Complex Numbers
Complex Number
nta_pyq_2025_jan
Grade 11
Question:
Let $\alpha,\beta$ be the roots of the equation $x^{2}-ax-b=0$ with $\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)$. Let $P_{n}=\alpha^{n}-\beta^{n}$. If $P_{3}=-5\sqrt{7}\,i$, $P_{4}=-3\sqrt{7}\,i$, $P_{5}=11\sqrt{7}\,i$ and $P_{6}=45\sqrt{7}\,i$, then $\bigl|\alpha^{4}+\beta^{4}\bigr|$ is equal to \rule{2cm}{0.4pt}\,.
Step-by-Step Solution
Key Concept: Newton's recurrence: if $\alpha,\beta$ are roots of $x^{2}-ax-b=0$, then $P_{n+1}=a\,P_{n}+b\,P_{n-1}$. Two consecutive applications determine $a$ and $b$.
From the recurrence $P_{n+1}=aP_{n}+bP_{n-1}$:
$$P_{6} = a P_{5}+b P_{4}\ \Longrightarrow\ 45=11a-3b,$$
$$P_{5} = a P_{4}+b P_{3}\ \Longrightarrow\ 11=-3a-5b.$$
Solve: $a=3,\ b=-4$, so $\alpha+\beta=a=3$ and $\alpha\beta=-b=4$.
Now
$$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta = 9-8 = 1,$$
$$\alpha^{4}+\beta^{4}=(\alpha^{2}+\beta^{2})^{2}-2(\alpha\beta)^{2}=1-2\cdot 16=-31.$$
Hence $|\alpha^{4}+\beta^{4}|=31$.
Correct Answer: 31