If `z_(1)` and `z_(2)` are two `n^(th)` roots of unity, then arg `(z_(1)/z_(2))` is a multiple of

Options

A. `npi`
B. `(3pi)/n`
C. `(2pi)/(n)`
D. none of these

Explanation

To solve the problem, we need to find the argument of the quotient of two \( n^{th} \) roots of unity, \( z_1 \) and \( z_2 \).

### Step-by-Step Solution:

1. **Definition of \( n^{th} \) Roots of Unity**:
The \( n^{th} \) roots of unity are given by the formula:
\[ z_k = e^{i \frac{2\pi k}{n}} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
where \( z_1 \) and \( z_2 \) can be represented as:
\[ z_1 = e^{i \frac{2\pi r}{n}} \quad \text{and} \quad z_2 = e^{i \frac{2\pi s}{n}} \]
Here, \( r \) and \( s \) are integers between \( 0 \) and \( n-1 \).

2. **Finding the Argument of \( \frac{z_1}{z_2} \)**:
The argument of the quotient of two complex numbers is given by:
\[ \text{arg}\left(\frac{z_1}{z_2}\right) = \text{arg}(z_1) - \text{arg}(z_2) \]
Substituting the values of \( z_1 \) and \( z_2 \):
\[ \text{arg}\left(\frac{z_1}{z_2}\right) = \text{arg}\left(e^{i \frac{2\pi r}{n}}\right) - \text{arg}\left(e^{i \frac{2\pi s}{n}}\right) \]

3. **Calculating the Arguments**:
Since the argument of \( e^{i\theta} \) is simply \( \theta \):
\[ \text{arg}(z_1) = \frac{2\pi r}{n} \quad \text{and} \quad \text{arg}(z_2) = \frac{2\pi s}{n} \]
Therefore:
\[ \text{arg}\left(\frac{z_1}{z_2}\right) = \frac{2\pi r}{n} - \frac{2\pi s}{n} \]

4. **Simplifying the Expression**:
This can be simplified to:
\[ \text{arg}\left(\frac{z_1}{z_2}\right) = \frac{2\pi (r - s)}{n} \]

5. **Conclusion**:
Since \( r \) and \( s \) are integers, \( r - s \) is also an integer. Thus, we can conclude that:
\[ \text{arg}\left(\frac{z_1}{z_2}\right) \text{ is a multiple of } \frac{2\pi}{n} \]

### Final Answer:
The argument \( \text{arg}\left(\frac{z_1}{z_2}\right) \) is a multiple of \( \frac{2\pi}{n} \).

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