Complex Numbers
Recurrence for Power Differences of Complex Roots
nta_pyq_2025_apr
Grade 11
Question:
Let $\alpha$ and $\beta$ be the roots of $x^2-ax-b=0$ with $\text{Im}(\alpha)<\text{Im}(\beta)$. 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 $|\alpha^4+\beta^4|$ is equal to
Step-by-Step Solution
Key Concept: Use the recurrence $P_n = aP_{n-1}+bP_{n-2}$ (derived from the characteristic equation) to solve for $a$ and $b$, then compute $\alpha^4+\beta^4$ via $\alpha+\beta$ and $\alpha\beta$.
The recurrence $P_n = aP_{n-1}+bP_{n-2}$ gives two equations (dividing through by $\sqrt{7}\,i$):
$P_6=aP_5+bP_4:\;45=11a+(-3)b\cdot... $ (stripping $\sqrt{7}i$): $45=11a-3b$ and $P_5=aP_4+bP_3:\; 11 = -3a-5b$.
Solving: from $11=-3a-5b$ and $45=11a-3b$: multiply first by $11/3$ and add $\Rightarrow b=-4$, $a=3$.
So $\alpha+\beta=a=3$ and $\alpha\beta=-b=4$.
$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=9-8=1$.
$\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2=1-2(16)=-31$.
$|\alpha^4+\beta^4|=31$.
Correct Answer: 31