Let $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

Question

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

Professor PC Thomas Classes · Answer

(1680) : We have,\ $\frac{z^{2}+8 i z-15}{z^{2}-3 i z-2}=\frac{z^{2}-3 i z-2+11 i z-13}{z^{2}-3 i z-2}$\ $=1+\left(\frac{11 i z-13}{z^{2}-3 i z-2}ight)=1+\frac{11 i\left(z+\frac{13 i}{11}ight)}{z^{2}-3 i z-2} \in R$\ Put $z=\alpha-\frac{13 i}{11}$\ Thus, $z^{2}-3 i z-2$ is imaginary\ Put $z=x+i y$\ $\Rightarrow \quad(x+i y)^{2}-3 i(x+i y)-2$ is imaginary.\ $\Rightarrow \quad x^{2}-y^{2}+2 x y i-3 i x+3 y-2$ is imaginary.\ $\Rightarrow \quad \operatorname{Re}\left(x^{2}-y^{2}+3 y-2+(2 x y-3 x) iight)=0$\ $\Rightarrow x^{2}-y^{2}+3 y-2=0 \Rightarrow x^{2}=y^{2}-3 y+2$\ $\Rightarrow \quad x^{2}=(y-1)(y-2)$\ Put $x=\alpha$ and $y=\frac{-13}{11}$, we get\ $\alpha^{2}=\left(\frac{-13}{11}-1ight)\left(\frac{-13}{11}-2ight)$ $$ \begin{aligned} & =\left(\frac{-13-11}{11}ight)\left(\frac{-13-22}{11}ight) \ & =\left(\frac{-24}{11}ight)\left(\frac{-35}{11}ight)=\frac{24 	imes 35}{121} \end{aligned} $$ $	herefore \quad 242 \alpha^{2}=\frac{24 	imes 35 	imes 242}{121}=1680$\