If the set \ Re ( z - ¯ z + z ¯ z 2 - 3z + 5 ¯ z ): z ∈ ℂ, Re z =...

Given that the set$\left\{Re \left(\dfrac{z - \overset{¯}{z} + z\overset{¯}{z}}{2 - 3z + 5\overset{¯}{z}}\right): z \in \mathbb{C}, Re z = 3\right\}$ is equal to the interval $( \alpha, \beta ]$.

Let $z = x + iy$

$= Re \left(\dfrac{x + iy - \left(x - iy\right) + {x}^{2} + {y}^{2}}{2 - 3\left(x + iy\right) + 5\left(x - iy\right)}\right)$

$= Re \left(\dfrac{{x}^{2} + {y}^{2} + i\left(2y\right)}{2 + 2x - 8iy}\right)$

$= Re \left(\dfrac{\left({x}^{2} + {y}^{2} + 2yi\right)\left(2\left(1 + x\right) + 8iy\right)}{{\left(2\left(1 + x\right)\right)}^{2} + {\left(8y\right)}^{2}}\right)$

$= \dfrac{2\left({x}^{2} + {y}^{2}\right)\left(1 + x\right) - 16{y}^{2}}{4{\left(1 + x\right)}^{2} + {\left(8y\right)}^{2}}$

Now using, $Re\left(z\right) = 3$$⇒ x = 3$ we get,

$= \dfrac{8\left(9 + {y}^{2}\right) - 16{y}^{2}}{64 + 64{y}^{2}}$

$⇒ f\left(y\right) = \dfrac{1}{8}\dfrac{\left(9 - {y}^{2}\right)}{1 + {y}^{2}}$

Let $t = \dfrac{1}{8}\dfrac{\left(9 - {y}^{2}\right)}{1 + {y}^{2}}$

$⇒ 8t + 8t{y}^{2} = 9 - {y}^{2}$

$⇒ {y}^{2}\left(8t + 1\right) + 8t - 9 = 0$

Now $D \ge 0$

$⇒ {0}^{2} - 4\left(8t + 1\right)\left(8t - 9\right) \ge 0$

$⇒ \left(t - \left(\dfrac{- 1}{8}\right)\right)\left(t - \dfrac{9}{8}\right) \le 0$

$⇒ t \in ( \dfrac{- 1}{8}, \dfrac{9}{8} ]$ as $t \ne \dfrac{- 1}{8}$

Range of $f\left(y\right) = ( - 0 . 125, 1 . 125 ]$

$⇒ \alpha = - 0 . 125$

$⇒ \beta = 1 . 125$

$⇒ \beta - \alpha = 1 . 25$

$⇒ 24\left(\beta - \alpha\right) = 30$

Hence this is the required solution.