JEE Main 2025 — 24 January, Morning Shift — Question 6

$2 z^{2}-3z-2 i=0$

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

Similarly

$\Rightarrow \frac{49}{16}+9 i=\beta^{4}+\frac{1}{\beta^{4}}$

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

$16. \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot\operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=441$

Ans. $441$