If the set Re z - z ¯ + z z ¯ 2 - 3 z + 5 z ¯ : z ∈ ℂ ,   Re z = 3 is equal…

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

Let $z=x+iy$

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

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

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

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

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

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

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

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

$\Rightarrow 8t+8t{y}^2=9-y^2$

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

Now $D\ge 0$

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

$\Rightarrow \left(t-\left(\frac{-1}{8}\right)\right)\left(t-\frac{9}{8}\right)\le 0$

$\Rightarrow t\in (\frac{-1}{8},\frac{9}{8}]$ as $t\ne \frac{-1}{8}$

Range of $f\left(y\right)=(-0.125,1.125]$

$\Rightarrow \alpha =- 0.125$

$\Rightarrow \beta =1.125$

$\Rightarrow \beta -\alpha =1.25$

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

Hence this is the required solution.