Suppose a, b, c `in C, and |a| = |b| = |c| = 1 and abc = a + b + c`, then |bc + ca + ab| is equal to

Options

A. 0
B. -1
C. 1
D. None of these

Answer

To solve the problem, we need to find the value of \(|bc + ca + ab|\) given that \(a, b, c \in \mathbb{C}\) (the set of complex numbers), \(|a| = |b| = |c| = 1\), and \(abc = a + b + c\).

### Step-by-Step Solution:

1. **Given Conditions**: We know that \(|a| = |b| = |c| = 1\). This means that \(a\), \(b\), and \(c\) lie on the unit circle in the complex plane.

2. **Using the Given Equation**: We have the equation \(abc = a + b + c\). We will take the modulus of both sides:
\[ |abc| = |a + b + c| \]

3. **Calculate \(|abc|\)**: Since \(|a| = |b| = |c| = 1\), we have:
\[ |abc| = |a| \cdot |b| \cdot |c| = 1 \cdot 1 \cdot 1 = 1 \]
Therefore, we can write:
\[ |a + b + c| = 1 \]

4. **Using the Identity**: We know that:
\[ |a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2 \text{Re}(ab^* + ac^* + bc^*) \]
Since \(|a|^2 = |b|^2 = |c|^2 = 1\), we have:
\[ |a + b + c|^2 = 1 + 1 + 1 + 2 \text{Re}(ab^* + ac^* + bc^*) = 3 + 2 \text{Re}(ab^* + ac^* + bc^*) \]

5. **Setting Up the Equation**: Since we know \(|a + b + c| = 1\), we can square both sides:
\[ 1^2 = 3 + 2 \text{Re}(ab^* + ac^* + bc^*) \]
This simplifies to:
\[ 1 = 3 + 2 \text{Re}(ab^* + ac^* + bc^*) \]
Rearranging gives:
\[ 2 \text{Re}(ab^* + ac^* + bc^*) = 1 - 3 = -2 \]
Thus:
\[ \text{Re}(ab^* + ac^* + bc^*) = -1 \]

6. **Finding \(|bc + ca + ab|\)**: We can express \(bc + ca + ab\) in terms of \(a\), \(b\), and \(c\):
\[ |bc + ca + ab| = |(b+c)a + bc| \]
However, we can also use the identity:
\[ |bc + ca + ab|^2 = |bc|^2 + |ca|^2 + |ab|^2 + 2 \text{Re}(bc \cdot ca^* + ca \cdot ab^* + ab \cdot bc^*) \]
Since \(|bc| = |ca| = |ab| = 1\):
\[ |bc + ca + ab|^2 = 1 + 1 + 1 + 2 \text{Re}(bc \cdot ca^* + ca \cdot ab^* + ab \cdot bc^*) \]
This simplifies to:
\[ |bc + ca + ab|^2 = 3 + 2 \text{Re}(bc \cdot ca^* + ca \cdot ab^* + ab \cdot bc^*) \]

7. **Final Calculation**: Since we have established that \(\text{Re}(ab^* + ac^* + bc^*) = -1\), we can conclude that:
\[ |bc + ca + ab| = 1 \]

### Conclusion:
Thus, the value of \(|bc + ca + ab|\) is equal to \(1\).