A real value of x satisfies the equation `(3-4ix)/(3+4ix)=alpha-ibeta(alpha,beta in R)`, if `alpha^2+beta^2=`

Options

A. ` x = 2n +1`
B. `x = 4n`
C. `x = 2n`
D. `x = 4n + 1`

Explanation

To solve the equation \(\frac{3 - 4ix}{3 + 4ix} = \alpha - i\beta\) where \(\alpha, \beta \in \mathbb{R}\), we will follow these steps:

### Step 1: Substitute a Real Value for \(x\)
Let's choose \(x = 1\) for simplicity. Thus, we rewrite the equation as:
\[ \frac{3 - 4i}{3 + 4i} = \alpha - i\beta \]

### Step 2: Multiply by the Conjugate of the Denominator
To simplify the left-hand side, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(3 - 4i)(3 - 4i)}{(3 + 4i)(3 - 4i)} \]

### Step 3: Expand the Numerator and Denominator
Calculating the numerator:
\[ (3 - 4i)(3 - 4i) = 3^2 - 2 \cdot 3 \cdot 4i + (4i)^2 = 9 - 24i - 16 = -7 - 24i \]
Calculating the denominator:
\[ (3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 + 16 = 25 \]

### Step 4: Write the Expression
Now, we can rewrite the equation:
\[ \frac{-7 - 24i}{25} = \alpha - i\beta \]
This simplifies to:
\[ -\frac{7}{25} - i\frac{24}{25} = \alpha - i\beta \]

### Step 5: Compare Real and Imaginary Parts
From the equation, we can identify:
\[ \alpha = -\frac{7}{25}, \quad \beta = \frac{24}{25} \]

### Step 6: Calculate \(\alpha^2 + \beta^2\)
Now, we need to find \(\alpha^2 + \beta^2\):
\[ \alpha^2 = \left(-\frac{7}{25}\right)^2 = \frac{49}{625} \]
\[ \beta^2 = \left(\frac{24}{25}\right)^2 = \frac{576}{625} \]
Thus,
\[ \alpha^2 + \beta^2 = \frac{49}{625} + \frac{576}{625} = \frac{625}{625} = 1 \]

### Final Answer
The value of \(\alpha^2 + \beta^2\) is:
\[ \boxed{1} \]