An 11-gon with complex numbers Let $A_1 A_2 \dotsb A_{11}$ be a regular $11$-gon inscribed in a circle of radius $2$.
Let $P$ be a point, such that the distance from $P$ to the center of the circle is $3$.
Find
$[PA_1^2 + PA_2^2 + \dots + PA_{11}^2]$
Note: This can't be made a duplicate. There are no other answers for the other two posts. Please answer. 
 A: Hints:


*

*You can represent points of the plane as complex numbers, say $p$ for $P$ and $a_j$ for $A_j$. Also, you are free to choose the origin (=zero) to be the center of the circle. Then we suddenly know $|a_j|=2$ and $|p|=3$ where absolute value of a complex number is distance from origin.

*For any complex number $z$, we have $|z|^2=z\bar z$ where $\bar z$ is the conjugate of $z$ (the point $z$ reflected through the real line).

*$PA_j^2=|a_j-p|^2=(a_j-p)(\bar a_j-\bar p)=\dots$

*$a_1+a_2+\dots+a_{11}$ is a vector which remains the same when rotated by $360^\circ/11$, so it is zero.

A: Let the center of the circle be the origin $0$. Let $p$ be the complex number representing $P$. I will represent the vertex $A_i$ by $\displaystyle 2(e^{\frac{2\pi}{11}})^i = 2\omega^i$ for $i=1,2,3,...,11$.
I will give an outline of the proof.
$\displaystyle \sum_{i=1}^{11} |p - 2\omega^i|^2 = \sum_{i=1}^{11} (|p|^2 + |2\omega^i|^2 -4p\omega^i) = 11|p|^2 + 44 - 4p\sum_{i=1}^{11}\omega^i = 11(|p|^2 + 4).$
Since it is given that $|p| = 3$, the answer is $143$.
