Let z be a complex number with non-zero imaginary part. If 2 + 3z +...

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

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

$∵ \alpha \in R$ then $2z + \dfrac{1}{z} \in R$

$⇒ 2z + \dfrac{1}{z} = 2\overset{¯}{z} + 1\overset{¯}{z}$

$⇒ 2\left(z - \overset{¯}{z}\right) - \dfrac{z - \overset{¯}{z}}{{\left|z\right|}^{2}} = 0$

$⇒ \left(z - \overset{¯}{z}\right)\left(2 - \dfrac{1}{{\left|z\right|}^{2}}\right) = 0$

$∵ z \ne \overset{¯}{z}$ (given)

So ${\left|z\right|}^{2} = \dfrac{1}{2}$