Question

Let $z$ be a complex number with non-zero imaginary part. If
$\frac{2+3 z+4 z^2}{2-3 z+4 z^2}$
is a real number, then the value of $|z|^2$ is ___

Answer 1

Given that
$z \neq \overline{ z }$
Let $\alpha=\frac{2+3 z+4 z^2}{2-3 z+4 z^2}=\frac{\left(2-3 z+4 z^2\right)+6 z}{2-3 z+4 z^2}$
$\therefore \alpha=1+\frac{6 z}{2-3 z+4 z^2}$
If $\alpha$ is a real number, then
$ \alpha=\bar{\alpha} $
$ \Rightarrow \frac{z}{2-3 z+4 z^2}=\frac{\bar{z}}{2-3 \bar{z}+4 \bar{z}^2} $
$ \therefore 2(z-\bar{z})=4 z \bar{z}(z-\bar{z}) $
$\Rightarrow(z-\bar{z})(2-4 z \bar{z})=0$
As $z \neq \overline{ z }$ (Given)
$ \Rightarrow z \bar{z}=\frac{2}{4}=\frac{1}{2}$
$ \Rightarrow|z|^2=0.50$