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

$2{\mathrm{z}}^2-32-2\mathrm{i}=0$
$\begin{array}{l}2\left(\mathrm{z}-\frac{\mathrm{i}}{\mathrm{z}}\right)=3\\ \mathrm{\alpha }-\frac{\mathrm{i}}{\mathrm{\alpha }}=\frac{3}{2}\\ \Rightarrow {\mathrm{\alpha }}^2-\frac{1}{{\mathrm{\alpha }}^2}-2\mathrm{i}=\frac{9}{4}\\ \Rightarrow {\mathrm{\alpha }}^2-\frac{1}{{\mathrm{\alpha }}^2}-2\mathrm{i}=\frac{9}{4}\\ \Rightarrow \frac{9}{4}+2\mathrm{i}={\mathrm{\alpha }}^2-\frac{1}{{\mathrm{\alpha }}^2}\\ \Rightarrow \frac{81}{16}-4+9\mathrm{i}={\mathrm{\alpha }}^4+\frac{1}{{\mathrm{\alpha }}^4}-2\end{array}$
$\Rightarrow \frac{49}{16}+9\mathrm{i}={\mathrm{\alpha }}^4+\frac{1}{{\mathrm{\alpha }}^4}$
Similarly
$\begin{array}{l}\Rightarrow \frac{49}{16}+9\mathrm{i}={\mathrm{\beta }}^4+\frac{1}{{\mathrm{\beta }}^4}\\ \Rightarrow \frac{{\mathrm{\alpha }}^{19}+{\mathrm{\beta }}^{19}+{\mathrm{\alpha }}^{11}+{\mathrm{\beta }}^{11}}{{\mathrm{\alpha }}^{15}+{\mathrm{\beta }}^{15}}=\frac{{\mathrm{\alpha }}^{15}\left({\mathrm{\alpha }}^4+\frac{1}{{\mathrm{\alpha }}^4}\right)+{\mathrm{\beta }}^{15}\left({\mathrm{\beta }}^4+\frac{1}{{\mathrm{\beta }}^4}\right)}{{\mathrm{\alpha }}^{15}+{\mathrm{\beta }}^{15}}\\ =\frac{\left({\mathrm{\alpha }}^{15}+{\mathrm{\beta }}^{15}\right)\left(\frac{49}{16}+9\mathrm{i}\right)}{\left({\mathrm{\alpha }}^{15}+{\mathrm{\beta }}^{15}\right)}\\ \text{ Real }=\frac{49}{16}\\ \text{ Im }=9\end{array}$