Q. If $\alpha$ and $\beta$ are the roots of the equation $2z^2-3z-2i=0$, where $i=\sqrt{-1}$, then $16\cdot\operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\cdot\operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)$ is equal to
A) $441$
B) $398$
C) $312$
D) $409$
Explanation
[Solution by Dr. Abubacker] With $\alpha+\beta=\frac32$ and $\alpha\beta=-i$, the power sums $s_k=\alpha^k+\beta^k$ satisfy $s_k=\frac32s_{k-1}+i\,s_{k-2}$. Evaluating $s_{11},s_{15},s_{19}$ by this recurrence gives $\frac{s_{19}+s_{11}}{s_{15}}=\frac{49}{16}+9i$. Hence $16\cdot\frac{49}{16}\cdot9=441$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.