If `x^(2) + y^(2) = 1 and x ne -1` then `(1+y+ix)/(1+y-ix)` equals

Options

A. 1
B. x + iy
C. 2
D. y + ix

Answer

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

### Step-by-step Solution:

1. **Identify the Expression**:
We start with the expression:
\[ \frac{1+y+ix}{1+y-ix} \]

2. **Multiply by the Conjugate**:
To simplify this expression, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(1+y+ix)(1+y+ix)}{(1+y-ix)(1+y+ix)} \]

3. **Calculate the Denominator**:
The denominator becomes:
\[ (1+y)^2 - (ix)^2 = (1+y)^2 - (-x^2) = (1+y)^2 + x^2 \]
Expanding \((1+y)^2\):
\[ (1+y)^2 = 1 + 2y + y^2 \]
Therefore, the denominator is:
\[ 1 + 2y + y^2 + x^2 \]
Using the condition \(x^2 + y^2 = 1\), we have:
\[ 1 + 2y + 1 = 2 + 2y \]

4. **Calculate the Numerator**:
The numerator becomes:
\[ (1+y+ix)(1+y+ix) = (1+y)^2 + 2(1+y)(ix) + (ix)^2 \]
Expanding this:
\[ = (1+y)^2 + 2ix(1+y) - x^2 \]
Using \((1+y)^2 = 1 + 2y + y^2\) and substituting \(x^2\):
\[ = 1 + 2y + y^2 - x^2 + 2ix(1+y) = 1 + 2y + (y^2 - x^2) + 2ix(1+y) \]
Again, using \(x^2 + y^2 = 1\), we have \(y^2 - x^2 = -x^2 + y^2 = 1 - 2x^2\):
\[ = 1 + 2y + 1 - 2x^2 + 2ix(1+y) = 2 + 2y - 2x^2 + 2ix(1+y) \]

5. **Combine the Results**:
Now we have:
\[ \frac{2 + 2y - 2x^2 + 2ix(1+y)}{2 + 2y} \]
Simplifying this gives:
\[ = \frac{2(1 + y - x^2) + 2ix(1+y)}{2(1+y)} = \frac{1 + y - x^2 + ix(1+y)}{1+y} \]

6. **Final Simplification**:
Since \(x^2 + y^2 = 1\), we can substitute \(x^2 = 1 - y^2\):
\[ = \frac{1 + y - (1 - y^2) + ix(1+y)}{1+y} = \frac{y + y^2 + ix(1+y)}{1+y} \]
This simplifies to:
\[ = \frac{y(1+y) + ix(1+y)}{1+y} = y + ix \]

### Final Answer:
Thus, the final value of the expression is:
\[ y + ix \]