If `z_(1),z_(2),z_(3),…,z_(n-1)` are the roots of the equation `z^(n-1)+z^(n-2)+z^(n-3)+…+z+1=0`, where `n in N, n gt 2` and `omega` is the cube root of unity, then

Options

A. `omega^(n),omega^(2n)` are also the roots of the given equation
B. `omega^(1//n),omega^(2//n)` are also the roots of the given equation
C. `z_(1),z_(2),…,z_(n-1)` form a geonetric progression
D. `a^((z_(r+1))/(z_( r )))` is constant for `a gt 0` and r = 1, 2, 3, …, n - 2.

Answer

To solve the problem, we need to analyze the roots of the equation:

\[ z^{n-1} + z^{n-2} + z^{n-3} + \ldots + z + 1 = 0 \]

This equation can be rewritten using the formula for the sum of a geometric series. The sum can be expressed as:

\[ \frac{z^n - 1}{z - 1} = 0 \]

This implies that:

\[ z^n - 1 = 0 \]

Thus, the roots of this equation are the \( n \)-th roots of unity, which are given by:

\[ z_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]

However, since we are interested in the roots of the original polynomial, we exclude \( z = 1 \) (which corresponds to \( k = 0 \)). Therefore, the roots of the equation \( z^{n-1} + z^{n-2} + \ldots + z + 1 = 0 \) are:

\[ z_1, z_2, \ldots, z_{n-1} = e^{2\pi i k/n} \quad \text{for } k = 1, 2, \ldots, n-1 \]

Next, we need to analyze the cube roots of unity, denoted as \( \omega \). The cube roots of unity are:

\[ \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{4\pi i / 3}, \quad \text{and } \omega^3 = 1 \]

Now, we need to determine if \( \omega^n \) or \( \omega^{2n} \) are roots of the original polynomial.

1. **Checking if \( \omega^n \) is a root:**
- Since \( \omega^3 = 1 \), we can express \( \omega^n \) as:
\[ \omega^n = \omega^{n \mod 3} \]
- Depending on \( n \mod 3 \), \( \omega^n \) can be \( 1, \omega, \) or \( \omega^2 \). None of these values are roots of the polynomial since they do not satisfy \( z^{n-1} + z^{n-2} + \ldots + z + 1 = 0 \) when \( n > 2 \).

2. **Checking if \( \omega^{2n} \) is a root:**
- Similarly, we can express \( \omega^{2n} \) as:
\[ \omega^{2n} = \omega^{2(n \mod 3)} \]
- Again, depending on \( n \mod 3 \), \( \omega^{2n} \) can be \( 1, \omega^2, \) or \( \omega \). None of these values are roots of the polynomial for the same reasons as above.

3. **Conclusion:**
- Since neither \( \omega^n \) nor \( \omega^{2n} \) are roots of the polynomial, we conclude that the roots \( z_1, z_2, \ldots, z_{n-1} \) are distinct \( n \)-th roots of unity excluding \( 1 \).

### Final Answer:
The roots \( z_1, z_2, \ldots, z_{n-1} \) are in geometric progression (GP) and \( \omega^n \) and \( \omega^{2n} \) are not roots of the equation.