If `|a+ib|=1` then the simplified form of `(1+b+ai)/(1+b-ai)` is

Options

A. b+ai
B. a+bi
C. `(1+b)^(2)+a^(2)`
D. ai

Answer

To solve the problem, we need to simplify the expression \((1 + b + ai)/(1 + b - ai)\) given that \(|a + ib| = 1\).

### Step-by-Step Solution:

1. **Understanding the modulus condition**:
Since \(|a + ib| = 1\), we know that:
\[ \sqrt{a^2 + b^2} = 1 \]
Squaring both sides gives:
\[ a^2 + b^2 = 1 \]

2. **Set up the expression**:
We need to simplify:
\[ \frac{1 + b + ai}{1 + b - ai} \]

3. **Multiply by the conjugate**:
To simplify the fraction, multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 + b + ai\):
\[ \frac{(1 + b + ai)(1 + b + ai)}{(1 + b - ai)(1 + b + ai)} \]

4. **Calculate the denominator**:
The denominator becomes:
\[ (1 + b)^2 - (ai)^2 = (1 + b)^2 - a^2(-1) = (1 + b)^2 + a^2 \]
Expanding \((1 + b)^2\):
\[ (1 + b)^2 = 1 + 2b + b^2 \]
Therefore, the denominator simplifies to:
\[ 1 + 2b + b^2 + a^2 \]
Using \(a^2 + b^2 = 1\), we substitute \(a^2\):
\[ 1 + 2b + b^2 + (1 - b^2) = 2 + 2b = 2(1 + b) \]

5. **Calculate the numerator**:
The numerator becomes:
\[ (1 + b + ai)^2 = (1 + b)^2 + 2(1 + b)(ai) + (ai)^2 \]
Expanding this gives:
\[ (1 + b)^2 + 2ai(1 + b) - a^2 \]
Substituting \((1 + b)^2\) and \(ai^2 = -a^2\):
\[ (1 + 2b + b^2) + 2ai(1 + b) - a^2 = 1 + 2b + b^2 - a^2 + 2ai(1 + b) \]
Using \(a^2 + b^2 = 1\):
\[ 1 + 2b + b^2 - (1 - b^2) + 2ai(1 + b) = 2b + 2b^2 + 2ai(1 + b) \]

6. **Combine the results**:
Now we have:
\[ \frac{2b + 2b^2 + 2ai(1 + b)}{2(1 + b)} \]
Simplifying this gives:
\[ \frac{2(b + b^2 + ai(1 + b))}{2(1 + b)} = \frac{b + b^2 + ai(1 + b)}{1 + b} \]

7. **Final simplification**:
Factoring out \(1 + b\):
\[ = \frac{b + ai}{1 + b} \]

Thus, the simplified form of \(\frac{1 + b + ai}{1 + b - ai}\) is:
\[ \frac{b + ai}{1 + b} \]