341For the first one, you have to first note that the circumcircle of the triangle formed will be the unit circle.
Using some easy geometry, you can see that the intersection is always at (-1,0) i.e. at \omega + \bar{\omega}
4 Answers
62question 2
look at the roots of xn-1=0
they are αr's that you have given...
xn-1= (x-α1)(x-α2)(x-α3)..... (x-αn)
what you want is the product of (|2-α1|)(|2-α2|)(|2-α3|)..... (|2-αn|)
whichi is 2n-1 (value of the function at x=2)
341
341So now if you call the intersection point P, and join it to ω or ω2 it forms an angle of 60° at A, which is same as angle formed by the chord joining ω or ω2 to (-1,0)
