If `Z_r=cos((2rpi)/5)+isin((2rpi)/5),r=0,1,2,3,4,...` then `z_1z_2z_3z_4z_5` is equal to

Options

A. `-1`
B. 0
C. 1
D. none of these

Explanation

To solve the problem, we need to find the product \( z_1 z_2 z_3 z_4 z_5 \) where

\[ Z_r = \cos\left(\frac{2r\pi}{5}\right) + i \sin\left(\frac{2r\pi}{5}\right) \]

for \( r = 0, 1, 2, 3, 4 \).

### Step-by-Step Solution:

1. **Identify the values of \( Z_r \)**:
- For \( r = 0 \):
\[ Z_0 = \cos(0) + i \sin(0) = 1 + 0i = 1 \]
- For \( r = 1 \):
\[ Z_1 = \cos\left(\frac{2\pi}{5}\right) + i \sin\left(\frac{2\pi}{5}\right) = e^{i\frac{2\pi}{5}} \]
- For \( r = 2 \):
\[ Z_2 = \cos\left(\frac{4\pi}{5}\right) + i \sin\left(\frac{4\pi}{5}\right) = e^{i\frac{4\pi}{5}} \]
- For \( r = 3 \):
\[ Z_3 = \cos\left(\frac{6\pi}{5}\right) + i \sin\left(\frac{6\pi}{5}\right) = e^{i\frac{6\pi}{5}} \]
- For \( r = 4 \):
\[ Z_4 = \cos\left(\frac{8\pi}{5}\right) + i \sin\left(\frac{8\pi}{5}\right) = e^{i\frac{8\pi}{5}} \]

2. **Write the product \( Z_1 Z_2 Z_3 Z_4 Z_5 \)**:
\[ Z_1 Z_2 Z_3 Z_4 Z_5 = Z_0 Z_1 Z_2 Z_3 Z_4 = 1 \cdot e^{i\frac{2\pi}{5}} \cdot e^{i\frac{4\pi}{5}} \cdot e^{i\frac{6\pi}{5}} \cdot e^{i\frac{8\pi}{5}} \]

3. **Combine the exponents**:
\[ Z_1 Z_2 Z_3 Z_4 Z_5 = e^{i\left(\frac{2\pi}{5} + \frac{4\pi}{5} + \frac{6\pi}{5} + \frac{8\pi}{5}\right)} \]

4. **Calculate the sum of the angles**:
\[ \frac{2\pi}{5} + \frac{4\pi}{5} + \frac{6\pi}{5} + \frac{8\pi}{5} = \frac{20\pi}{5} = 4\pi \]

5. **Substitute back into the exponential**:
\[ Z_1 Z_2 Z_3 Z_4 Z_5 = e^{i(4\pi)} \]

6. **Evaluate \( e^{i(4\pi)} \)**:
\[ e^{i(4\pi)} = \cos(4\pi) + i \sin(4\pi) = 1 + 0i = 1 \]

### Final Answer:
\[ Z_1 Z_2 Z_3 Z_4 Z_5 = 1 \]