Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$. 
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
  The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.

At first, I tried to evaluate it directly. And the LHS equals to
\begin{align}
\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6} 
 & = \frac{q}{1+q^2}+\frac{q^2}{1+q^4}\cdot\frac{q^3}{q^3}+\frac{q^3}{1+q^6}\cdot\frac{q}{q} \\
 & = \frac{q}{1+q^2}+\frac{q^5}{1+q^3}+\frac{q^4}{1+q} \\
 & = q\cdot\frac{(1+q)(1+q^3)+q^4(1+q)(1+q^2)+q^3(1+q^2)(1+q^3)}{(1+q)(1+q^2)(1+q^3)} \\ 
 & = q\cdot\frac{1+q+q^3+q^4+q^4+q^5+q^6+1+q^3+q^5+q^6+q}{(1+q)(1+q^2)(1+q^3)} \\
 & = \frac{-2q^3}{(1+q)(1+q^2)(1+q^3)} \\ 
\end{align}
And 
$$(x-q)(x-q^2)(x-q^3)(x-q^4)(x-q^5)(x-q^6)=x^6+x^5+x^4+x^3+x^2+x+1$$
Let $x=-1$ I get that
$$(1+q)(1+q^2)(1+q^3)(1+q^4)(1+q^5)(1+q^6)=1$$
and
$$(1+q)(1+q^2)(1+q^3)\cdot q^4(q^3+1)\cdot q^5(q^2+1)\cdot q^6(q+1)=1$$
therefore
$$\left[(1+q)(1+q^2)(1+q^3)\right]^2=\frac{1}{q^{15}}=\frac{1}{q}$$
hence
$$\left[\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}\right]^2=\frac{q}{1}\cdot 4q^6=4$$
$$\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}=\pm 2$$
And I try for a solution as a polar-form method$.\\$Suppose $q=\cos\frac{2j\pi}{7}+i\sin\frac{2j\pi}{7}$
$$\frac{q^k}{1+q^{2k}}=\frac{\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}}{2\cos\frac{2jk\pi}{7}\left(\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}\right)}=\frac{1}{2\cos\frac{2jk\pi}{7}}$$
Am I going to the right direction? How I finish it? And please help to figure out what's wrong with my calculation at the first part. I appreciate for your help.

You may continue with
$
\frac{q^k}{1+q^{2k}}= \frac{1}{2\cos k\alpha} 
  $, where $\alpha=\frac{2\pi j}7,\>j=1,2,...,6$, and write the expression as,
$$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}
=\frac{\cos\alpha+ \cos2\alpha+ \cos3\alpha}{2\cos\alpha\cos2\alpha\cos3\alpha}=\frac ND\tag1\\$$
where we used $2\cos x\cos y=\cos(x+y)+\cos(x-y)$ and $\cos 4\alpha= \cos 3\alpha$, $\cos 5\alpha= \cos 2\alpha$. For the denominator, apply  $\sin 2x = 2\sin x \cos x$,
$$ 4\sin\alpha \cdot D =4 \sin 2\alpha\cos 2\alpha\cos 3\alpha=2 \sin 4\alpha\cos 4\alpha= \sin 8\alpha= \sin\alpha\tag2 $$ 
For the numerator, use $2\sin x\cos y=\sin(x+y)+\sin(x-y)$ and $\sin 3\alpha= -\sin 4\alpha$,
\begin{align}2\sin\alpha \cdot N &=2\sin \alpha\cos\alpha+ 2\sin \alpha\cos2\alpha+ 2\sin \alpha\cos3\alpha\\
&= \sin 2\alpha+ (\sin 3\alpha- \sin \alpha) + (\sin 4\alpha- \sin 2\alpha)= -\sin \alpha\tag3\\
\end{align}
From (2) and (3), we have $D=\frac14$ and $N = -\frac12$. Plug into (1) to obtain
$$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6} =-2$$
