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hi just visiting not an aryan intellectual
a is 0, 2pi/5, 4pi/5, 6pi/5, 8pi/5
cos theta + i sin theta is r=1, angle=theta in polar coordinates: the point on the unit circle with angle theta (bc cos^2 t + sin^2 t = 1)
Complex multiplication adds the angles and multiplies the radii, so the fifth roots of 1 are 1/5 around the unit circle, 2/5 around the unit circle, etc. When you take the fifth power they end up 0 around the unit circle
Now I'm not sure how you get b algebraically (update post if you find out), but I did it kind of geometrically?
We know (z+1)/z needs to have radius 1 and angle 2kpi/5 because those are the solutions to the fifth root of 1. Because it has radius 1, z+1 needs to have the same radius as z. Since their imaginary parts (y coordinates) are the same, they need to have opposite real parts: Re(z + 1) = - Re(z). We're flipping z around the y axis. Thus Re(z)=-1/2 and Re(z+1)=1/2.
Because the angle of (z+1)/z must be 2kpi/5, the difference in their angles must be 2kpi/5. However, we already discussed how adding 1 to z flips it around the y axis. This means <(z+1)=pi/2-(<(z)-pi/2)=pi-<(z). Thus <(z+1)-<(z) = pi-2<(z)= 2kpi/5, so <(z)=(5-2k)pi/10. Since Re(z) is negative, <(z) is strictly between pi/2 and 3pi/2, so it must be 7pi/10, 9pi/10, 11pi/10, 13pi/10.
This means that the valid solutions to z are the points of intersection of x=-1/2 and the lines through the origin at those angles, y/x=-2y=tan(<(z)). If we look at the triangles of those lines and the y axis instead, we can also describe them as y/x=cot(<(z)-pi), or y=Im(z)=-1/2 * cot(n*pi/10) where n is 2, 4, 6, or 8. Simplifying, this is -1/2 * cot(k*pi/5) where k is 1, 2, 3, or 4. Thus we have found the imaginary part and the answer is -1/2(1+ i cot(k*pi/5))
