Question

\( (i + \sqrt{3})^{100} + (i - \sqrt{3})^{100} + 2^{100} \) is equal to:

• A. 0
• B. 1
• C. -1
• D. $2^{101}$

Hint:

Use symmetry in conjugate powers and simplify via cancellation or De Moivre's theorem.

Answer 1

The Correct Option is A

The expression given is \( (i + \sqrt{3})^{100} + (i - \sqrt{3})^{100} + 2^{100} \), and we need to find its value.

Step 1: Analyze the powers of complex numbers

Let \( z_1 = i + \sqrt{3} \) and \( z_2 = i - \sqrt{3} \). Notice that these are conjugates of each other, and the sum of their powers will yield a real number since the imaginary parts cancel each other out when added. Let’s express both \( z_1^{100} \) and \( z_2^{100} \) in polar form.

The modulus of \( z_1 \) and \( z_2 \) is:

\[ |z_1| = |z_2| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2. \] The argument of \( z_1 = i + \sqrt{3} \) is: \[ \theta_1 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}. \] The argument of \( z_2 = i - \sqrt{3} \) is: \[ \theta_2 = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}. \] So, we can write the polar forms of \( z_1 \) and \( z_2 \) as: \[ z_1 = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right), \] and \[ z_2 = 2 \left( \cos \left(-\frac{\pi}{6}\right) + i \sin \left(-\frac{\pi}{6}\right) \right). \] Using De Moivre's Theorem, we raise these to the power of 100: \[ z_1^{100} = 2^{100} \left( \cos \frac{100\pi}{6} + i \sin \frac{100\pi}{6} \right), \] and \[ z_2^{100} = 2^{100} \left( \cos \frac{-100\pi}{6} + i \sin \frac{-100\pi}{6} \right). \] Since \( \cos \frac{100\pi}{6} = \cos \frac{-100\pi}{6} \) and \( \sin \frac{100\pi}{6} = -\sin \frac{-100\pi}{6} \), adding these two expressions cancels out the imaginary parts, leaving only the real parts: \[ z_1^{100} + z_2^{100} = 2^{100} \left( 2 \cos \frac{100\pi}{6} \right). \] Now, simplifying the angle: \[ \frac{100\pi}{6} = \frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}. \] So, \[ \cos \frac{50\pi}{3} = \cos \frac{2\pi}{3} = -\frac{1}{2}. \] Thus, \[ z_1^{100} + z_2^{100} = 2^{100} \left( 2 \times -\frac{1}{2} \right) = -2^{100}. \]

Step 2: Add \( 2^{100} \)

Now, the full expression is: \[ (i + \sqrt{3})^{100} + (i - \sqrt{3})^{100} + 2^{100} = -2^{100} + 2^{100}. \] Finally, the sum is: \[ 0. \]

Final Answer:

The value of \( (i + \sqrt{3})^{100} + (i - \sqrt{3})^{100} + 2^{100} \) is \( \boxed{0} \).