If \(\alpha\) and \(\beta\) are the roots of the equation \(2{z}^{2}-3z-2i=0,\text{ where }i=\sqrt{-1}\) then \(\text{ 1…

\(\text{Given},2{z}^{2}-3z-2i=0.......\left(i\right)\\ \text{Divide equation by z, we get,}\\ 2\left(z-\frac{i}{z}\right)=3\\ \text{ As }\alpha ,\beta \text{ are roots of (i) }\\ \alpha -\frac{i}{\alpha }=\frac{3}{2}\\ \text{on squaring both sides, we get}\\ \Rightarrow {\alpha }^{2}-\frac{1}{{\alpha }^{2}}=\frac{9}{4}+2i\\ \text{ Again squaring both sides , we get}\\ \Rightarrow {\alpha }^{4}+\frac{1}{{\alpha }^{4}}=\frac{49}{16}+9i.........\left(ii\right)\\ \text{Similarly, for another root β}\\ {\beta }^{4}+\frac{1}{{\beta }^{4}}=\frac{49}{16}+9i...........\left(ii\right)\\ Now,\\ \frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}=\frac{{\alpha }^{19}+{\alpha }^{11}+{\beta }^{19}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\\ =\frac{{\alpha }^{15}\left({\alpha }^{4}+\frac{1}{{\alpha }^{4}}\right)+{\beta }^{15}\left({\beta }^{4}+\frac{1}{{\beta }^{4}}\right)}{{\alpha }^{15}+{\beta }^{15}}\\ =\frac{49}{16}+9i\\ Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)=\frac{49}{16}\\ lm\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)=9\\ \Rightarrow 16Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\cdot lm\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\\ =16\times \frac{49}{16}\times 9=441\)