Q. Let `z_1` and `z_2` be nth roots of unity which subtend a right angle at the origin, then n must be the form of .

Options

A. `4k+1`
B. `4k+2`
C. `4k+3`
D. 4k

Answer

To solve the problem, we need to analyze the conditions given for the nth roots of unity, \( z_1 \) and \( z_2 \), which subtend a right angle at the origin.

### Step-by-Step Solution:

1. **Understanding nth Roots of Unity**:
The nth roots of unity are given by the formula:
\[ z_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
This means \( z_1 = e^{2\pi i k_1/n} \) and \( z_2 = e^{2\pi i k_2/n} \) for some integers \( k_1 \) and \( k_2 \).

**Hint**: Remember that the nth roots of unity are evenly spaced points on the unit circle in the complex plane.

2. **Condition of Right Angle**:
For \( z_1 \) and \( z_2 \) to subtend a right angle at the origin, the angle between them must be \( 90^\circ \) or \( \frac{\pi}{2} \) radians. This can be expressed in terms of their arguments:
\[ \text{arg}(z_2) - \text{arg}(z_1) = \frac{\pi}{2} \quad \text{(mod } 2\pi\text{)} \]

**Hint**: Use the property of arguments in complex numbers to find the difference between the angles.

3. **Expressing the Argument Condition**:
We can write:
\[ \frac{2\pi k_2}{n} - \frac{2\pi k_1}{n} = \frac{\pi}{2} + 2\pi m \quad \text{for some integer } m \]
Simplifying gives:
\[ \frac{2\pi (k_2 - k_1)}{n} = \frac{\pi}{2} + 2\pi m \]
Dividing through by \( \pi \):
\[ \frac{2(k_2 - k_1)}{n} = \frac{1}{2} + 2m \]

**Hint**: Isolate \( n \) to find a relationship involving \( k_2 - k_1 \).

4. **Finding n**:
Rearranging gives:
\[ n = \frac{4(k_2 - k_1)}{1 + 4m} \]
This indicates that \( n \) must be a multiple of 4.

**Hint**: Look for integer values of \( k_2 - k_1 \) and \( m \) that keep \( n \) an integer.

5. **Conclusion**:
Since \( n \) must be a multiple of 4, we can express \( n \) in the form:
\[ n = 4k \quad \text{where } k \text{ is an integer} \]

**Final Answer**: The value of \( n \) must be of the form \( 4k \), where \( k \) is an integer.