Complex Numbers
High-Power Symmetric Functions of Complex Roots
nta_pyq_2025_apr
Grade 11
Question:
If $\alpha$ and $\beta$ are the roots of $2z^2-3z-2i=0$, where $i=\sqrt{-1}$, then $16\cdot\operatorname{Re}\!\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\cdot\operatorname{Im}\!\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)$ is equal to
Step-by-Step Solution
Key Concept: Use $\alpha\beta=-i$ so $(\alpha\beta)^8=1$; factor out $\alpha^{11}+\beta^{11}$ from the numerator as $(\alpha^8+1)\alpha^{11}+(\beta^8+1)\beta^{11}$ divided by... simplify to $\alpha^4+\beta^4+\text{Re/Im}$ using $\alpha^8\beta^8=(\alpha\beta)^8=1$.
By Vieta's: $\alpha+\beta=\dfrac{3}{2}$, $\alpha\beta=-i$.
Since $(\alpha\beta)^8=(-i)^8=1$, we have $\beta^8=\dfrac{1}{\alpha^8}$. Let $S_n=\alpha^n+\beta^n$.
$\dfrac{S_{19}+S_{11}}{S_{15}} = \dfrac{S_{15}\cdot(\alpha^4+\beta^4) - \alpha^{15}\beta^4-\beta^{15}\alpha^4 + S_{11}}{S_{15}}$. Because $\alpha\beta=-i$ and $(\alpha\beta)^4=1$, one finds $\alpha^{15}\beta^4+\beta^{15}\alpha^4=(\alpha\beta)^4(\alpha^{11}+\beta^{11})=S_{11}$. So the ratio $= \alpha^4+\beta^4$.
$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\dfrac{9}{4}+2i$.
$\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2=\left(\dfrac{9}{4}+2i\right)^2-2(-1)=\dfrac{81}{16}+9i-4+2=\dfrac{49}{16}+9i$.
$\operatorname{Re}=\dfrac{49}{16}$, $\operatorname{Im}=9$. $16\times\dfrac{49}{16}\times9 = 49\times9 = 441$.
Correct Answer: 1