Let S be the set of all complex numbers z satisfying the equation z^2 + iz + z = 0 . If the points representing the elements of S form a triangle in the complex plane, then the circumradius of this triangle is equal to

Step-by-step solution

Let z = x + iy . Substituting this into the given equation: (x+iy)^2 + i(x+iy) + (x-iy) = 0 (x^2 - y^2 + 2ixy) + (ix - y) + (x - iy) = 0 (x^2 - y^2 + x - y) + i(2xy + x - y) = 0 Equating real and imaginary parts to zero, we get: x^2 - y^2 + x - y = 0 (x-y)(x+y+1) = 0 2xy + x - y = 0 Case 1: x = y Substituting into the second equation: 2x^2 + x - x = 0 2x^2 = 0 x = 0, y = 0 . So, z₁ = 0 . Case 2: x + y + 1 = 0 y = -x - 1 Substituting into the second equation: 2x(-x-1) + x - (-x-1) = 0 -2x^2 - 2x + 2x + 1 = 0 2x^2 =