Question

If (√3 + i)¹⁰⁰ = 2⁹⁹ (a + ib), then a² + b² is equal to:

• A. \( \sqrt{2} \)
• B. 4
• C. \( \sqrt{3} \)
• D. None of these

Hint:

Use De Moivre’s Theorem to compute high powers of complex numbers in polar form, and remember that \( a^2 + b^2 \) gives the modulus squared.

Answer 1

The Correct Option is B

We are given that \( (\sqrt{3} + i)^{100} = 2^{99} (a + ib) \). Step 1: Convert to polar form The complex number \( \sqrt{3} + i \) can be written in polar form as: \[ \sqrt{3} + i = 2 \left( \cos \theta + i \sin \theta \right) \] where \( \theta = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \). Step 2: Apply De Moivre’s Theorem Using De Moivre's Theorem: \[ (\sqrt{3} + i)^{100} = 2^{100} \left( \cos \frac{100\pi}{6} + i \sin \frac{100\pi}{6} \right) \] Simplifying: \[ \frac{100\pi}{6} = 16\pi + \frac{4\pi}{3} \] Thus, the angle is \( \frac{4\pi}{3} \). Step 3: Find \( a^2 + b^2 \) Since \( a + ib = 2^{99} \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) \), we know that \( a^2 + b^2 = 2^{198} \), which simplifies to 4. Thus, the correct answer is 4.