If `z = x+ iy` lies in the third quadrant, then `( bar(z))/( z)` also lies in the third quadrant, if

Options

A. `x gt y gt 0`
B. `x lt y lt 0`
C. ` y lt x lt 0`
D. `y gt x gt 0`

Answer

To solve the problem, we need to analyze the conditions under which the complex number \( z = x + iy \) lies in the third quadrant and how that affects the expression \( \frac{\bar{z}}{z} \).

### Step-by-Step Solution:

1. **Understanding the Quadrants**:
- In the complex plane, the third quadrant is characterized by both the real part \( x \) and the imaginary part \( y \) being negative. Therefore, we have:
\[ x < 0 \quad \text{and} \quad y < 0 \]

2. **Finding the Conjugate**:
- The conjugate of \( z \) is given by:
\[ \bar{z} = x - iy \]

3. **Calculating \( \frac{\bar{z}}{z} \)**:
- We can express \( \frac{\bar{z}}{z} \) as follows:
\[ \frac{\bar{z}}{z} = \frac{x - iy}{x + iy} \]

4. **Multiplying by the Conjugate**:
- To simplify this expression, multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{\bar{z}}{z} = \frac{(x - iy)(x - iy)}{(x + iy)(x - iy)} = \frac{x^2 - 2i xy - y^2}{x^2 + y^2} \]

5. **Separating Real and Imaginary Parts**:
- The expression simplifies to:
\[ \frac{\bar{z}}{z} = \frac{x^2 - y^2}{x^2 + y^2} - \frac{2xy}{x^2 + y^2} i \]
- Here, the real part is \( \frac{x^2 - y^2}{x^2 + y^2} \) and the imaginary part is \( -\frac{2xy}{x^2 + y^2} \).

6. **Conditions for the Third Quadrant**:
- For \( \frac{\bar{z}}{z} \) to lie in the third quadrant, both the real and imaginary parts must be negative:
1. **Real Part**:
\[ \frac{x^2 - y^2}{x^2 + y^2} < 0 \implies x^2 - y^2 < 0 \implies x^2 < y^2 \]
2. **Imaginary Part**:
\[ -\frac{2xy}{x^2 + y^2} < 0 \implies 2xy > 0 \implies xy > 0 \]

7. **Analyzing the Conditions**:
- From \( x^2 < y^2 \), we can conclude that \( |x| < |y| \).
- From \( xy > 0 \), since both \( x \) and \( y \) are negative in the third quadrant, this condition holds true.

8. **Final Conclusion**:
- Therefore, the conditions for \( \frac{\bar{z}}{z} \) to lie in the third quadrant are:
\[ x^2 < y^2 \quad \text{and} \quad xy > 0 \]
- This means \( x \) is less than \( y \) in absolute value, but both are negative.