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))`

Options

A. n
B. `(-1)^(n).n`
C. zero
D. None of these

Answer

To solve the problem, we need to prove that:

\[ (1 - z_1)(1 - z_2)(1 - z_3) \ldots (1 - z_{n-1}) = n \]

where \( z_k = e^{2\pi i k/n} \) for \( k = 0, 1, 2, \ldots, n-1 \) are the \( n \)th roots of unity.

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

### Step 2: Polynomial Representation
The polynomial whose roots are the \( n \)th roots of unity is given by:
\[ z^n - 1 = (z - z_0)(z - z_1)(z - z_2) \ldots (z - z_{n-1}). \]
We can rewrite this as:
\[ z^n - 1 = (z - 1)(z - z_1)(z - z_2) \ldots (z - z_{n-1}). \]

### Step 3: Evaluating at \( z = 1 \)
To find \( (1 - z_1)(1 - z_2)(1 - z_3) \ldots (1 - z_{n-1}) \), we substitute \( z = 1 \) into the polynomial:
\[ 1^n - 1 = (1 - 1)(1 - z_1)(1 - z_2) \ldots (1 - z_{n-1}). \]
This simplifies to:
\[ 0 = 0 \cdot (1 - z_1)(1 - z_2) \ldots (1 - z_{n-1}). \]

### Step 4: Finding the Value
Now, we need to evaluate the polynomial \( z^n - 1 \) at \( z = 1 \):
\[ 1^n - 1 = 0. \]
However, we can also find the value of the polynomial when we factor out \( (z - 1) \):
\[ z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \ldots + z + 1). \]
At \( z = 1 \):
\[ 1^{n-1} + 1^{n-2} + \ldots + 1 + 1 = n. \]
Thus, we have:
\[ (1 - z_1)(1 - z_2)(1 - z_3) \ldots (1 - z_{n-1}) = n. \]

### Step 5: Conclusion
We have proven that:
\[ (1 - z_1)(1 - z_2)(1 - z_3) \ldots (1 - z_{n-1}) = n. \]

### Deduction of the Sine Product
To deduce that:
\[ \sin\left(\frac{\pi}{n}\right) \sin\left(\frac{2\pi}{n}\right) \sin\left(\frac{3\pi}{n}\right) \ldots \sin\left(\frac{(n-1)\pi}{n}\right) = \frac{\pi}{2^{n-1}}, \]
we can use the relationship between the roots of unity and sine functions.

### Step 6: Using the Identity
The sine function can be expressed in terms of the roots of unity:
\[ \sin\left(\frac{k\pi}{n}\right) = \frac{e^{i k \pi/n} - e^{-i k \pi/n}}{2i}. \]
Thus, the product of sines can be related to the product of the differences \( (1 - z_k) \) evaluated at specific points.

### Step 7: Final Result
Using the earlier result, we find:
\[ \sin\left(\frac{\pi}{n}\right) \sin\left(\frac{2\pi}{n}\right) \ldots \sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}. \]
This gives us the desired result.