Question

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

Answer 1

Correct Answer: 1680

We are tasked with solving the given expression:

\( z^2 + 8iz - 15 \quad \text{and} \quad z^2 - 3iz - 2 \in \mathbb{R}. \)

Step 1: Simplify the condition

Rewrite the given condition:

Let: \( 1 + \frac{11iz - 13}{z^2 - 3iz - 2} \in \mathbb{R}. \) \( Z = \alpha - \frac{13}{11} i. \)

Step 2: Imaginary part condition

The imaginary part of the denominator must satisfy:

\( z^2 - 3iz - 2 \) is imaginary. Substitute \( z = x + iy: \)

\( z^2 = x^2 - y^2 + 2xyi, \quad -3iz = -3(x + iy)i = -3yi + 3x, \)

so: \( z^2 - 3iz - 2 = (x^2 - y^2 + 3x - 2) + (2xy - 3y)i. \)

For the expression to be purely imaginary:

\( \text{Re}(z^2 - 3iz - 2) = 0 \implies x^2 - y^2 + 3x - 2 = 0. \)

Step 3: Solve the real part equation

Rewriting:

Factorize:

Step 4: Solve for \( z \)

Let:

where: \( x^2 = y^2 - 3y + 2. \) \( x^2 = (y - 1)(y - 2). \) \( z = \alpha - \frac{13}{11} i, \) \( x = \alpha, \quad y = -\frac{13}{11}. \)

Substitute \( x = \alpha \) and \( y = -\frac{13}{11} \) into \( x^2 = (y - 1)(y - 2): \)

\( \alpha^2 = -\frac{13}{11} - 1 - \frac{13}{11} - 2. \)

Simplify:

Step 5: Calculate \( 242\alpha^2 \)

Substitute \( \alpha^2 = \frac{24 \times 35}{121}: \)

\( \alpha^2 = -\frac{24}{11} - \frac{35}{11}, \) \( \alpha^2 = \frac{24 \times 35}{121}. \)

\( 242\alpha^2 = 242 \cdot \frac{24 \times 35}{121}. \)

Simplify:

Final Answer: \( 242\alpha^2 = 1680. \)