If one of the vertices of the square circumscribing the circle `|z - 1| = sqrt2` is `2+ sqrt3 iota`. Find the other vertices of square

Explanation

The given circle is `|z-1|=sqrt(2),` where `z_(0)=1` is the center and `sqrt(2)` is radius of the circle. `z_(1)` is one of the vertices of the square inscribed in the given circle.

Clearly, `z_(2)` can be obtained by rotating `z_(1)` by an angle of `90^(@)` in anticlockwise sense about center `z_(0).` Thus,
`z_(2)-z_(0)=(z_(1)-z_(0))e^(ipi//2)`
`impliesz_(2)-1=(2+isqrt(3)-1)i`
`impliesz_(2)=i-sqrt(3)+1`
`impliesz_(2)=(1-sqrt(3))+i`
Now `z_(0)` is midpoint of `z_(1)` and `z_(3)` and `z_(4)`
`:." "(z_(1)+z_(3))/(2)=z_(0)implies(2+isqrt(3)+z_(3))/(2)=1`
`impliesz_(3)=-isqrt(3)`
and `(z_(2)+z_(4))/(2)=z_(0)implies((1-sqrt(3))+i+z_(4))/(2)=1`
`impliesz_(4)=(sqrt(3)+1)-i`