Complex Numbers
Sequences involving complex numbers
Grade 11
Question:
<p>Let the set of complex numbers \((a_1, b_1), (a_2, b_2), (a_3, b_3)\ldots\) 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: Recognize the recurrence relation as a complex number multiplication z_{n+1} = (√3 + i)·z_n, then use the property that z_100 = (√3 + i)^99·z_1 to work backwards. Convert √3 + i to polar form 2e^(iπ/6) to simplify the 99th power calculation.
<p><strong>Step 1:</strong> Convert the recurrence to complex form. Let z_n = a_n + ib_n. The relation (a_{n+1}, b_{n+1}) = (√3·a_n - b_n, √3·b_n + a_n) becomes:</p><p>z_{n+1} = (√3 + i)z_n</p><p><strong>Step 2:</strong> Express z_100 in terms of z_1:</p><p>z_100 = (√3 + i)^99·z_1</p><p><strong>Step 3:</strong> Convert √3 + i to polar form. |√3 + i| = √(3 + 1) = 2, and arg(√3 + i) = π/6, so √3 + i = 2e^(iπ/6)</p><p><strong>Step 4:</strong> Calculate (√3 + i)^99 = 2^99·e^(i·99π/6) = 2^99·e^(i·33π/2) = 2^99·e^(iπ/2) = 2^99·i</p><p><strong>Step 5:</strong> Given z_100 = 2 + 4i, we have:</p><p>2 + 4i = 2^99·i·z_1</p><p>z_1 = (2 + 4i)/(2^99·i) = (2 + 4i)·(-i)/2^99 = (-2i + 4)/2^99 = (4 - 2i)/2^99 = 2(2 - i)/2^99 = (2 - i)/2^98</p><p><strong>Step 6:</strong> Therefore a_1 + b_1 = Re(z_1) + Im(z_1) = 2/2^98 + (-1)/2^98 = 1/2^98</p><p>∴ Answer: C</p>
Correct Answer: C