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 ${\left|z\right|}^2$ is _____________.

The correct answer is 0.50.

Let $w=\frac{2+3z+4z^2}{2-3z+4z^2}$.

We are given that $w$ is a real number, and the imaginary part of $z$ is non-zero ($z\ne \overline{z}$).

To simplify $w$, we can divide the numerator and the denominator by $z$ (since $z\ne 0$):

$w=\frac{\frac{2}{z}+3+4z}{\frac{2}{z}-3+4z}=\frac{\left(4z+\frac{2}{z}\right)+3}{\left(4z+\frac{2}{z}\right)-3}$

For $w$ to be a purely real number, the term $4z+\frac{2}{z}$ must also be a real number. Let $u=4z+\frac{2}{z}$.

Since $u$ is a real number, it must be equal to its complex conjugate, i.e., $u=\overline{u}$:

$4z+\frac{2}{z}=4\overline{z}+\frac{2}{\overline{z}}$

Rearranging the terms, we get:

$4\left(z-\overline{z}\right)+2\left(\frac{1}{z}-\frac{1}{\overline{z}}\right)=0$

$4\left(z-\overline{z}\right)+2\left(\frac{\overline{z}-z}{z\overline{z}}\right)=0$

Factoring out $\left(z-\overline{z}\right)$:

$\left(z-\overline{z}\right)\left(4-\frac{2}{z\overline{z}}\right)=0$

Since $z$ has a non-zero imaginary part, $z-\overline{z}\ne 0$. Therefore, the second factor must be equal to zero:

$4-\frac{2}{z\overline{z}}=0$

Recall that $z\overline{z}={\left|z\right|}^2$:

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

${\left|z\right|}^2=\frac{2}{4}=0.50$

Thus, the value of ${\left|z\right|}^2$ is 0.50.