If $$z = x + iy$$, $$xy \neq 0$$, satisfies the equation $$z^2 + i\bar{z} = 0$$, then $$|z^2|$$ is equal to :

Given $$z = x + iy$$, $$xy \neq 0$$, and $$z^2 + i\bar{z} = 0$$.

We have $$\bar{z} = x - iy$$, and $$z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$$.

Substituting into the equation:

$$ (x^2 - y^2 + 2ixy) + i(x - iy) = 0 $$

$$ (x^2 - y^2 + 2ixy) + (ix + y) = 0 $$

$$ (x^2 - y^2 + y) + i(2xy + x) = 0 $$

Equating real and imaginary parts to zero:

Real: $$x^2 - y^2 + y = 0$$ ... (1)

Imaginary: $$2xy + x = 0 \Rightarrow x(2y + 1) = 0$$ ... (2)

Since $$xy \neq 0$$, we have $$x \neq 0$$, so from (2): $$2y + 1 = 0 \Rightarrow y = -\frac{1}{2}$$.

Substituting in (1):

$$ x^2 - \frac{1}{4} - \frac{1}{2} = 0 \Rightarrow x^2 = \frac{3}{4} $$

Now, $$|z|^2 = x^2 + y^2 = \frac{3}{4} + \frac{1}{4} = 1$$.

Therefore, $$|z^2| = |z|^2 = 1$$.