Question

If \((\sqrt{3} + i)^{100} = 2^{99}(p + iq)\), then p and q are roots of the equation :

• A. \(x^2 + (\sqrt{3}-1)x - \sqrt{3} = 0\)
• B. \(x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0\)
• C. \(x^2 - (\sqrt{3}+1)x + \sqrt{3} = 0\)
• D. \(x^2 + (\sqrt{3}+1)x + \sqrt{3} = 0\)

Hint:

When dealing with high powers of complex numbers, converting to polar or exponential form (\(re^{i\theta}\)) is almost always the most efficient method. Remember De Moivre's theorem and how to find the principal argument of a complex number.

Answer 1

The Correct Option is B

To solve the problem, we need to find the roots, \( p \) and \( q \), which satisfy the given equation. The equation involves manipulating complex numbers and using De Moivre's theorem.

1. The given expression is \((\sqrt{3} + i)^{100} = 2^{99}(p + iq)\).
2. The complex number \( \sqrt{3} + i \) can be expressed in polar form. The modulus of \( \sqrt{3} + i \) is: |\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2.
3. The argument \(\theta\) of the complex number is calculated as: \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\), which results in \(\theta = \frac{\pi}{6}.
4. In polar form, this can be written as: 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right).
5. Using De Moivre’s Theorem, \((\sqrt{3} + i)^{100}\) becomes: 2^{100} \left(\cos \frac{100\pi}{6} + i\sin \frac{100\pi}{6}\right).
6. Simplifying the angle: \(\frac{100\pi}{6} = \frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}\). Thus, the angle simplifies to \(\frac{2\pi}{3}\) due to the \(2\pi\) periodicity.
7. Applying this to the polar form gives: 2^{100} \left(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right) = 2^{100} \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right).
8. Replacing in the original equation gives: 2^{99}(p + iq) = 2^{100} \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right).
9. Divide both sides by \(2^{99}\) to isolate \((p + iq)\): p + iq = 2\left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}.
10. This gives us \(p = -1\) and \(q = \sqrt{3}\).
11. Now, we need to verify which polynomial \(p\) and \(q\) satisfy:
    • Substitute \(p = -1\) and \(q = \sqrt{3}\) in the polynomial \(x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0\):
    • Verify the roots: (-1)^2 - (\sqrt{3}-1)(-1) - \sqrt{3} = 1 + \sqrt{3} - 1 - \sqrt{3} = 0.
12. Therefore, \(p\) and \(q\) are the roots of \((x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0)\).

Thus, the correct answer is \(x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0\).