Let 1, `z_(1),z_(2),z_(3),…., z_(n-1)` be the nth roots of unity. Then prove that `(1-z_(1))(1 - z_(2)) …. (1-z_(n-1))= n`. Also,deduce that `sin .(pi)/(n) sin.(2pi)/(pi)sin.(3pi)/(n)...sin.((n-1)pi)/(n) = (pi)/(2^(n-1))`

Explanation

To prove that \((1 - z_1)(1 - z_2) \cdots (1 - z_{n-1}) = n\), where \(z_1, z_2, \ldots, z_{n-1}\) are the \(n\)th roots of unity, we can follow these steps:

### Step 1: Understand the \(n\)th Roots of Unity
The \(n\)th roots of unity are the solutions to the equation \(z^n = 1\). These roots can be expressed as:
\[ z_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
This means \(z_0 = 1\), \(z_1 = e^{2\pi i / n}\), \(z_2 = e^{4\pi i / n}\), and so on.
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