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

Options

A. (a)circle
B. (b)parabola
C. (c)St. line
D. (d)Pair of ST. lines

Explanation

To find the locus of the complex number \( z \) given that \( \text{Re} \left( \frac{2z - 1}{z + 1} \right) = 1 \), we can follow these steps:

### Step 1: Substitute \( z = x + iy \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The expression becomes:
\[ \frac{2(x + iy) - 1}{(x + iy) + 1} = \frac{(2x - 1) + 2iy}{(x + 1) + iy} \]

### Step 2: Simplify the expression
The expression can be rewritten as:
\[ \frac{(2x - 1) + 2iy}{(x + 1) + iy} \]

### Step 3: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((2x - 1) + 2iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)} \]
The denominator simplifies to:
\[ (x + 1)^2 + y^2 \]

### Step 4: Expand the numerator
Now, expand the numerator:
\[ (2x - 1)(x + 1) - (2y^2) + i \left( 2y(x + 1) - y(2x - 1) \right) \]
This gives us:
\[ (2x^2 + 2x - x - 1 - 2y^2) + i(2yx + 2y - 2xy + y) \]
Thus, the numerator becomes:
\[ (2x^2 + x - 1 - 2y^2) + i(3y) \]

### Step 5: Set the real part equal to 1
From the problem statement, we know that the real part of the fraction equals 1:
\[ \frac{2x^2 + x - 1 - 2y^2}{(x + 1)^2 + y^2} = 1 \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 2x^2 + x - 1 - 2y^2 = (x + 1)^2 + y^2 \]
Expanding the right-hand side:
\[ 2x^2 + x - 1 - 2y^2 = x^2 + 2x + 1 + y^2 \]
Rearranging terms gives:
\[ x^2 - x - 2y^2 - y^2 - 2 = 0 \]
This simplifies to:
\[ x^2 - x - 3y^2 - 2 = 0 \]

### Step 7: Rearranging to identify the locus
Rearranging gives:
\[ x^2 - x + 3y^2 + 2 = 0 \]
This represents a conic section. Completing the square for \( x \):
\[ \left( x - \frac{1}{2} \right)^2 + 3y^2 = \frac{1}{4} - 2 \]
This leads to:
\[ \left( x - \frac{1}{2} \right)^2 + 3y^2 = -\frac{7}{4} \]
Since the right side is negative, this indicates that there are no real solutions, implying that the locus is empty.

### Final Answer
Thus, the locus of \( z \) is an empty set.