If α and β are the roots of the equation 2 z^2-3 z-2 i=0, where i=√-1, then 16 · Re ( α^19+β^19+α^11+β^11 α^15+β^15 ) · lm ( α^19+β^19+α^11+β^11…

$2 z^2-3 z-2 i=0$2z2−3z−2i=0 ...(i) $2\left(z-\frac{i}{z}\right)=3$2(z−zi​)=3 As $\alpha, \beta$α,β are roots of (i) $\begin{aligned}
& \alpha-\frac{i}{\alpha}=\frac{3}{2} \
& \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \
& \Rightarrow \alpha^2-\frac{1}{\alpha^2}=\frac{9}{4}+2 i
\end{aligned}$Squaring both sides$Squaringbothsides\begin{aligned}
& \Rightarrow \alpha^4+\frac{1}{\alpha^4}-2=\frac{81}{16}-4+9 i \
& \Rightarrow \alpha^4+\frac{1}{\alpha^4}=\frac{49}{16}+9 i
\end{aligned}$Similarly,$Similarly,\beta^4+\frac{1}{\beta^4}=\frac{49}{16}+9 i$$\begin{aligned}
& \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{49}{16}+9 i \
& \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=\frac{49}{16} \
& \operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=9 \
& \Rightarrow 16 \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) \
& =16 \times \frac{49}{16} \times 9 \
& =441
\end{aligned}$