Animated Solution for Mathematics - Complex Numbers: If the set $\{\text{Re}(\frac{z-\bar{z}+z\bar{z}}{2-3z+5\bar{z}}):z\in \mathbb{C},\text{Re }z=3\}$ is equal to the interval $(\alpha ,\beta ]$, then $24(\beta -\alpha )$ is equal to

Imagine you are standing on the complex plane, looking at the vertical line defined by $\text{Re}(z)=3$. This is not just a line; it is the entire universe of possibilities for our complex number $z$.

Because the real part is locked at $3$, we can describe any point on this line with the simple, elegant form $z=3+iy$, where $y$ is any real number. Consequently, its conjugate is $\bar{z}=3-iy$. This substitution is our key to unlocking the problem.

Now, let us turn our attention to the expression:

First, consider the numerator. We have $z-\bar{z}=(3+iy)-(3-iy)=2iy$, and $z\bar{z}=|z{|}^2=3^2+y^2=9+y^2$. Thus, the numerator is $N=9+y^2+2iy$.

Next, consider the denominator:

We have successfully reduced the complex fraction into:

To isolate the real part, we must clear the imaginary unit from the denominator. We multiply the numerator and denominator by the conjugate $(1+iy)$.

The denominator becomes $8(1-iy)(1+iy)=8(1+y^2)$. The numerator becomes:

We only care about the real part, so:

We are now in the home stretch. Let $f(y)=\frac{1}{8}\left(\frac{9-y^2}{1+y^2}\right)$. Notice that $y$ only appears as $y^2$.

Let $t=y^2$. Since $y$ is real, $t\ge 0$. Our function becomes:

As $t$ increases from $0$ to $\infty$, the term $\frac{10}{1+t}$ decreases from $10$ to $0$. Thus, $g(t)$ decreases from $\frac{1}{8}(10-1)=\frac{9}{8}$ to $\frac{1}{8}(0-1)=-\frac{1}{8}$.

The range of $\text{Re}(w)$ is $(-\frac{1}{8},\frac{9}{8}]$.

We have identified $\alpha =-\frac{1}{8}$ and $\beta =\frac{9}{8}$. The problem asks for $24(\beta -\alpha )$.

Calculating the difference:

Finally, $24\times \frac{5}{4}=6\times 5=30$. We have navigated the complex plane, simplified the algebra, and found the range with precision.

The final answer is 30.