The question goes:
The complex numbers $z_1, z_2, z_3 \ldots z_n, \sum_{i=1}^{n} z_i \neq 0$ represent the vertices of a regular polygon of $n$ sides in order, inscribed in a circle of unit radius such that $z_3 + z_n =A\cdot z_1 + \bar A\cdot z_2$ Find $\lfloor |A|\rfloor$ when $n=4,6,8,12$.
In my attempt I tried to generalize over $n$. My attempt:
Suppose $z_1=e^{i\phi}$. Now, $$z_j=e^{i((j-1)\frac{2\pi}{n} + \phi)}$$
On substitution, I obtained $$e^{i\frac{4\pi}{n}}+e^{i(n-1)\frac{2\pi}{n}}=A+\bar A\cdot e^{i\frac{2\pi}{n}}=A+i\bar A $$
Then I put $A=x +iy$ and $n=4$ to end up with $-1-i=(x+y)+i(x+y)$ giving me only $x+y=-1$.
If I'm going right, how should I proceed? Thanks in advance.
Edit: The answers are $2,2,1,1$ respectively.
Edit 2: The original question. 
Answer is A-R,B-R,C-Q,D-Q.
Edit 3: The solution as given by the problem setter.
Let $n \geq 4$. We are given that $z_1, \dotsc, z_n$ are the vertices of a regular $n$-gon inscribed in a circle in the complex plane, with $\sum_i z_i \neq 0$. So, letting $w := x_0 + iy_0$ denote the centre of the circle, we have $w \neq 0$.
Let $\theta \in [0,2\pi)$ such that $z_1 = w + e^{i \theta}$. Then, $$z_j = w + e^{i(\theta + 2\pi(j-1)/n)} = \left[x_0 + \cos\left(\theta + \frac{2\pi(j-1)}{n}\right)\right] + i\left[y_0 + \sin\left(\theta + \frac{2\pi(j-1)}{n}\right)\right]$$ for all $j = 1, \dotsc, n$. Hence, $$ z_3 + z_n = \left[2x_0 + \cos\left(\theta + \frac{4\pi}{n}\right) + \cos\left(\theta + \frac{2(n-1)\pi}{n}\right)\right] + i\left[2y_0 + \sin\left(\theta + \frac{4\pi}{n}\right)+\sin\left(\theta + \frac{2(n-1)\pi}{n}\right)\right]. $$ Let $A = p + iq$. Then, $$ Az_1 + \bar{A}z_2 = \left\{\left[ 2x_0 + \cos( \theta ) + \cos\left( \theta + \frac{2\pi}{n} \right)\right]p + \left[-\sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right) \right]q \right\} + i \left\{\left[ 2y_0 + \sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right)\right]p + \left[\cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right) \right]q\right\}. $$ We are given that $z_3 + z_n = Az_1 + \bar{A}z_2$. So, equating the real and imaginary parts of the two sides, we get a system of two linear equations in two unknowns: $$ \begin{bmatrix} 2x_0 + \cos( \theta ) + \cos\left( \theta + \frac{2\pi}{n} \right) & -\sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right)\\ 2y_0 + \sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right) & \cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right) \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} 2x_0 + \cos\left(\theta + \frac{4\pi}{n}\right) + \cos\left(\theta + \frac{2(n-1)\pi}{n}\right) \\ 2y_0 + \sin\left(\theta + \frac{4\pi}{n}\right)+\sin\left(\theta + \frac{2(n-1)\pi}{n}\right) \end{bmatrix} $$ Now, the determinant of the matrix of coefficients is $$ \Delta = 4 \sin \left( \frac{\pi}{n} \right) \left[ -y_0 \cos\left( \theta + \frac{\pi}{n} \right) + x_0 \cos\left( \theta + \frac{\pi}{n} \right) \right], $$ so $\Delta = 0 \iff w$ lies on the line with slope $\tan\left(\theta + \frac{\pi}{n} \right)$. Note: here we mean that $w$ lies on the imaginary axis if $\theta + \pi/n = \pi/2$ or $3\pi/2$.
Suppose that $\Delta \neq 0$. Then, left multiplying by the inverse of the matrix of coefficients, we get $$ \begin{bmatrix} p\\ q \end{bmatrix} = \frac{1}{\Delta} \begin{bmatrix} 2x_0\left[ \cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right) \right] + 2y_0\left[ \sin(\theta) - \sin\left( \theta + \frac{2\pi}{n} \right) \right] \\ \\ 2x_0\left[ - \sin(\theta) - \sin\left( \theta + \frac{2\pi}{n} \right) + \sin\left( \theta + \frac{4\pi}{n} \right) + \sin\left( \theta + \frac{2(n-1)\pi}{n} \right)\right] + 2y_0\left[ \cos(\theta) + \cos\left( \theta + \frac{2\pi}{n} \right) -\cos\left( \theta + \frac{4\pi}{n} \right) - \cos\left( \theta + \frac{2(n-1)\pi}{n} \right)\right] \end{bmatrix} = \frac{1}{\Delta} \begin{bmatrix} \Delta\\ -2\Delta\sin\left( \frac{2\pi}{n} \right) \end{bmatrix}. $$ Hence, $$A = 1 - 2i\sin\left( \frac{2\pi}{n} \right).$$ Since $\sin(\pi - x) = \sin(x)$ for all $x$, and $\pi - 2\pi/n$ is the interior angle of the polygon, our solution for $A$ matches the one in the "official" solution.
Now, the given solution is damn clever and makes my calculations look caveman-like. But, just so I can salvage some of my previous attempt, I will note that the "official" solution actually does not make use of the fact that $\sum_i z_i \neq 0$ anywhere! This suggests that they are implicitly dealing with the non-degenerate case only. Since we've come so far, let's complete the degenerate case as well. :)
Firstly, our analysis tells us that the correct criterion for degeneracy is $\Delta = 0$. This is the condition that $w$ lies on the line with slope $\tan\left( \theta + \frac{\pi}{n} \right)$. This is stronger than the given condition, which just says that $w$ must not be the origin.
So, let's assume that $\Delta = 0$. In this case, the system of linear equations looks like: $$ \begin{bmatrix} \pm 2\lvert w \rvert \cos\left( \theta + \frac{\pi}{n} \right) + \cos( \theta ) + \cos\left( \theta + \frac{2\pi}{n} \right) & -\sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right)\\ \pm 2\lvert w \rvert \sin\left( \theta + \frac{\pi}{n} \right) + \sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right) & \cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right) \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} \pm 2\lvert w \rvert \cos\left( \theta + \frac{\pi}{n} \right) + \cos\left(\theta + \frac{4\pi}{n}\right) + \cos\left(\theta + \frac{2(n-1)\pi}{n}\right) \\ \pm 2\lvert w \rvert \sin\left( \theta + \frac{\pi}{n} \right) + \sin\left(\theta + \frac{4\pi}{n}\right)+\sin\left(\theta + \frac{2(n-1)\pi}{n}\right) \end{bmatrix} $$ which can be simplified to $$ \begin{bmatrix} 2 \cos\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left( \frac{3\pi}{n} \right)\right] & 2 \sin\left( \frac{\pi}{n} \right) \cos\left( \theta + \frac{\pi}{n} \right) \\ 2 \sin\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left( \frac{3\pi}{n} \right)\right] & 2 \sin\left( \frac{\pi}{n} \right) \sin\left( \theta + \frac{\pi}{n} \right) \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} 2 \cos\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left(\frac{3\pi}{n}\right)\right] \\ 2 \sin\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left(\frac{3\pi}{n}\right)\right] \\ \end{bmatrix}. $$ If $\theta = \pi - \pi/n$ or $2\pi - \pi/n$, then the second equation is identically zero; if $\theta = \pi/2 - \pi/n$ or $3\pi/2 - \pi/n$, then the first equation is identically zero; otherwise, we can subtract $\tan\left( \theta + \frac{\pi}{n} \right)$ times the first row from the second row. In each case, we see that the system is consistent and there is one nontrivial equation.
So, every complex number on the line determined by that equation is a valid choice for $A$, and $\lfloor \lvert A \rvert \rfloor$ is not uniquely determined!
Lastly, one can check that the point $1 - 2i\sin\left( \frac{2\pi}{n} \right)$ lies on this degenerate line so there is no contradiction in the given solution. It is only that the problem setter missed some additional solutions in this case.
I made a visualization of the solutions in GeoGebra:
I give some screenshots below: