If ` (a _(1) + i b_(2)) ( a_(2) + i b _(2)) . . . ( a_(n) + i b_(n)) = A + i B , " then " ( a_(1)^(2) + b_(1)^(2)) ( a_(2)^(2) + b_(2)^(2)) . . . ( a_(n)^(2) + b_(n)^(2))` is equal to

Options

A. 1
B. `A^(2) + B^(2)`
C. A + B
D. `(1)/(A^(2)) + (1)/(B^(2))`

Answer

To solve the problem, we need to find the value of the product \((a_1^2 + b_1^2)(a_2^2 + b_2^2) \ldots (a_n^2 + b_n^2)\) given that \((a_1 + ib_1)(a_2 + ib_2) \ldots (a_n + ib_n) = A + iB\).

### Step-by-Step Solution:

1. **Understanding the Given Expression**:
We have the product of complex numbers:
\[ (a_1 + ib_1)(a_2 + ib_2) \ldots (a_n + ib_n) = A + iB \]

2. **Taking Modulus on Both Sides**:
We take the modulus of both sides:
\[ |(a_1 + ib_1)(a_2 + ib_2) \ldots (a_n + ib_n)| = |A + iB| \]

3. **Using the Property of Modulus**:
The modulus of a product of complex numbers is the product of their moduli:
\[ |a_1 + ib_1| \cdot |a_2 + ib_2| \cdots |a_n + ib_n| = |A + iB| \]

4. **Calculating the Modulus of Each Complex Number**:
The modulus of a complex number \(a + ib\) is given by:
\[ |a + ib| = \sqrt{a^2 + b^2} \]
Therefore, we can express the moduli as:
\[ |a_1 + ib_1| = \sqrt{a_1^2 + b_1^2}, \quad |a_2 + ib_2| = \sqrt{a_2^2 + b_2^2}, \ldots, |a_n + ib_n| = \sqrt{a_n^2 + b_n^2} \]

5. **Substituting Back into the Modulus Equation**:
This gives us:
\[ \sqrt{a_1^2 + b_1^2} \cdot \sqrt{a_2^2 + b_2^2} \cdots \sqrt{a_n^2 + b_n^2} = \sqrt{A^2 + B^2} \]

6. **Squaring Both Sides**:
Squaring both sides to eliminate the square roots:
\[ (a_1^2 + b_1^2)(a_2^2 + b_2^2) \cdots (a_n^2 + b_n^2) = A^2 + B^2 \]

7. **Conclusion**:
Thus, we find that:
\[ (a_1^2 + b_1^2)(a_2^2 + b_2^2) \cdots (a_n^2 + b_n^2) = A^2 + B^2 \]

### Final Answer:
The value of \((a_1^2 + b_1^2)(a_2^2 + b_2^2) \ldots (a_n^2 + b_n^2)\) is equal to \(A^2 + B^2\).