If \(z=x+ i y, x y \neq 0\), satisfies the equation \(z^2+ i \bar{z}=0\), then \(\left| z ^2\right|\) is equal to : [JEE…

Given \(z=x+iy\)

\(\overline{z}=x-iy\)

Substituting into

\(z^2+i\overline{z}=0\)

\((x+iy)^2+i(x-iy)=0\)

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

Comparing real and imaginary parts,

\(x^2-y^2+y=0\) and \(2xy+x=0\)

From \(x(2y+1)=0\)

and \(x\ne0\) \(2y+1=0\)

\(y=-\dfrac12\)

Substituting into

\(x^2-y^2+y=0\)

\(x^2-\dfrac14-\dfrac12=0\)

\(x^2=\dfrac34\)

Hence \(|z|^2=x^2+y^2\)

\(=\dfrac34+\dfrac14=1\)

Therefore

\(|z^2|=|z|^2=1\)