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QJEEM23Ma83d0: JEE Main 2023 Mathematics - Class 11 JEEM23Ma83d0 Solution: Let . If , then is equal to _________.

Let S={z∈C−{i,2i}:z2+8iz−15z2−3iz−2∈R}\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 α−1311i∈S,α∈R−{0}\alpha-\frac{13}{11} i \in \mathrm{S}, \alpha \in \mathbb{R}-\{0\} , then 242α2242 \alpha^{2} is equal to _________.

Step-by-Step Verified Solution

Put z=x+iyz=x+i y

amp;lm⁡(z2+8iz−15z2−3iz−2)=0amp;⇒−(x2−y2−8y−15)(2xy−3x)+(2xy+8x)(x2−y2+3y−2)=0amp;⇒(x2−y2)(2xy+8x−2xy+3x)+(8y+15)(2xy−3x)+(2xy+8x)(3y−2)=0amp;⇒11x3−11xy2+16xy2−24xy+30xy−45x+6xy2−4xy+24xy−16x=0amp;⇒11x3+11xy2+26xy−61x=0amp;⇒(11x2+11y2+26y−61)=0amp;∵α≠0,⇒x=0 (neglected) amp; Put y=−1311,x=αamp;11α2+11⋅132112−26⋅1311−61=0amp;⇒121α2=840amp;⇒242α2=1680\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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