What is the value of `[(i+sqrt(3))/(2)]^(2019)+[(i-sqrt(3))/(2)]^(2019)=`

Options

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

Answer

To solve the expression \(\left(\frac{i + \sqrt{3}}{2}\right)^{2019} + \left(\frac{i - \sqrt{3}}{2}\right)^{2019}\), we can follow these steps:

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

### Step 2: Express \(z_1\) and \(z_2\) in polar form
We can rewrite \(z_1\) and \(z_2\) in polar form. First, we find the modulus and argument of \(z_1\):
\[ |z_1| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]
The argument \(\theta_1\) can be calculated as:
\[ \theta_1 = \tan^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
Thus, we can express \(z_1\) as:
\[ z_1 = e^{i\frac{\pi}{6}} \]

Now for \(z_2\):
\[ |z_2| = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{1} = 1 \]
The argument \(\theta_2\) is:
\[ \theta_2 = \tan^{-1}\left(\frac{1/2}{-\sqrt{3}/2}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \]
Thus, we can express \(z_2\) as:
\[ z_2 = e^{-i\frac{\pi}{6}} \]

### Step 3: Raise \(z_1\) and \(z_2\) to the power of 2019
Now we calculate:
\[ z_1^{2019} = \left(e^{i\frac{\pi}{6}}\right)^{2019} = e^{i\frac{2019\pi}{6}} = e^{i(336\pi + \frac{3\pi}{6})} = e^{i(336\pi + \frac{\pi}{2})} = e^{i\frac{\pi}{2}} = i \]
\[ z_2^{2019} = \left(e^{-i\frac{\pi}{6}}\right)^{2019} = e^{-i\frac{2019\pi}{6}} = e^{-i(336\pi + \frac{3\pi}{6})} = e^{-i(336\pi + \frac{\pi}{2})} = e^{-i\frac{\pi}{2}} = -i \]

### Step 4: Combine the results
Now we add \(z_1^{2019}\) and \(z_2^{2019}\):
\[ z_1^{2019} + z_2^{2019} = i + (-i) = 0 \]

### Final Answer
Thus, the value of \(\left(\frac{i + \sqrt{3}}{2}\right)^{2019} + \left(\frac{i - \sqrt{3}}{2}\right)^{2019}\) is:
\[ \boxed{0} \]