Key Concepts and Formulas

• A complex number $w$ is real if and only if $w = \bar{w}$, where $\bar{w}$ is the complex conjugate of $w$. Equivalently, $\text{Im}(w)=0$.
• For a complex number $z = x + iy$, where $x$ and $y$ are real numbers, $z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$.
• If $\frac{A+iB}{C+iD} \in \mathbb{R}$, then $BC - AD = 0$.

Step-by-Step Solution

Step 1: Define z and set up the problem

We are given that $S = \left\{z \in \mathbb{C}-\{i, 2 i\}: \frac{z^{2}+8 i z-15}{z^{2}-3 i z-2} \in \mathbb{R}\right\}$. Let $z = x+iy$, where $x, y \in \mathbb{R}$. We want to find the condition for $\frac{z^{2}+8 i z-15}{z^{2}-3 i z-2}$ to be a real number.

Step 2: Express the numerator in terms of x and y

Let $N = z^2 + 8iz - 15$. Substituting $z = x+iy$, we get $N = (x+iy)^2 + 8i(x+iy) - 15 = x^2 + 2ixy - y^2 + 8ix - 8y - 15 = (x^2 - y^2 - 8y - 15) + i(2xy + 8x).$ Thus, $A = x^2 - y^2 - 8y - 15$ and $B = 2xy + 8x$.

Step 3: Express the denominator in terms of x and y

Let $D = z^2 - 3iz - 2$. Substituting $z = x+iy$, we get $D = (x+iy)^2 - 3i(x+iy) - 2 = x^2 + 2ixy - y^2 - 3ix + 3y - 2 = (x^2 - y^2 + 3y - 2) + i(2xy - 3x).$ Thus, $C = x^2 - y^2 + 3y - 2$ and $D = 2xy - 3x$.

Step 4: Apply the condition for the expression to be real

For $\frac{N}{D}$ to be real, we require $BC - AD = 0$. Substituting the expressions for $A, B, C, D$, we have $(2xy + 8x)(x^2 - y^2 + 3y - 2) - (x^2 - y^2 - 8y - 15)(2xy - 3x) = 0.$

Step 5: Simplify the equation

Expanding the equation, we get $2x^3y - 2xy^3 + 6xy^2 - 4xy + 8x^3 - 8xy^2 + 24xy - 16x - (2x^3y - 3x^3 - 2xy^3 + 3xy^2 - 16xy^2 + 24xy - 30xy + 45x) = 0$ $2x^3y - 2xy^3 + 6xy^2 - 4xy + 8x^3 - 8xy^2 + 24xy - 16x - 2x^3y + 3x^3 + 2xy^3 - 3xy^2 + 16xy^2 - 24xy + 30xy - 45x = 0$ Combining like terms, we get $11x^3 + 11xy^2 + 26xy - 61x = 0.$ Factoring out $x$, we have $x(11x^2 + 11y^2 + 26y - 61) = 0.$

Step 6: Use the given information

We are given that $\alpha - \frac{13}{11}i \in S$, where $\alpha \in \mathbb{R}-\{0\}$. Thus, $x = \alpha$ and $y = -\frac{13}{11}$. Since $\alpha \neq 0$, we have $x \neq 0$, so we must have $11x^2 + 11y^2 + 26y - 61 = 0.$

Step 7: Substitute and solve for alpha

Substituting $x = \alpha$ and $y = -\frac{13}{11}$, we get $11\alpha^2 + 11\left(-\frac{13}{11}\right)^2 + 26\left(-\frac{13}{11}\right) - 61 = 0$ $11\alpha^2 + 11\left(\frac{169}{121}\right) - \frac{338}{11} - 61 = 0$ $11\alpha^2 + \frac{169}{11} - \frac{338}{11} - 61 = 0$ $11\alpha^2 - \frac{169}{11} - 61 = 0$ $11\alpha^2 = \frac{169}{11} + 61 = \frac{169 + 671}{11} = \frac{840}{11}$ $\alpha^2 = \frac{840}{121}$

Step 8: Calculate the final value

We want to find $242\alpha^2$. We have $242\alpha^2 = 242 \cdot \frac{840}{121} = 2 \cdot 840 = 1680.$

Common Mistakes & Tips

• Be careful with signs when expanding and simplifying the equation.
• Remember the condition for a complex number to be real.
• Pay attention to the given information, especially the condition $\alpha \neq 0$.

Summary

We used the condition for a complex number to be real to set up an equation involving $x$ and $y$. We then used the given information to substitute for $x$ and $y$ and solve for $\alpha^2$. Finally, we calculated $242\alpha^2$.

The final answer is $\boxed{1680}$.