Mathematics - Complex Number Question with Solution | TestHub

Let $\alpha =\frac{4z^2+3z+2}{4z^2-3z+2}=1+\frac{6z}{4z^2-3z+2}$

$\Rightarrow \alpha =1+\frac{6}{2\left(2z+\frac{1}{z}\right)-3}$

$\because \alpha \in R$ then $2z+\frac{1}{z}\in R$

$\Rightarrow 2z+\frac{1}{z}=2\overline{z}+\frac{1}{\overline{z}}$

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

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

$\because z\ne \overline{z}$  (given)

So ${\left|z\right|}^2=\frac{1}{2}$