2024/033) If z is complex number and imaginary part of z is non zero and $\frac{2+3z-4z^2}{2 - 3z+ 4z^2} \in R$ then find $|z|^2$

 We have

 $\frac{2+3z-4z^2}{2 - 3z+ 4z^2} \in R$

So  $\frac{2+3z-4z^2}{2 - 3z+ 4z^2} -1  \in R$

Or  $\frac{6z}{2 - 3z+ 4z^2}  \in R$

Or  $\frac{2 - 3z+ 4z^2}{6z}  \in R$

Or $\frac{2 - 3z+ 4z^2}{z}  \in R$

 Or $\frac{2}{z} -3 + 4z \in R$

  Or $\frac{2}{z} + 4z \in R$

Let $z = x + iy$ and $y\ne 0$

So we have $\frac{2}{x+iy} + 4(x+iy) \in R$

Or  $\frac{2(x-iy)}{x^2+y^2} + 4(x+iy) \in R$

Or   $\frac{-2y}{x^2+y^2} + 4y = 0$ as imaginary part has to be zero

So as we have y is not zero dividing by y we get $x^2+y^2 = \frac{1}{2}$ or $|z|^2=\frac{1}{2}$