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 $|z|^= \frac{1}{2}$  

2023/030) Solve in real x $6^x + 9^x =2^(2x+1)$

As we see above power of 2 and 3 ( 9 is $3^2$) and 6 are invloved

Let $3^x = a$ and $2^x=b$

We get $ab + a^2 = 2 b^2$

Or $a^2 + ab - 2b^2 = 0$

 or $(a-b)(a+2b) = 0$

$a=b$ or $a+2b=0$

as a and b are  positive s a = b (a+2b=0 is inadmissible)

or $3^x = 2^x$ or x = 0