The fourth root of `1/2 +isqrt3/2` is

Options

A. `cis ""pi/(12)`
B. `"cis "pi/2`
C. `"cis "pi/6`
D. `"cis "pi/3`

Answer

To find the fourth root of the complex number \( \frac{1}{2} + i\frac{\sqrt{3}}{2} \), we can follow these steps:

### Step 1: Convert the complex number to polar form
The complex number can be expressed in polar form as \( r(\cos \theta + i \sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument.

1. **Calculate the modulus \( r \)**:
\[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

2. **Calculate the argument \( \theta \)**:
\[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

Thus, we can express the complex number in polar form:
\[ \frac{1}{2} + i\frac{\sqrt{3}}{2} = 1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = e^{i \frac{\pi}{3}} \]

### Step 2: Find the fourth root
To find the fourth root of \( z = e^{i \frac{\pi}{3}} \), we use the formula for the \( n \)-th roots of a complex number:
\[ z^{\frac{1}{n}} = r^{\frac{1}{n}} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right) \right) \]
where \( k = 0, 1, 2, \ldots, n-1 \).

For our case:
- \( n = 4 \)
- \( r = 1 \)
- \( \theta = \frac{\pi}{3} \)

Thus, the fourth roots are given by:
\[ z^{\frac{1}{4}} = 1^{\frac{1}{4}} \left( \cos\left(\frac{\frac{\pi}{3} + 2k\pi}{4}\right) + i \sin\left(\frac{\frac{\pi}{3} + 2k\pi}{4}\right) \right) \]

### Step 3: Calculate the roots for \( k = 0, 1, 2, 3 \)
1. For \( k = 0 \):
\[ z_0 = \cos\left(\frac{\frac{\pi}{3}}{4}\right) + i \sin\left(\frac{\frac{\pi}{3}}{4}\right) = \cos\left(\frac{\pi}{12}\right) + i \sin\left(\frac{\pi}{12}\right) \]

2. For \( k = 1 \):
\[ z_1 = \cos\left(\frac{\frac{\pi}{3} + 2\pi}{4}\right) + i \sin\left(\frac{\frac{\pi}{3} + 2\pi}{4}\right) = \cos\left(\frac{7\pi}{12}\right) + i \sin\left(\frac{7\pi}{12}\right) \]

3. For \( k = 2 \):
\[ z_2 = \cos\left(\frac{\frac{\pi}{3} + 4\pi}{4}\right) + i \sin\left(\frac{\frac{\pi}{3} + 4\pi}{4}\right) = \cos\left(\frac{13\pi}{12}\right) + i \sin\left(\frac{13\pi}{12}\right) \]

4. For \( k = 3 \):
\[ z_3 = \cos\left(\frac{\frac{\pi}{3} + 6\pi}{4}\right) + i \sin\left(\frac{\frac{\pi}{3} + 6\pi}{4}\right) = \cos\left(\frac{19\pi}{12}\right) + i \sin\left(\frac{19\pi}{12}\right) \]

### Final Answer
The fourth roots of \( \frac{1}{2} + i\frac{\sqrt{3}}{2} \) are:
- \( z_0 = \cos\left(\frac{\pi}{12}\right) + i \sin\left(\frac{\pi}{12}\right) \)
- \( z_1 = \cos\left(\frac{7\pi}{12}\right) + i \sin\left(\frac{7\pi}{12}\right) \)
- \( z_2 = \cos\left(\frac{13\pi}{12}\right) + i \sin\left(\frac{13\pi}{12}\right) \)
- \( z_3 = \cos\left(\frac{19\pi}{12}\right) + i \sin\left(\frac{19\pi}{12}\right) \)