Least positive argument of the `4^(th)` root of the complex number `2-isqrt12` is :

Options

A. `pi/6`
B. `pi/12`
C. `(5pi)/12`
D. `(7pi)/12`

Answer

To find the least positive argument of the fourth root of the complex number \( z = 2 - i\sqrt{12} \), we will follow these steps:

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

1. Calculate the modulus \( r \):
\[ r = |z| = \sqrt{(2)^2 + (-\sqrt{12})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \]

2. Calculate the argument \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{-\sqrt{12}}{2}\right) = \tan^{-1}\left(-\sqrt{3}\right) \]
Since the point \( (2, -\sqrt{12}) \) is in the fourth quadrant, we have:
\[ \theta = 2\pi - \frac{\pi}{3} = \frac{6\pi}{3} - \frac{\pi}{3} = \frac{5\pi}{3} \]

### Step 2: Find the fourth root of the complex number
The fourth root of a complex number in polar form is given by:
\[ z^{1/4} = r^{1/4} \left( \cos\left(\frac{\theta + 2k\pi}{4}\right) + i \sin\left(\frac{\theta + 2k\pi}{4}\right) \right) \]
where \( k = 0, 1, 2, 3 \).

1. Calculate \( r^{1/4} \):
\[ r^{1/4} = 4^{1/4} = 2^{1/2} = \sqrt{2} \]

2. Calculate the arguments for \( k = 0, 1, 2, 3 \):
\[ \text{For } k = 0: \frac{5\pi/3 + 0}{4} = \frac{5\pi}{12} \]
\[ \text{For } k = 1: \frac{5\pi/3 + 2\pi}{4} = \frac{5\pi + 6\pi}{12} = \frac{11\pi}{12} \]
\[ \text{For } k = 2: \frac{5\pi/3 + 4\pi}{4} = \frac{5\pi + 12\pi}{12} = \frac{17\pi}{12} \]
\[ \text{For } k = 3: \frac{5\pi/3 + 6\pi}{4} = \frac{5\pi + 18\pi}{12} = \frac{23\pi}{12} \]

### Step 3: Identify the least positive argument
Now we need to find the least positive argument among \( \frac{5\pi}{12}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{23\pi}{12} \).

- \( \frac{5\pi}{12} \) is approximately \( 1.308 \) (in radians)
- \( \frac{11\pi}{12} \) is approximately \( 2.879 \) (in radians)
- \( \frac{17\pi}{12} \) is approximately \( 4.49 \) (in radians)
- \( \frac{23\pi}{12} \) is approximately \( 5.93 \) (in radians)

The least positive argument is:
\[ \frac{5\pi}{12} \]

### Final Answer
The least positive argument of the fourth root of the complex number \( 2 - i\sqrt{12} \) is:
\[ \frac{5\pi}{12} \]