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

Question

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

Accepted Answer

$$\begin{aligned}
& \text { Sol. } 2 z^2-32-2 i=0 \
& 2\left(z-\frac{i}{z}\right)=3 \
& \alpha-\frac{i}{\alpha}=\frac{3}{2} \
& \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \
& \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \
& \Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2} \
& \Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2 \
& \Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4} \
& \text { Similarly } \
& \Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4}
\end{aligned}$$

$$\begin{aligned}
& \Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \
& =\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 \mathrm{i}\right)}{\left(\alpha^{15}+\beta^{15}\right)} \
& \text { Real }=\frac{49}{16} \
& \text { Im }=9 \
& \text { Ans. } 441
\end{aligned}$$

Solutions · 1