Complex Numbers
Recursion with complex numbers
Grade 11
Question:
<p>Let the set of complex numbers \((a_1, b_1), (a_2, b_2), (a_3, b_3)\)........ denoting the points on the complex plane satisfying \((a_{n+1}, b_{n+1}) = (\sqrt{3}\, a_n - b_n,\, \sqrt{3}\, b_n + a_n)\) for \(n = 1, 2, 3, \ldots\) Suppose \((a_{100}, b_{100}) = (2, 4)\), then the value of \((a_1 + b_1)\) is equal to:</p>
<p>(a) \(\dfrac{1}{2^{96}}\)</p>
<p>(b) \(\dfrac{1}{2^{97}}\)</p>
<p>(c) \(\dfrac{1}{2^{98}}\)</p>
<p>(d) \(\dfrac{1}{2^{99}}\)</p>
Step-by-Step Solution
Key Concept: Interpret the transformation as complex number multiplication in the form z_{n+1} = (√3 + i)·z_n, then recognize this as rotation and scaling. Use the property that z_{100} = (√3 + i)^99 · z_1 to find z_1.
<p><strong>Step 1:</strong> Convert the recurrence to complex number form. Let z_n = a_n + ib_n. Then:</p><p>a_{n+1} + ib_{n+1} = (√3·a_n - b_n) + i(√3·b_n + a_n) = (√3 + i)(a_n + ib_n)</p><p>Thus: <strong>z_{n+1} = (√3 + i)·z_n</strong></p><p><strong>Step 2:</strong> Express (√3 + i) in polar form. |√3 + i| = √(3 + 1) = 2, and arg(√3 + i) = π/6</p><p>So: √3 + i = 2e^{iπ/6} = 2(cos π/6 + i sin π/6)</p><p><strong>Step 3:</strong> Apply the recurrence 99 times:</p><p>z_{100} = (√3 + i)^{99}·z_1 = 2^{99}·e^{i·99π/6}·z_1</p><p><strong>Step 4:</strong> Simplify the angle: 99π/6 = 16π + 3π/6 = 16π + π/2 ≡ π/2 (mod 2π)</p><p>Therefore: z_{100} = 2^{99}·e^{iπ/2}·z_1 = 2^{99}·i·z_1</p><p><strong>Step 5:</strong> Given z_{100} = 2 + 4i, we have:</p><p>2^{99}·i·z_1 = 2 + 4i</p><p>z_1 = (2 + 4i)/(2^{99}·i) = (2 + 4i)·(-i)/(2^{99}) = (-2i + 4)/(2^{99}) = (4 - 2i)/(2^{99})</p><p><strong>Step 6:</strong> Therefore:</p><p>a_1 + b_1 = 4/(2^{99}) + (-2)/(2^{99}) = 2/(2^{99}) = <strong>1/(2^{98})</strong></p><p>∴ Answer: C</p>
Correct Answer: C