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 \( A^n = 1 \) and \( (A + 1)^n = 1 \). This implies that both \( A \) and \( A + 1 \) are \( n \)-th roots of unity.

2. **Expressing \( A \)**:
Let \( A = x + iy \), where \( x \) and \( y \) are real numbers. The condition \( A^n = 1 \) implies that the modulus of \( A \) is 1:
\[ |A| = 1 \implies x^2 + y^2 = 1 \tag{1} \]

3. **Finding \( A + 1 \)**:
Now, we compute the modulus of \( A + 1 \):
\[ A + 1 = (x + 1) + iy \]
The modulus is given by:
\[ |A + 1| = \sqrt{(x + 1)^2 + y^2} = 1 \]
Squaring both sides, we get:
\[ (x + 1)^2 + y^2 = 1 \tag{2} \]

4. **Expanding Equation (2)**:
Expanding equation (2):
\[ (x^2 + 2x + 1 + y^2) = 1 \]
Using equation (1) \( x^2 + y^2 = 1 \):
\[ 1 + 2x + 1 = 1 \implies 2x + 2 = 1 \implies 2x = -1 \implies x = -\frac{1}{2} \]

5. **Finding \( y \)**:
Substitute \( x = -\frac{1}{2} \) back into equation (1):
\[ \left(-\frac{1}{2}\right)^2 + y^2 = 1 \implies \frac{1}{4} + y^2 = 1 \implies y^2 = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, \( y = \pm \frac{\sqrt{3}}{2} \).

6. **Identifying \( A \)**:
Therefore, the possible values for \( A \) are:
\[ A = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \quad \text{and} \quad A = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \]
These correspond to the cube roots of unity \( \omega \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \).

7. **Finding \( n \)**:
Since \( A = \omega \) or \( A = \omega^2 \), we have:
\[ A^n = 1 \implies \omega^n = 1 \quad \text{and} \quad (A + 1)^n = (-\omega^2)^n = 1 \]
This means \( n \) must be a multiple of 3, and since \( A + 1 = -\omega^2 \), \( n \) must also be even.

8. **Finding the Least Value of \( n \)**:
The least common multiple of the conditions (multiple of 3 and even) gives us the smallest \( n \) as 6.

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