If z = x + iy lies in III quadrant, then `bar z / z` also lies in III 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`

Explanation

To solve the problem, we need to show that if \( z = x + iy \) lies in the third quadrant, then \( \frac{\bar{z}}{z} \) also lies in the third quadrant.

### Step-by-Step Solution:

1. **Understanding the Quadrants**:
- In the third quadrant, both the real part \( x \) and the imaginary part \( y \) of the complex number \( z \) are negative. Thus, we have:
\[ x < 0 \quad \text{and} \quad y < 0 \]

2. **Finding \( \bar{z} \)**:
- 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. **Rationalizing the Denominator**:
- To simplify \( \frac{\bar{z}}{z} \), we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{\bar{z}}{z} = \frac{(x - iy)(x - iy)}{(x + iy)(x - iy)} = \frac{x^2 - 2xyi - y^2}{x^2 + y^2} \]
- This simplifies to:
\[ \frac{x^2 - y^2 - 2xyi}{x^2 + y^2} \]

5. **Identifying Real and Imaginary Parts**:
- The real part \( \text{Re} \left( \frac{\bar{z}}{z} \right) \) is \( \frac{x^2 - y^2}{x^2 + y^2} \) and the imaginary part \( \text{Im} \left( \frac{\bar{z}}{z} \right) \) is \( \frac{-2xy}{x^2 + y^2} \).

6. **Analyzing the Real Part**:
- For \( \frac{\bar{z}}{z} \) to lie in the third quadrant, we need:
\[ \frac{x^2 - y^2}{x^2 + y^2} < 0 \]
- This implies:
\[ x^2 - y^2 < 0 \quad \Rightarrow \quad x^2 < y^2 \]

7. **Analyzing the Imaginary Part**:
- We also need:
\[ \frac{-2xy}{x^2 + y^2} < 0 \]
- Since \( x^2 + y^2 > 0 \) (as both \( x \) and \( y \) are non-zero), we require:
\[ -2xy < 0 \quad \Rightarrow \quad xy > 0 \]

8. **Conclusion**:
- From \( x^2 < y^2 \) and \( xy > 0 \), we conclude that both \( x \) and \( y \) must be negative, and \( |x| < |y| \). Thus, \( \frac{\bar{z}}{z} \) lies in the third quadrant.