`(( - 1 + sqrt""(-3))/(2))^(100) + ((- 1 - sqrt""(-3))/( 2))^(100)` equals

Options

A. 2
B. 0
C. `-1`
D. 1

Answer

To solve the expression \(\left( \frac{-1 + \sqrt{-3}}{2} \right)^{100} + \left( \frac{-1 - \sqrt{-3}}{2} \right)^{100}\), we can follow these steps:

### Step 1: Identify the complex numbers
Let:
\[ z_1 = \frac{-1 + \sqrt{-3}}{2} \]
\[ z_2 = \frac{-1 - \sqrt{-3}}{2} \]

### Step 2: Express the complex numbers in polar form
We can rewrite \(z_1\) and \(z_2\) in terms of \(i\) (the imaginary unit):
\[ z_1 = \frac{-1 + i\sqrt{3}}{2}, \quad z_2 = \frac{-1 - i\sqrt{3}}{2} \]

### Step 3: Find the modulus and argument
The modulus of \(z_1\) and \(z_2\) is:
\[ |z_1| = |z_2| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

Next, we find the arguments:
\[ \text{arg}(z_1) = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \tan^{-1}(-\sqrt{3}) = \frac{2\pi}{3} \quad (\text{in the second quadrant}) \]
\[ \text{arg}(z_2) = \tan^{-1}\left(\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \quad (\text{in the third quadrant}) \]

### Step 4: Write in exponential form
Thus, we can express \(z_1\) and \(z_2\) as:
\[ z_1 = e^{i \frac{2\pi}{3}}, \quad z_2 = e^{i \frac{4\pi}{3}} \]

### Step 5: Raise to the power of 100
Now we compute:
\[ z_1^{100} = \left(e^{i \frac{2\pi}{3}}\right)^{100} = e^{i \frac{200\pi}{3}} = e^{i \left(66\pi + \frac{2\pi}{3}\right)} = e^{i \frac{2\pi}{3}} \quad (\text{since } e^{i 66\pi} = 1) \]
\[ z_2^{100} = \left(e^{i \frac{4\pi}{3}}\right)^{100} = e^{i \frac{400\pi}{3}} = e^{i \left(133\pi + \frac{1\pi}{3}\right)} = e^{i \frac{4\pi}{3}} \quad (\text{since } e^{i 133\pi} = -1) \]

### Step 6: Add the two results
Now we add:
\[ z_1^{100} + z_2^{100} = e^{i \frac{2\pi}{3}} + e^{i \frac{4\pi}{3}} \]

### Step 7: Use Euler's formula
Using Euler's formula:
\[ e^{i \frac{2\pi}{3}} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}, \quad e^{i \frac{4\pi}{3}} = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \]
Adding these gives:
\[ z_1^{100} + z_2^{100} = \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = -1 \]

### Final Result
Thus, the final answer is:
\[ \left( \frac{-1 + \sqrt{-3}}{2} \right)^{100} + \left( \frac{-1 - \sqrt{-3}}{2} \right)^{100} = -1 \]