If `(a_(1)+ib_(1))(a_(2)+ib_(2))………………(a_(n)+ib_(n))=A+iB`, then `sum_(i=1)^(n) tan^(-1)(b_(i)/a_(i))` is equal to

Options

A. `B/A`
B. `tan(B/A)`
C. `tan^(-1)(B/A)`
D. `tan^(-1)(A/B)`

Answer

To solve the given problem, we need to find the value of the expression:

\[ \sum_{i=1}^{n} \tan^{-1}\left(\frac{b_i}{a_i}\right) \]

given that:

\[ (a_1 + ib_1)(a_2 + ib_2) \cdots (a_n + ib_n) = A + iB \]

### Step-by-Step Solution:

1. **Understanding the Product of Complex Numbers**:
The product of the complex numbers can be expressed as:
\[ (a_1 + ib_1)(a_2 + ib_2) \cdots (a_n + ib_n) = A + iB \]

2. **Using the Argument Property**:
We know that the argument of a product of complex numbers is the sum of their arguments:
\[ \text{arg}(z_1 z_2) = \text{arg}(z_1) + \text{arg}(z_2) \]
Therefore, we have:
\[ \text{arg}((a_1 + ib_1)(a_2 + ib_2) \cdots (a_n + ib_n)) = \text{arg}(A + iB) \]

3. **Expressing Arguments**:
The argument of a complex number \(a + ib\) can be expressed as:
\[ \text{arg}(a + ib) = \tan^{-1}\left(\frac{b}{a}\right) \]
Thus, we can write:
\[ \text{arg}(a_1 + ib_1) + \text{arg}(a_2 + ib_2) + \cdots + \text{arg}(a_n + ib_n) = \text{arg}(A + iB) \]

4. **Substituting the Arguments**:
Substituting the expressions for the arguments, we get:
\[ \tan^{-1}\left(\frac{b_1}{a_1}\right) + \tan^{-1}\left(\frac{b_2}{a_2}\right) + \cdots + \tan^{-1}\left(\frac{b_n}{a_n}\right) = \tan^{-1}\left(\frac{B}{A}\right) \]

5. **Final Expression**:
Therefore, we can express the sum as:
\[ \sum_{i=1}^{n} \tan^{-1}\left(\frac{b_i}{a_i}\right) = \tan^{-1}\left(\frac{B}{A}\right) \]

### Conclusion:
The final result is:
\[ \sum_{i=1}^{n} \tan^{-1}\left(\frac{b_i}{a_i}\right) = \tan^{-1}\left(\frac{B}{A}\right) \]