Let $$S = \{z \in \mathbb{C} - \{i, 2i\}: \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}\}$$. $$\alpha - \frac{13}{11}i \in S$$, $$\alpha \in \mathbb{R} - \{0\}$$, then $$242\alpha^2$$ is equal to _______

Given $$S = \left\{z \in \mathbb{C} - \{i, 2i\} : \dfrac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}\right\}$$ and $$\alpha - \dfrac{13}{11}i \in S$$ with $$\alpha \in \mathbb{R} - \{0\}$$.

Factor the expressions.

$$z^2 + 8iz - 15 = (z + 3i)(z + 5i)$$

$$z^2 - 3iz - 2 = (z - i)(z - 2i)$$

Substitute $$z = \alpha - \frac{13}{11}i$$.

Computing the numerator $$N = z^2 + 8iz - 15$$:

$$z^2 = \alpha^2 - \frac{26\alpha i}{11} - \frac{169}{121}$$

$$8iz = 8\alpha i + \frac{104}{11}$$

$$N = \left(\alpha^2 - \frac{169}{121} + \frac{104}{11} - 15\right) + i\left(-\frac{26\alpha}{11} + 8\alpha\right)$$

$$= \left(\alpha^2 - \frac{840}{121}\right) + i \cdot \frac{62\alpha}{11}$$

Computing the denominator $$D = z^2 - 3iz - 2$$:

$$-3iz = -3\alpha i - \frac{39}{11}$$

$$D = \left(\alpha^2 - \frac{169}{121} - \frac{39}{11} - 2\right) + i\left(-\frac{26\alpha}{11} - 3\alpha\right)$$

$$= \left(\alpha^2 - \frac{840}{121}\right) + i \cdot \left(-\frac{59\alpha}{11}\right)$$

Apply the condition for $$N/D$$ to be real.

$$\frac{N}{D} \in \mathbb{R} \iff \text{Im}(N \cdot \overline{D}) = 0$$

Let $$R = \alpha^2 - \frac{840}{121}$$. Then $$N = R + i\frac{62\alpha}{11}$$ and $$\overline{D} = R + i\frac{59\alpha}{11}$$.

$$\text{Im}(N \cdot \overline{D}) = R \cdot \frac{59\alpha}{11} + \frac{62\alpha}{11} \cdot R = R \cdot \frac{121\alpha}{11} = 11R\alpha$$

Setting this to zero: since $$\alpha \neq 0$$, we need $$R = 0$$:

$$\alpha^2 = \frac{840}{121}$$

Compute the answer.

$$242\alpha^2 = 242 \times \frac{840}{121} = 2 \times 840 = 1680$$

The answer is $$1680$$.