If Real `((2z-1)/(z+1))`=1, then locus of z is , where z=x+iy and `i=sqrt(-1)`

Options

A. circle
B. parabola
C. Sr.line
D. Pair of St. lines

Explanation

To solve the problem, we need to find the locus of \( z \) given that the real part of \( \frac{2z - 1}{z + 1} = 1 \), where \( z = x + iy \).

### Step-by-Step Solution:

1. **Substitute \( z \)**:
\[ z = x + iy \]
Substitute this into the expression:
\[ \frac{2z - 1}{z + 1} = \frac{2(x + iy) - 1}{(x + iy) + 1} \]

2. **Simplify the expression**:
\[ = \frac{2x + 2iy - 1}{x + 1 + iy} = \frac{(2x - 1) + 2iy}{(x + 1) + iy} \]

3. **Multiply by the conjugate of the denominator**:
To separate the real and imaginary parts, multiply the numerator and denominator by the conjugate of the denominator:
\[ = \frac{((2x - 1) + 2iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)} \]

4. **Calculate the denominator**:
\[ = (x + 1)^2 + y^2 \]

5. **Calculate the numerator**:
Expanding the numerator:
\[ = (2x - 1)(x + 1) - (2y^2) + i((2y)(x + 1) - (2x - 1)y) \]
This simplifies to:
\[ = (2x^2 + 2x - x - 1 - 2y^2) + i(2xy + 2y - 2xy + y) \]
\[ = (2x^2 + x - 1 - 2y^2) + i(y + 2y) \]

6. **Separate real and imaginary parts**:
The real part is:
\[ \frac{2x^2 + x - 1 - 2y^2}{(x + 1)^2 + y^2} \]
The imaginary part is:
\[ \frac{(y + 2y)}{(x + 1)^2 + y^2} \]

7. **Set the real part equal to 1**:
From the problem statement:
\[ \frac{2x^2 + x - 1 - 2y^2}{(x + 1)^2 + y^2} = 1 \]

8. **Cross-multiply**:
\[ 2x^2 + x - 1 - 2y^2 = (x + 1)^2 + y^2 \]

9. **Expand and rearrange**:
Expanding the right-hand side:
\[ 2x^2 + x - 1 - 2y^2 = x^2 + 2x + 1 + y^2 \]
Rearranging gives:
\[ 2x^2 - x^2 + x - 2x - 1 - 1 + 2y^2 - y^2 = 0 \]
This simplifies to:
\[ x^2 - x + y^2 - 2 = 0 \]

10. **Rearranging to standard form**:
\[ x^2 + y^2 - x - 2 = 0 \]
Completing the square for \( x \):
\[ (x - \frac{1}{2})^2 + y^2 = \frac{9}{4} \]

### Conclusion:
The locus of \( z \) is a circle with center \( \left(\frac{1}{2}, 0\right) \) and radius \( \frac{3}{2} \).