Let S={z ∈ ℂ: z̄=i(z²+Re(z̄))}. Then Σz ∈ S|z|² is equal to :

Solution

Maths sir · 4 steps

Put z = x + iy, compare real and imaginary parts, and keep every case when you factor.

1. Expand both sides. With z = x + iy, the left side is z̄ = x - iy. Also z² = x² - y² + 2ixy and Re(z̄) = x, so the right side is i(x² - y² + x + 2ixy) = -2xy + i(x² - y² + x).
2. Compare parts. Real parts give x = -2xy, that is x(1 + 2y) = 0. Imaginary parts give -y = x² - y² + x.
3. Two cases. If x = 0, then -y = -y², so y = 0 or y = 1. If y = -1/2, then x² + x - 3/4 = 0, so x = 1/2 or x = -3/2.

   Case	z	|z|²
   x = 0	0	0
   x = 0	i	1
   y = -1/2	1/2 - 1/2i	1/4 + 1/4 = 1/2
   y = -1/2	-3/2 - 1/2i	9/4 + 1/4 = 5/2

4. So. The sum is 0 + 1 + 1/2 + 5/2 = 4, which is option (b).

The idea to keep. For any equation in z and z̄, write z = x + iy and equate real and imaginary parts. When you get a product equal to zero, solve each factor separately.

Answer from the NTA final answer key. Solution by PrepWiser.