Question

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

Answer 1

Answer: 0.50

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