Let `arg(z_(k))=((2k+1)pi)/(n)` where `k=1,2,………n`. If `arg(z_(1),z_(2),z_(3),………….z_(n))=pi`, then `n` must be of form `(m in z)`

Options

A. `4m`
B. `2m-1`
C. `2m`
D. None of these

Explanation

To solve the problem, we need to analyze the given expression for the arguments of the complex numbers \( z_k \) and find the conditions under which their product equals \( \pi \).

### Step-by-Step Solution:

1. **Understand the given argument**:
The argument of \( z_k \) is given by:
\[ \text{arg}(z_k) = \frac{(2k + 1) \pi}{n} \]
for \( k = 1, 2, \ldots, n \).

2. **Sum the arguments**:
The total argument of the product \( z_1 z_2 \cdots z_n \) is:
\[ \text{arg}(z_1 z_2 \cdots z_n) = \text{arg}(z_1) + \text{arg}(z_2) + \cdots + \text{arg}(z_n) \]
Substituting the expression for \( \text{arg}(z_k) \):
\[ \text{arg}(z_1 z_2 \cdots z_n) = \sum_{k=1}^{n} \frac{(2k + 1) \pi}{n} \]

3. **Calculate the sum**:
We can factor out \( \frac{\pi}{n} \):
\[ \text{arg}(z_1 z_2 \cdots z_n) = \frac{\pi}{n} \sum_{k=1}^{n} (2k + 1) \]
The sum \( \sum_{k=1}^{n} (2k + 1) \) can be simplified:
\[ \sum_{k=1}^{n} (2k + 1) = 2\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 = 2 \cdot \frac{n(n + 1)}{2} + n = n(n + 1) + n = n^2 + 2n \]

4. **Substituting back into the argument**:
Now substituting this back:
\[ \text{arg}(z_1 z_2 \cdots z_n) = \frac{\pi}{n} (n^2 + 2n) = \pi \left(1 + \frac{2}{n}\right) \]

5. **Setting the argument equal to \( \pi \)**:
According to the problem, we have:
\[ \text{arg}(z_1 z_2 \cdots z_n) = \pi \]
Therefore, we set:
\[ \pi \left(1 + \frac{2}{n}\right) = \pi \]

6. **Solving for \( n \)**:
Dividing both sides by \( \pi \):
\[ 1 + \frac{2}{n} = 1 \]
This simplifies to:
\[ \frac{2}{n} = 0 \]
which is not possible. Hence we need to consider the periodic nature of the argument.

We can write:
\[ 1 + \frac{2}{n} = 1 + 2m \quad \text{for some integer } m \]
Rearranging gives:
\[ \frac{2}{n} = 2m \implies n = \frac{1}{m} \]

7. **Conclusion**:
Thus, \( n \) must be of the form \( 2m - 1 \) for \( m \in \mathbb{Z} \).