2023/31) Given $\frac{2+3z+4z^2}{2-3z+4z^2} \in \mathbb{R}$ and imaginary part of x is not zero find $|z|^2$

We have $\frac{2+3z+4z^2}{2-3z+4z^2}$ real 

Both numerator and denominator are expression with 3 terms 

Subtracting 1 fron the expression we shall have it real and numerator is simpler

Or   $\frac{2+3z+4z^2}{2-3z+4z^2}-1$ is real

Or$\frac{-6z}{2-3z+4z^2}$ is real

As imaglinary part of x is not zero so x is not zero so inverting 

 $\frac{2-3z+4z^2}{-6z}$ is real

Or $\frac{2-3z+4z^2}{z}$ is real

Or $\frac{2}{z}-3+4z$ is real 

And adding 3 we get   $\frac{2}{z}+4z$ is real 

Or $\frac{1}{z}+2z$ is real

Now let $z= x+ iy$

So $\frac{1}{x+iy}+2(x+iy)$ is real

Or $frac{x-iy}{x^+y^2} + 2(x+iy)|$ is real

Or $\frac{-y}{x^+y^2} + 2y)=0$

as y is non zero $\frac{1}{x^2+y^2} -2=0$ or $|z|^= \frac{1}{2}$