Suppose `A` is a complex number and `n in N ,` such that `A^n=(A+1)^n=1,` then the least value of `n` is `3` b. `6` c. `9` d. `12`

Options

A. `3`
B. `6`
C. `9`
D. `12`

Explanation

To solve the problem, we need to find the least value of \( n \) such that \( A^n = (A + 1)^n = 1 \) for a complex number \( A \).

### Step-by-Step Solution

1. **Understanding the Conditions**:
We have two conditions:
\[ A^n = 1 \quad \text{and} \quad (A + 1)^n = 1 \]
This implies that both \( A \) and \( A + 1 \) are \( n \)-th roots of unity.

2. **Expressing \( A \) in Polar Form**:
Let \( A \) be expressed in polar form as:
\[ A = re^{i\theta} \]
Since \( A^n = 1 \), we have \( r^n = 1 \) which implies \( r = 1 \). Thus, \( A \) lies on the unit circle:
\[ A = e^{i\theta} \]

3. **Finding \( A + 1 \)**:
We can express \( A + 1 \) as:
\[ A + 1 = e^{i\theta} + 1 \]
We need to find the modulus of \( A + 1 \):
\[ |A + 1| = |e^{i\theta} + 1| = \sqrt{(1 + \cos \theta)^2 + \sin^2 \theta} \]
Simplifying this gives:
\[ |A + 1| = \sqrt{2 + 2\cos \theta} = 2\cos\left(\frac{\theta}{2}\right) \]

4. **Setting the Condition for \( (A + 1)^n = 1 \)**:
Since \( |A + 1|^n = 1 \), we must have:
\[ |A + 1| = 1 \implies 2\cos\left(\frac{\theta}{2}\right) = 1 \implies \cos\left(\frac{\theta}{2}\right) = \frac{1}{2} \]
This occurs when:
\[ \frac{\theta}{2} = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad \frac{\theta}{2} = -\frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
Thus:
\[ \theta = \frac{2\pi}{3} + 4k\pi \quad \text{or} \quad \theta = -\frac{2\pi}{3} + 4k\pi \]

5. **Finding the Values of \( n \)**:
The values of \( A \) can be \( e^{i\frac{2\pi}{3}} \) or \( e^{-i\frac{2\pi}{3}} \). The corresponding values of \( A \) are the cube roots of unity, which means:
\[ A = \omega \quad \text{or} \quad A = \omega^2 \]
where \( \omega = e^{i\frac{2\pi}{3}} \).

6. **Finding \( n \)**:
Since \( A^n = 1 \) and \( (A + 1)^n = 1 \), both \( A \) and \( A + 1 \) must be \( n \)-th roots of unity. The least common multiple of the orders of \( A \) and \( A + 1 \) must be calculated. The order of \( A \) is 3 (since \( \omega^3 = 1 \)), and the order of \( A + 1 \) is also 3.

Therefore, the least value of \( n \) that satisfies both conditions is:
\[ n = 3 \times 2 = 6 \]

### Conclusion:
Thus, the least value of \( n \) is **6**.