`( i + sqrt(3) )^(100) + (i - sqrt(3))^(100) + 2 ^(100) = `

Options

A. 1
B. `-1`
C. 0
D. none

Answer

To solve the expression \(( i + \sqrt{3} )^{100} + ( i - \sqrt{3} )^{100} + 2^{100}\), we can use the polar form of complex numbers and De Moivre's theorem.

### Step-by-Step Solution:

1. **Convert to Polar Form**:
- For \( z_1 = i + \sqrt{3} \):
- The modulus is \( |z_1| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2 \).
- The argument (angle) is \( \theta_1 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \).
- Thus, \( z_1 = 2 \left( \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} \right) \).

- For \( z_2 = i - \sqrt{3} \):
- The modulus is \( |z_2| = \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2 \).
- The argument (angle) is \( \theta_2 = \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) = \frac{5\pi}{6} \).
- Thus, \( z_2 = 2 \left( \cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6} \right) \).

2. **Apply De Moivre's Theorem**:
- Using De Moivre's theorem, we have:
\[ (z_1)^{100} = (2)^{100} \left( \cos\left(100 \cdot \frac{\pi}{6}\right) + i \sin\left(100 \cdot \frac{\pi}{6}\right) \right) \]
\[ (z_2)^{100} = (2)^{100} \left( \cos\left(100 \cdot \frac{5\pi}{6}\right) + i \sin\left(100 \cdot \frac{5\pi}{6}\right) \right) \]

3. **Calculate the Angles**:
- For \( z_1^{100} \):
\[ 100 \cdot \frac{\pi}{6} = \frac{100\pi}{6} = \frac{50\pi}{3} = 16\pi + \frac{2\pi}{3} \quad (\text{since } 16\pi \text{ is a multiple of } 2\pi) \]
Thus, \( \cos\left(\frac{50\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \) and \( \sin\left(\frac{50\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \).

- For \( z_2^{100} \):
\[ 100 \cdot \frac{5\pi}{6} = \frac{500\pi}{6} = \frac{250\pi}{3} = 83\pi + \frac{1\pi}{3} \quad (\text{since } 83\pi \text{ is a multiple of } 2\pi) \]
Thus, \( \cos\left(\frac{250\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{250\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).

4. **Combine the Results**:
- Now substituting back:
\[ (z_1)^{100} = 2^{100} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -2^{99} + i 2^{99} \sqrt{3} \]
\[ (z_2)^{100} = 2^{100} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 2^{99} + i 2^{99} \sqrt{3} \]

5. **Final Calculation**:
- Adding \( (z_1)^{100} + (z_2)^{100} + 2^{100} \):
\[ (-2^{99} + i 2^{99} \sqrt{3}) + (2^{99} + i 2^{99} \sqrt{3}) + 2^{100} \]
The real parts cancel out:
\[ 0 + i 2^{100} \sqrt{3} + 2^{100} = 2^{100} + i 2^{100} \sqrt{3} \]

Thus, the final answer is:
\[ 2^{100} + i 2^{100} \sqrt{3} \]