Complex Numbers
Complex Number
nta_pyq_2025_jan
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: Factor the numerator: $\alpha^{19}+\alpha^{11}=\alpha^{15}\!\left(\alpha^{4}+\dfrac{1}{\alpha^{4}}\right)$ and similarly for $\beta$. The ratio collapses because $\alpha^{4}+\dfrac{1}{\alpha^{4}}=\beta^{4}+\dfrac{1}{\beta^{4}}$ (both equal a single complex constant).
From $2\alpha^{2}-3\alpha-2i=0$ divide by $2\alpha$:
$$\alpha-\frac{i}{\alpha}=\frac{3}{2}.$$
Square: $\alpha^{2}-2i\cdot\dfrac{\alpha}{\alpha}+\dfrac{i^{2}}{\alpha^{2}}=\dfrac{9}{4}$, i.e.
$$\alpha^{2}-\frac{1}{\alpha^{2}}=\frac{9}{4}+2i.$$
Square again:
$$\alpha^{4}-2+\frac{1}{\alpha^{4}}=\left(\frac{9}{4}+2i\right)^{2}=\frac{81}{16}+9i-4=\frac{17}{16}+9i,$$
$$\alpha^{4}+\frac{1}{\alpha^{4}}=\frac{17}{16}+9i+2 = \frac{49}{16}+9i.$$
By symmetry the same value holds for $\beta$:
$$\beta^{4}+\frac{1}{\beta^{4}}=\frac{49}{16}+9i.$$
Now
$$\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}} = \frac{\alpha^{15}\!\left(\alpha^{4}+\frac{1}{\alpha^{4}}\right)+\beta^{15}\!\left(\beta^{4}+\frac{1}{\beta^{4}}\right)}{\alpha^{15}+\beta^{15}} = \frac{49}{16}+9i.$$
Therefore
$$16\cdot \operatorname{Re}\cdot\operatorname{Im} = 16\cdot\frac{49}{16}\cdot 9 = 49\cdot 9 = 441.$$
Correct Answer: 1