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  $4{\left|z\right|}^2$ is ___________

Let  $X=\frac{4z^2+3z+2}{4z^2-3z+2}=1+\frac{6z}{4z^2-3z+2}$ 
$=1+\frac{6}{4z}-3+\frac{2}{z}=1+\frac{6}{2\left(2z+\frac{1}{z}\right)-3}$ 
Since  $X\in R$, then $2z+\frac{1}{z}\in R\Rightarrow 2z+\frac{1}{z}=2\overline{z}+\frac{1}{\overline{z}}$
$2\left(z-\overline{z}\right)-\frac{z-\overline{z}}{z\overline{z}}=0$

$\left(z-\overline{z}\right)\left(2-\frac{1}{{\left|z\right|}^2}\right)=0$

$2-\frac{1}{{\left|z\right|}^2}=0\left(\because z\ne \overline{z}\right)$

${\left|z\right|}^2=\frac{1}{2}\Rightarrow 4{\left|z\right|}^2=2$