Let `Z=x+iy` is a complex number, such that `x^(2)+y^(2)=1.` In which of the following cases `(Z)/(1-Z)" "("for "x ne1)` lies in the `"II"^("nd")` quadrant? `(AA x,y in R, i^(2)=-1)`

Options

A. `xgt0`
B. `xlt0`
C. `ygt0`
D. `ylt0`

Answer

To solve the problem step by step, we need to analyze the given complex number \( Z = x + iy \) under the condition \( x^2 + y^2 = 1 \) and determine when the expression \( \frac{Z}{1 - Z} \) lies in the second quadrant.

### Step 1: Define the complex number and the condition
We have:
\[ Z = x + iy \]
with the condition:
\[ x^2 + y^2 = 1 \]
This means that \( Z \) lies on the unit circle in the complex plane.

### Step 2: Calculate \( \frac{Z}{1 - Z} \)
We need to compute:
\[ \frac{Z}{1 - Z} = \frac{x + iy}{1 - (x + iy)} = \frac{x + iy}{1 - x - iy} \]

### Step 3: Multiply by the conjugate
To simplify, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{Z}{1 - Z} = \frac{(x + iy)(1 - x + iy)}{(1 - x)^2 + y^2} \]

### Step 4: Expand the numerator
Expanding the numerator:
\[ (x + iy)(1 - x + iy) = x(1 - x) + x(iy) + iy(1 - x) + (iy)(iy) \]
\[ = x(1 - x) + ixy + iy(1 - x) - y^2 \]
\[ = (x - x^2 - y^2) + i(y - xy) \]

### Step 5: Simplify the denominator
The denominator simplifies to:
\[ (1 - x)^2 + y^2 = 1 - 2x + x^2 + y^2 = 1 - 2x + 1 = 2 - 2x \]

### Step 6: Combine the results
Thus, we have:
\[ \frac{Z}{1 - Z} = \frac{(x - x^2 - y^2) + i(y - xy)}{2 - 2x} \]

### Step 7: Identify real and imaginary parts
Let:
- Real part \( R = \frac{x - x^2 - y^2}{2 - 2x} \)
- Imaginary part \( I = \frac{y - xy}{2 - 2x} \)

### Step 8: Conditions for the second quadrant
For the expression to lie in the second quadrant:
1. The real part \( R < 0 \)
2. The imaginary part \( I > 0 \)

### Step 9: Analyze the conditions
1. **Imaginary part condition**:
\[ y - xy > 0 \implies y(1 - x) > 0 \]
Since \( y > 0 \) (to be in the second quadrant), we need \( 1 - x > 0 \) or \( x < 1 \).

2. **Real part condition**:
\[ x - x^2 - y^2 < 0 \]
Using \( y^2 = 1 - x^2 \):
\[ x - x^2 - (1 - x^2) < 0 \implies x - 1 < 0 \implies x < 1 \]

### Conclusion
Both conditions imply that \( x < 1 \) and \( y > 0 \). Therefore, the expression \( \frac{Z}{1 - Z} \) lies in the second quadrant when \( y > 0 \) and \( x < 1 \).