If and are the roots of the equation 2z^2-3z-2i=0, where i=-1, then 16 Re( ^19+ ^19+ ^11+ …

Question

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

Accepted Answer

$\begin{gathered} 2{z}^{2}-32-2i=0 \ 2(z-\frac{i}{z})=3 \ \alpha -\frac{i}{\alpha }=\frac{3}{2} \ \Rightarrow {\alpha }^{2}-\frac{1}{{\alpha }^{2}}-2i=\frac{9}{4} \ \Rightarrow {\alpha }^{2}-\frac{1}{{\alpha }^{2}}-2i=\frac{9}{4} \ \Rightarrow \frac{9}{4}+2i={\alpha }^{2}-\frac{1}{{\alpha }^{2}} \ \Rightarrow \frac{81}{16}-4+9i={\alpha }^{4}+\frac{1}{{\alpha }^{4}}-2 \ \Rightarrow \frac{49}{16}+9i={\alpha }^{4}+\frac{1}{{\alpha }^{4}} \ \Rightarrow \frac{49}{16}+9i={\beta }^{4}+\frac{1}{{\beta }^{4}} \end{gathered}$

$\begin{gathered} \Rightarrow \frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}=\frac{{\alpha }^{15}({\alpha }^{4}+\frac{1}{{\alpha }^{4}})+{\beta }^{15}({\beta }^{4}+\frac{1}{{\beta }^{4}})}{{\alpha }^{15}+{\beta }^{15}} \ =\frac{({\alpha }^{15}+{\beta }^{15})(\frac{49}{16}+9i)}{({\alpha }^{15}+{\beta }^{15})} \ =\frac{49}{16} \ =9 \ 441 \end{gathered}$

Solutions · 1