If `a+i b=(c+i)/(c-i)` , where `c` is real, then :`a^2+b^2`

Options

A. 0
B. 1
C. 2
D. c

Explanation

To solve the problem \( a + ib = \frac{c + i}{c - i} \) where \( c \) is real, we need to express \( a \) and \( b \) in terms of \( c \) and then find \( a^2 + b^2 \).

### Step-by-Step Solution:

1. **Multiply by the Conjugate**:
We start with the equation:
\[ a + ib = \frac{c + i}{c - i} \]
To simplify the right-hand side, we multiply the numerator and denominator by the conjugate of the denominator:
\[ a + ib = \frac{(c + i)(c + i)}{(c - i)(c + i)} \]

2. **Calculate the Denominator**:
The denominator simplifies as follows:
\[ (c - i)(c + i) = c^2 - i^2 = c^2 + 1 \]

3. **Calculate the Numerator**:
The numerator simplifies as follows:
\[ (c + i)(c + i) = c^2 + 2ci + i^2 = c^2 + 2ci - 1 = (c^2 - 1) + 2ci \]

4. **Combine the Results**:
Now we can write:
\[ a + ib = \frac{(c^2 - 1) + 2ci}{c^2 + 1} \]
This can be separated into real and imaginary parts:
\[ a = \frac{c^2 - 1}{c^2 + 1}, \quad b = \frac{2c}{c^2 + 1} \]

5. **Find \( a^2 + b^2 \)**:
Now we need to calculate \( a^2 + b^2 \):
\[ a^2 = \left(\frac{c^2 - 1}{c^2 + 1}\right)^2 = \frac{(c^2 - 1)^2}{(c^2 + 1)^2} \]
\[ b^2 = \left(\frac{2c}{c^2 + 1}\right)^2 = \frac{4c^2}{(c^2 + 1)^2} \]
Adding these together:
\[ a^2 + b^2 = \frac{(c^2 - 1)^2 + 4c^2}{(c^2 + 1)^2} \]

6. **Simplify the Numerator**:
Expanding the numerator:
\[ (c^2 - 1)^2 + 4c^2 = c^4 - 2c^2 + 1 + 4c^2 = c^4 + 2c^2 + 1 = (c^2 + 1)^2 \]

7. **Final Result**:
Thus, we have:
\[ a^2 + b^2 = \frac{(c^2 + 1)^2}{(c^2 + 1)^2} = 1 \]

### Conclusion:
The result is:
\[ a^2 + b^2 = 1 \]