QJEEM25Ma9295: JEE Main 2025 Mathematics - Class 11 JEEM25Ma9295 Solution

$\begin{aligned}& \text { Sol. } 2 z^2-32-2 i=0 \\& 2\left(z-\frac{i}{z}\right)=3 \\& \alpha-\frac{i}{\alpha}=\frac{3}{2} \\& \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\& \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\& \Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2} \\& \Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2 \\& \Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4} \\& \text { Similarly } \\& \Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4}\end{aligned}$

$\begin{aligned}& \Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \\& =\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 \mathrm{i}\right)}{\left(\alpha^{15}+\beta^{15}\right)} \\& \text { Real }=\frac{49}{16} \\& \text { Im }=9 \\& \text { Ans. } 441\end{aligned}$