For a complex number `Z,|Z|=1 and "arg "(Z)=theta`. If `(Z)(Z^(2))(Z^(3))…(Z^(n))=1`, then the value of `theta` is

Options

A. `(4mpi)/(n(n+1)), m in I`
B. `(2mpi)/(n(n+1)), m in I`
C. `(mpi)/(n(n+1)), m in I`
D. None of these

Answer

To solve the problem, we start with the given conditions for the complex number \( Z \):

1. **Given Conditions**:
- \( |Z| = 1 \)
- \( \text{arg}(Z) = \theta \)
- The product \( Z \cdot Z^2 \cdot Z^3 \cdots Z^n = 1 \)

2. **Expressing \( Z \)**:
Since \( |Z| = 1 \), we can express \( Z \) in exponential form:
\[ Z = e^{i\theta} \]

3. **Calculating the Product**:
We need to calculate the product:
\[ Z \cdot Z^2 \cdot Z^3 \cdots Z^n = e^{i\theta} \cdot e^{2i\theta} \cdot e^{3i\theta} \cdots e^{ni\theta} \]
This can be rewritten as:
\[ e^{i\theta(1 + 2 + 3 + \ldots + n)} \]

4. **Sum of the First \( n \) Natural Numbers**:
The sum of the first \( n \) natural numbers is given by the formula:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Therefore, we can substitute this into our product:
\[ Z \cdot Z^2 \cdot Z^3 \cdots Z^n = e^{i\theta \cdot \frac{n(n + 1)}{2}} \]

5. **Setting the Product Equal to 1**:
We know that this product equals 1:
\[ e^{i\theta \cdot \frac{n(n + 1)}{2}} = 1 \]
The exponential function equals 1 when its exponent is an integer multiple of \( 2\pi \):
\[ i\theta \cdot \frac{n(n + 1)}{2} = i \cdot 2m\pi \quad \text{for some integer } m \]

6. **Canceling \( i \)**:
We can cancel \( i \) from both sides:
\[ \theta \cdot \frac{n(n + 1)}{2} = 2m\pi \]

7. **Solving for \( \theta \)**:
Rearranging gives us:
\[ \theta = \frac{4m\pi}{n(n + 1)} \]

8. **Conclusion**:
The value of \( \theta \) is:
\[ \theta = \frac{4m\pi}{n(n + 1)} \]
where \( m \) is any integer.