If $\alpha$ and $\beta$ are the roots of the equation $2 z^2-3 z-2 i=0$, where $i=\sqrt{-1}$, then $16 \cdot \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)$ is equal to

<p>$$\begin{aligned} & \text { Sol. } 2 z^2-32-2 i=0 \\ & 2\left(z-\frac{i}{z}\right)=3 \\ & \alpha-\frac{i}{\alpha}=\frac{3}{2} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2} \\ & \Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2 \\ & \Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4} \\ & \text { Similarly } \\ & \Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4} \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \\ & =\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 \mathrm{i}\right)}{\left(\alpha^{15}+\beta^{15}\right)} \\ & \text { Real }=\frac{49}{16} \\ & \text { Im }=9 \\ & \text { Ans. } 441 \end{aligned}$$</p>