If ` alpha and beta` are complex numbers then the maximum value of `(alpha barbeta+baralphabeta)/(|alpha beta|)=` (A) 1 (B) 2 (C) gt2 (D) lt1

Options

A. 1
B. 2
C. greater than 2
D. less than 1

Answer

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

\[ \frac{\alpha \bar{\beta} + \bar{\alpha} \beta}{|\alpha \beta|} \]

where \(\alpha\) and \(\beta\) are complex numbers.

### Step-by-Step Solution:

1. **Express \(\alpha\) and \(\beta\) in polar form**:
Let \(\alpha = r_1 e^{i\theta_1}\) and \(\beta = r_2 e^{i\theta_2}\), where \(r_1\) and \(r_2\) are the magnitudes of \(\alpha\) and \(\beta\) respectively, and \(\theta_1\) and \(\theta_2\) are their arguments.

2. **Find \(\bar{\alpha}\) and \(\bar{\beta}\)**:
The conjugates are given by:
\[ \bar{\alpha} = r_1 e^{-i\theta_1}, \quad \bar{\beta} = r_2 e^{-i\theta_2} \]

3. **Calculate \(\alpha \bar{\beta}\) and \(\bar{\alpha} \beta\)**:
\[ \alpha \bar{\beta} = (r_1 e^{i\theta_1})(r_2 e^{-i\theta_2}) = r_1 r_2 e^{i(\theta_1 - \theta_2)} \]
\[ \bar{\alpha} \beta = (r_1 e^{-i\theta_1})(r_2 e^{i\theta_2}) = r_1 r_2 e^{-i(\theta_1 - \theta_2)} \]

4. **Combine the two expressions**:
\[ \alpha \bar{\beta} + \bar{\alpha} \beta = r_1 r_2 \left(e^{i(\theta_1 - \theta_2)} + e^{-i(\theta_1 - \theta_2)}\right) \]
Using Euler's formula, this simplifies to:
\[ = r_1 r_2 \cdot 2 \cos(\theta_1 - \theta_2) \]

5. **Calculate the modulus \(|\alpha \beta|\)**:
\[ |\alpha \beta| = |\alpha| |\beta| = r_1 r_2 \]

6. **Substitute back into the original expression**:
\[ \frac{\alpha \bar{\beta} + \bar{\alpha} \beta}{|\alpha \beta|} = \frac{r_1 r_2 \cdot 2 \cos(\theta_1 - \theta_2)}{r_1 r_2} \]
This simplifies to:
\[ 2 \cos(\theta_1 - \theta_2) \]

7. **Determine the maximum value**:
The maximum value of \(\cos(\theta_1 - \theta_2)\) is 1, which occurs when \(\theta_1 - \theta_2 = 0\) (i.e., \(\theta_1 = \theta_2\)). Thus, the maximum value of the expression is:
\[ 2 \cdot 1 = 2 \]

### Final Answer:
The maximum value of \(\frac{\alpha \bar{\beta} + \bar{\alpha} \beta}{|\alpha \beta|}\) is \(2\).