Let S = { z- { i , 2 i } : z 2 + 8 i z - 15 z 2 - 3 i z - 2 } . If - 13 11 i S , { 0 } , then 242 2 is equal to __________. [2023]

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Solution

Q.
$Let S = {z \in ℂ - {i, 2 i} : \frac{z^{2} + 8 i z - 15}{z^{2} - 3 i z - 2} \in ℝ} .$

$If α - \frac{13}{11} i \in S, α \in ℝ - {0}, then 242 α^{2} is equal to$ __________. [2023]

Ans.
(1680)

We have, $\frac{z^{2} + 8 i z - 15}{z^{2} - 3 i z - 2} = \frac{z^{2} - 3 i z - 2 + 11 i z - 13}{z^{2} - 3 i z - 2}$

$= 1 + (\frac{11 i z - 13}{z^{2} - 3 i z - 2}) = 1 + \frac{11 i (z + \frac{13 i}{11})}{z^{2} - 3 i z - 2} \in R$

Put $z = α - \frac{13 i}{11}$

Thus, $z^{2} - 3 i z - 2$ is imaginary.

Put $z = x + i y$

$\Rightarrow {(x + i y)}^{2} - 3 i (x + i y) - 2 is imaginary.$

$\Rightarrow x^{2} - y^{2} + 2 x y i - 3 i x + 3 y - 2 is imaginary.$

$\Rightarrow Re (x^{2} - y^{2} + 3 y - 2 + (2 x y - 3 x) i) = 0$

$\Rightarrow x^{2} - y^{2} + 3 y - 2 = 0 \Rightarrow x^{2} = y^{2} - 3 y + 2$

$\Rightarrow x^{2} = (y - 1) (y - 2)$

Put $x = α$ and $y = \frac{- 13}{11}$ , we get $α^{2} = (\frac{- 13}{11} - 1) (\frac{- 13}{11} - 2)$

$= (\frac{- 13 - 11}{11}) (\frac{- 13 - 22}{11}) = (\frac{- 24}{11}) (\frac{- 35}{11}) = \frac{24 \times 35}{121}$

$∴$ $242 α^{2} = \frac{24 \times 35 \times 242}{121} = 1680$

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