If z=x+i y, x y ≠ 0, satisfies the equation z²+i z̄=0, then |z²| is equal to :

Solution

Maths sir · 4 steps

Move iz̄ across and take modulus on both sides; |z| falls out at once.

1. Rearrange. Write the equation as z² = -iz̄. Now take the modulus of both sides, using |-i| = 1 and |z̄| = |z|.

   |z|² = |z| ⇒ |z|(|z| - 1) = 0

2. Rule out zero. Since xy ≠ 0, z ≠ 0, so |z| = 1. That gives |z²| = |z|² = 1.
3. Check with parts. Putting z = x + iy gives real part x² - y² + y = 0 and imaginary part x(2y + 1) = 0. With x ≠ 0, y = -1/2 and x² = 3/4, so x² + y² = 1. Such a z really exists.
4. So. |z²| = 1, which is option (d).

The idea to keep. When an equation links zⁿ and z̄, take modulus first. Since |z̄| = |z| and |i| = 1, you often get |z| without ever finding x and y.

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