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, we need to analyze the expression \(\frac{Z}{1-Z}\) where \(Z = x + iy\) and \(x^2 + y^2 = 1\). We want to determine the conditions under which this expression lies in the second quadrant of the complex plane.

### Step-by-Step Solution:

1. **Substituting the Expression**:
Given \(Z = x + iy\), we can write:
\[ \frac{Z}{1-Z} = \frac{x + iy}{1 - (x + iy)} = \frac{x + iy}{1 - x - iy} \]

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

3. **Expanding the Numerator**:
Now, let's expand the numerator:
\[ (x + iy)(1 - x + iy) = x(1 - x) + x(iy) + iy(1 - x) + (iy)(iy) \]
This simplifies to:
\[ x(1 - x) + iy(1 - x) + iy - y^2 = x(1 - x) - y^2 + i(y(1 - x) + y) \]

4. **Combining Terms**:
Thus, we have:
\[ \frac{x(1 - x) - y^2 + i(y(1 - x) + y)}{2(1 - x)} \]
This can be separated into real and imaginary parts:
- Real part: \(\frac{x(1 - x) - y^2}{2(1 - x)}\)
- Imaginary part: \(\frac{y(1 - x) + y}{2(1 - x)} = \frac{y(2 - x)}{2(1 - x)}\)

5. **Condition for the Second Quadrant**:
For the complex number to lie in the second quadrant, the real part must be negative and the imaginary part must be positive:
- **Imaginary part**: \(\frac{y(2 - x)}{2(1 - x)} > 0\) implies \(y > 0\) and \(2 - x > 0\) (i.e., \(x < 2\)).
- **Real part**: \(\frac{x(1 - x) - y^2}{2(1 - x)} < 0\) implies \(x(1 - x) < y^2\).

6. **Using the Circle Equation**:
Since \(x^2 + y^2 = 1\), we can substitute \(y^2 = 1 - x^2\) into the inequality:
\[ x(1 - x) < 1 - x^2 \]
Rearranging gives:
\[ x + x^2 - x^2 < 1 \implies x < 1 \]

7. **Conclusion**:
Thus, for \(Z/(1-Z)\) to lie in the second quadrant, we require:
- \(y > 0\) (which is satisfied since \(x^2 + y^2 = 1\))
- \(x < 1\)

### Final Result:
The conditions under which \(\frac{Z}{1-Z}\) lies in the second quadrant are:
- \(x < 1\)
- \(y > 0\)