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 4th 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 \( z = 2 - i\sqrt{12} \) can be expressed in polar form as \( z = r(\cos \theta + i \sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument.

#### Calculation of modulus \( r \):
\[ r = |z| = \sqrt{a^2 + b^2} = \sqrt{2^2 + (-\sqrt{12})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \]

### Step 2: Calculate the argument \( \theta \)
The argument \( \theta \) can be calculated using:
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{-\sqrt{12}}{2}\right) = \tan^{-1}\left(-\sqrt{3}\right) \]
Since \( \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \), we have:
\[ \theta = -\frac{\pi}{3} \]

### Step 3: Express \( z \) in polar form
Now we can write:
\[ z = 4 \left( \cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right) \right) = 4 e^{-i\frac{\pi}{3}} \]

### Step 4: Find the 4th root of \( z \)
To find the 4th root of \( z \), we use the formula:
\[ z^{1/4} = r^{1/4} e^{i\frac{\theta + 2k\pi}{4}} \quad \text{for } k = 0, 1, 2, 3 \]
Substituting the values:
\[ r^{1/4} = 4^{1/4} = 2 \]
Now, substituting \( \theta \):
\[ z^{1/4} = 2 e^{i\frac{-\frac{\pi}{3} + 2k\pi}{4}} = 2 e^{i\left(-\frac{\pi}{12} + \frac{k\pi}{2}\right)} \]

### Step 5: Calculate the arguments for \( k = 0, 1, 2, 3 \)
1. For \( k = 0 \):
\[ \theta_0 = -\frac{\pi}{12} \]
2. For \( k = 1 \):
\[ \theta_1 = -\frac{\pi}{12} + \frac{\pi}{2} = -\frac{\pi}{12} + \frac{6\pi}{12} = \frac{5\pi}{12} \]
3. For \( k = 2 \):
\[ \theta_2 = -\frac{\pi}{12} + \pi = -\frac{\pi}{12} + \frac{12\pi}{12} = \frac{11\pi}{12} \]
4. For \( k = 3 \):
\[ \theta_3 = -\frac{\pi}{12} + \frac{3\pi}{2} = -\frac{\pi}{12} + \frac{18\pi}{12} = \frac{17\pi}{12} \]

### Step 6: Identify the least positive argument
The arguments we calculated are:
- \( -\frac{\pi}{12} \) (not positive)
- \( \frac{5\pi}{12} \) (positive)
- \( \frac{11\pi}{12} \) (positive)
- \( \frac{17\pi}{12} \) (positive)

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

### Final Answer
The least positive argument of the 4th root of the complex number \( 2 - i\sqrt{12} \) is \( \frac{5\pi}{12} \).

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