JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 19)
Let $$\mathrm{S}=\left\{z \in \mathbb{C}-\{i, 2 i\}: \frac{z^{2}+8 i z-15}{z^{2}-3 i z-2} \in \mathbb{R}\right\}$$. If $$\alpha-\frac{13}{11} i \in \mathrm{S}, \alpha \in \mathbb{R}-\{0\}$$, then
$$242 \alpha^{2}$$ is equal to _________.
Answer
1680
Explanation
Put $z=x+i y$
$$ \begin{aligned} & \operatorname{lm}\left(\frac{z^2+8 i z-15}{z^2-3 i z-2}\right)=0 \\\\ & \Rightarrow-\left(x^2-y^2-8 y-15\right)(2 x y-3 x)+(2 x y+8 x)\left(x^2-\right. \left.y^2+3 y-2\right)=0 \\\\ & \Rightarrow\left(x^2-y^2\right)(2 x y+8 x-2 x y+3 x)+(8 y+15)(2 x y- 3 x)+(2 x y+8 x)(3 y-2)=0 \\\\ & \Rightarrow 11 x^3-11 x y^2+16 x y^2-24 x y+30 x y-45 x+6 x y^2 -4 x y+24 x y-16 x=0 \\\\ & \Rightarrow 11 x^3+11 x y^2+26 x y-61 x=0 \\\\ & \Rightarrow\left(11 x^2+11 y^2+26 y-61)=0\right. \\\\ & \because \alpha \neq 0, \Rightarrow x=0 \text { (neglected) } \\\\ & \text { Put } y=-\frac{13}{11}, \quad x=\alpha \\\\ & 11 \alpha^2+11 \cdot \frac{13^2}{11^2}-26 \cdot \frac{13}{11}-61=0 \\\\ & \Rightarrow 121 \alpha^2=840 \\\\ & \Rightarrow 242 \alpha^2=1680 \end{aligned} $$
$$ \begin{aligned} & \operatorname{lm}\left(\frac{z^2+8 i z-15}{z^2-3 i z-2}\right)=0 \\\\ & \Rightarrow-\left(x^2-y^2-8 y-15\right)(2 x y-3 x)+(2 x y+8 x)\left(x^2-\right. \left.y^2+3 y-2\right)=0 \\\\ & \Rightarrow\left(x^2-y^2\right)(2 x y+8 x-2 x y+3 x)+(8 y+15)(2 x y- 3 x)+(2 x y+8 x)(3 y-2)=0 \\\\ & \Rightarrow 11 x^3-11 x y^2+16 x y^2-24 x y+30 x y-45 x+6 x y^2 -4 x y+24 x y-16 x=0 \\\\ & \Rightarrow 11 x^3+11 x y^2+26 x y-61 x=0 \\\\ & \Rightarrow\left(11 x^2+11 y^2+26 y-61)=0\right. \\\\ & \because \alpha \neq 0, \Rightarrow x=0 \text { (neglected) } \\\\ & \text { Put } y=-\frac{13}{11}, \quad x=\alpha \\\\ & 11 \alpha^2+11 \cdot \frac{13^2}{11^2}-26 \cdot \frac{13}{11}-61=0 \\\\ & \Rightarrow 121 \alpha^2=840 \\\\ & \Rightarrow 242 \alpha^2=1680 \end{aligned} $$
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