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

Step 1: Understanding the Question
We need to evaluate the complex number \((\sqrt{3} + i)^{100}\), find the values of \(p\) and \(q\), and then form a quadratic equation whose roots are \(p\) and \(q\).
Step 2: Key Formula or Approach
We will use De Moivre's Theorem, which states that \( (r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta)) \). To apply this, we first convert \(\sqrt{3} + i\) into its polar form.
Step 3: Detailed Explanation
Convert to Polar Form:
For \(z = \sqrt{3} + i\):
Modulus, \(r = |\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2\).
Argument, \(\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
So, \(\sqrt{3} + i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\).
Apply De Moivre's Theorem:
\[ (\sqrt{3} + i)^{100} = \left[2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\right]^{100} = 2^{100}\left(\cos\frac{100\pi}{6} + i\sin\frac{100\pi}{6}\right) \] \[ = 2^{100}\left(\cos\frac{50\pi}{3} + i\sin\frac{50\pi}{3}\right) \] We simplify the angle: \(\frac{50\pi}{3} = \frac{48\pi + 2\pi}{3} = 16\pi + \frac{2\pi}{3}\). Since \(\cos(2k\pi + \alpha) = \cos\alpha\) and \(\sin(2k\pi + \alpha) = \sin\alpha\), this is equivalent to \(\frac{2\pi}{3}\).
\[ (\sqrt{3} + i)^{100} = 2^{100}\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2^{100}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \] \[ = 2^{99}(-1 + i\sqrt{3}) \] Find p and q:
We are given \((\sqrt{3} + i)^{100} = 2^{99}(p + iq)\).
Comparing our result, \(2^{99}(-1 + i\sqrt{3})\), with the given expression, we get:
\(p = -1\) and \(q = \sqrt{3}\).
Form the Quadratic Equation:
The equation with roots \(p\) and \(q\) is given by \(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\).
Sum of roots = \(p+q = -1 + \sqrt{3} = \sqrt{3}-1\).
Product of roots = \(pq = (-1)(\sqrt{3}) = -\sqrt{3}\).
The equation is: \(x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0\).
Step 4: Final Answer
The required equation is \(x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0\).