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Step-by-Step Solution
Step 1: Represent the complex number z
Given that the real part of $z$ is 3, we can express $z$ as
$z = 3 + i\,y$ and its complex conjugate as $\bar{z} = 3 - i\,y$,
where $y \in \mathbb{R}$.
Step 2: Write the given expression in terms of y
We need the real part of
$$
\frac{z - \bar{z} + z\,\bar{z}}{2 - 3\,z + 5\,\bar{z}}.
$$
Substitute $z = 3 + i\,y$ and $\bar{z} = 3 - i\,y$ into the numerator first:
$z - \bar{z} = (3 + i\,y) - (3 - i\,y) = 2\,i\,y.$
$z \,\bar{z} = (3 + i\,y)\,(3 - i\,y) = 9 + y^2.$
Hence, the numerator becomes
$$
(2\,i\,y) + (9 + y^2) = 9 + y^2 + 2\,i\,y.
$$
Now, for the denominator:
$2 - 3\,z = 2 - 3\,(3 + i\,y) = 2 - 9 - 3\,i\,y = -7 - 3\,i\,y.$
$+5\,\bar{z} = 5\,(3 - i\,y) = 15 - 5\,i\,y.$
So the denominator is
$$
(-7 - 3\,i\,y) + (15 - 5\,i\,y) = ( -7 + 15 ) + ( -3\,i\,y - 5\,i\,y ) = 8 - 8\,i\,y.
$$
Step 3: Simplify the expression
Thus, the expression becomes
$$
\frac{9 + y^2 + 2\,i\,y}{8 - 8\,i\,y} = \frac{(9 + y^2) + 2\,i\,y}{8(1 - i\,y)}.
$$
Factor out $8$ from the denominator to get
$$
z_1 = \frac{(9 + y^2) + 2\,i\,y}{8\,\bigl(1 - i\,y\bigr)}.
$$
Step 4: Obtain the real part
We can find the real part by multiplying numerator and denominator by the complex conjugate of the denominator, $1 + i\,y$. Without showing every intermediate algebraic step here, the real part simplifies to:
$$
\operatorname{Re}\bigl(z_1\bigr) = \frac{9 - y^2}{8\,(1 + y^2)}.
$$
Step 5: Investigate the range of the real part
Let
$$
r(y) = \operatorname{Re}(z_1) = \frac{9 - y^2}{8\,(1 + y^2)}.
$$
We seek to find all possible values of $r(y)$ as $y$ ranges over all real numbers.
Rewrite:
$$
r(y) = \frac{1}{8}\Bigl(\frac{9 - y^2}{1 + y^2}\Bigr)
= \frac{1}{8}\Bigl(\frac{10 - (1 + y^2)}{1 + y^2}\Bigr)
= \frac{1}{8}\Bigl(\frac{10}{1 + y^2} - 1\Bigr).
$$
Since $1 + y^2 \in [1,\infty)$:
$\frac{1}{1 + y^2} \in (0, 1]$,
$\frac{10}{1 + y^2} \in (0, 10]$,
$\frac{10}{1 + y^2} - 1 \in (-1, 9]$.
Hence,
$$
r(y) \;\in\; \bigl(\tfrac{-1}{8},\, \tfrac{9}{8}\bigr].
$$
Thus, the set of all such real parts is $\bigl(\tfrac{-1}{8}, \tfrac{9}{8}\bigr]$, i.e., $(\alpha, \beta] = \bigl(\tfrac{-1}{8}, \tfrac{9}{8}\bigr]$.
Step 6: Calculate $24(\beta - \alpha)$
Identify $\alpha = -\tfrac{1}{8}$ and $\beta = \tfrac{9}{8}$. Then
$$
\beta - \alpha = \tfrac{9}{8} - \Bigl(-\tfrac{1}{8}\Bigr) = \tfrac{9}{8} + \tfrac{1}{8} = \tfrac{10}{8} = \tfrac{5}{4}.
$$
So,
$$
24(\beta - \alpha) = 24 \times \tfrac{5}{4} = 6 \times 5 = 30.
$$
Final Answer
The value of $24(\beta - \alpha)$ is 30.
