If `x^(2) + y^(2) =1`, then what is `( 1+x+ iy)/( 1+x-iy)` equal to ?

Options

A. `x-iy`
B. `x+iy`
C. `2x`
D. `-2iy`

Answer

To solve the problem, we need to simplify the expression \(\frac{1 + x + iy}{1 + x - iy}\) given that \(x^2 + y^2 = 1\).

### Step-by-Step Solution:

1. **Identify the Complex Number**:
We can express \(x\) and \(y\) in terms of trigonometric functions since \(x^2 + y^2 = 1\) suggests that \(x = \cos(\theta)\) and \(y = \sin(\theta)\) for some angle \(\theta\).

2. **Rewrite the Expression**:
Substitute \(x\) and \(y\) into the expression:
\[ \frac{1 + \cos(\theta) + i\sin(\theta)}{1 + \cos(\theta) - i\sin(\theta)} \]

3. **Use Euler's Formula**:
Recall that \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\) and \(e^{-i\theta} = \cos(\theta) - i\sin(\theta)\). Thus, we can rewrite the expression as:
\[ \frac{1 + e^{i\theta}}{1 + e^{-i\theta}} \]

4. **Multiply by the Conjugate**:
To simplify, we can multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(1 + e^{i\theta})(1 + e^{i\theta})}{(1 + e^{-i\theta})(1 + e^{i\theta})} \]
The denominator simplifies to:
\[ 1 + e^{i\theta} + e^{-i\theta} + 1 = 2 + 2\cos(\theta) = 2(1 + \cos(\theta)) \]

5. **Simplify the Numerator**:
The numerator becomes:
\[ (1 + e^{i\theta})^2 = 1 + 2e^{i\theta} + e^{2i\theta} \]

6. **Final Expression**:
Thus, we have:
\[ \frac{1 + 2e^{i\theta} + e^{2i\theta}}{2(1 + \cos(\theta))} \]

7. **Simplify Further**:
We can factor out \(e^{i\theta}\) from the numerator:
\[ = \frac{e^{i\theta}(e^{i\theta} + 2 + 1)}{2(1 + \cos(\theta))} \]
This simplifies to:
\[ = e^{i\theta} \]

8. **Convert Back to Trigonometric Form**:
Finally, we can express \(e^{i\theta}\) back in terms of \(x\) and \(y\):
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) = x + iy \]

### Conclusion:
Thus, we find that:
\[ \frac{1 + x + iy}{1 + x - iy} = e^{i\theta} = x + iy \]