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. `-2it`

Explanation

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 Expression**:
We start with the expression:
\[ \frac{1+x+iy}{1+x-iy} \]

2. **Rationalize the Denominator**:
To simplify this expression, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(1+x+iy)(1+x+iy)}{(1+x-iy)(1+x+iy)} \]

3. **Expand the Numerator**:
The numerator becomes:
\[ (1+x+iy)(1+x+iy) = (1+x)^2 + 2(1+x)(iy) + (iy)^2 \]
Simplifying this:
\[ = (1+x)^2 + 2(1+x)iy - y^2 \]
\[ = (1 + 2x + x^2 - y^2) + 2(1+x)iy \]

4. **Expand the Denominator**:
The denominator becomes:
\[ (1+x)^2 - (iy)^2 = (1+x)^2 - (-y^2) = (1+x)^2 + y^2 \]
\[ = (1 + 2x + x^2 + y^2) \]

5. **Substitute \(x^2 + y^2\)**:
Since we know \(x^2 + y^2 = 1\), we can substitute this into our expression:
\[ = (1 + 2x + 1) + 2(1+x)iy = (2 + 2x) + 2(1+x)iy \]
The denominator simplifies to:
\[ = 1 + 2x + 1 = 2 + 2x \]

6. **Combine the Results**:
Now we have:
\[ \frac{(2 + 2x) + 2(1+x)iy}{2 + 2x} \]
We can separate this into real and imaginary parts:
\[ = 1 + \frac{2(1+x)iy}{2 + 2x} \]
\[ = 1 + \frac{(1+x)iy}{1+x} \]
This simplifies to:
\[ = 1 + iy \]

### Final Result:
Thus, the expression \(\frac{1+x+iy}{1+x-iy}\) simplifies to:
\[ 1 + iy \]