+\frac1{z^r}\right)^2
=&\sum_{r=1}^{99}\left(z^{2}\right)^r+\sum_{r=1}^{99}+\left(\frac1{z^2}\right)^r+2\cdot99\\
=&\frac{z^2-(z^2)^{100}}{1-z^2}+\frac{z^{-2}-(z^{-2})^{100}}{1-z^{-2}}+198\\
=&z^2\frac{1-z^{198}}{1-z^2}+\frac1{z^{198}}\frac{1-z^{198}}{1-z^{2}}+198\\
\end{align*}
but $z+1/z=-1$ holds iff $z^2+z+1=0$, that is $z=e^
{\pm i\frac23\pi}$, thus $z^3=1$ (you can see this also as follows: supposing $z\neq1$, you can multiply this last one by $z-1$ to get $z^3-1=0$).
Hence since $3|198$ we have that $1-z^{198}=0$ thus the first two summands of the last line are thrown away and the only survivor os $198$.
A: 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 $$\sum_{r=1}^{99} f_r^2(z) = \sum_{k=1}^{33} f_{3k-2}^2(z) + f_{3k-1}^2(z) + f_{3k}^2(z) = \sum_{k=1}^{33} (-1)^2 + (-1)^2 + 2^2 = 33(1 + 1 + 4) = 198.$$
