Sum of squares of modulus of all the complex numbers z satisfying z̅ = iz² + z² - z is equal to ___________.

Solution

Maths sir · 5 steps

Move z across: z + z̅ = 2x is real, so (1+i)z² must be real too.

1. Rearrange first. Add z to both sides to get z + z̅ = (1+i)z². With z = x + iy, the left side is 2x. Now expand (1+i)(x² - y² + 2ixy) and compare real and imaginary parts.

   Real: 2x = x² - y² - 2xy, Imaginary: 0 = x² - y² + 2xy

2. Combine the two. The imaginary equation gives x² - y² = -2xy. Put this into the real equation: 2x = -4xy, so 2x(1 + 2y) = 0. That means x = 0 or y = -1/2.
3. Case by case. If x = 0, the imaginary equation gives y² = 0, so z = 0 with |z|² = 0. If y = -1/2, it becomes x² - x - 1/4 = 0, so x = (1 ± √2)/2, which gives two roots.
4. Add the squares. For these two roots, |z|² = x² + 1/4. Since x² = x + 1/4, this is |z|² = x + 1/2. Summing over both roots gives (x₁ + x₂) + 1 = 1 + 1 = 2, using the sum of roots x₁ + x₂ = 1.
5. So. The values are 0, (2+√2)/2 and (2-√2)/2. Their sum is 2.

The idea to keep. When an equation has both z and z̅, collect z + z̅ or z - z̅ on one side first. Then split into real and imaginary parts, and factor before you solve.

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