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 

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 $198$.
