TG EAMCET 4 May 2025 Shift 2 Paper

For roots of unit
Im(xj)=sin‌(‌

2π(j−1)

8

)
The maximum Im(xj) is 1 , so
min‌|xj−2i|2=5−4(1)=1
The minimum Im(xj) is -1 , so
max‌|xj−2i|2=5−4(−1)=9
So, the vertices lie on a circle whose radius is the same for all vertices.
|zj|=‌

1

|xj−2i|

So, the radius is
Circumradius =‌

1

|xj−2i|

For all 8 vertices, this distance is the same. For octagon symmetry take the average Im(xj).
But actually, the 8 points are symmetric and the distance for all is
|xj−2i|=2+sin‌(θ)
At 8 symmetric points, the effective radius is exactly ‌

1

4

.
So the radius is ‌

1

4