If `"Im"(2z+1)/(iz+1)=-2`, then locus of z, is

Options

A. a circle
B. a parabola
C. a straight line
D. none of these

Answer

To solve the problem, we need to find the locus of the complex number \( z \) given that the imaginary part of \( \frac{2z + 1}{iz + 1} = -2 \).

### Step-by-Step Solution:

1. **Express \( z \) in terms of \( x \) and \( y \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

2. **Substitute \( z \) into the equation**:
Substitute \( z \) into the expression:
\[ 2z + 1 = 2(x + iy) + 1 = 2x + 1 + 2iy \]
\[ iz + 1 = i(x + iy) + 1 = ix - y + 1 = 1 - y + ix \]

3. **Form the fraction**:
Now, we can write:
\[ \frac{2z + 1}{iz + 1} = \frac{2x + 1 + 2iy}{1 - y + ix} \]

4. **Multiply numerator and denominator by the conjugate of the denominator**:
The conjugate of the denominator \( 1 - y + ix \) is \( 1 - y - ix \). Thus, we have:
\[ \frac{(2x + 1 + 2iy)(1 - y - ix)}{(1 - y)^2 + x^2} \]

5. **Expand the numerator**:
Expanding the numerator:
\[ (2x + 1)(1 - y) + (2iy)(1 - y) - (2x + 1)(ix) - (2iy)(ix) \]
This gives:
\[ (2x + 1 - 2xy - y) + i(2y - 2x - 2y^2) \]

6. **Separate real and imaginary parts**:
The real part is \( 2x + 1 - 2xy - y \) and the imaginary part is \( 2y - 2x - 2y^2 \).

7. **Set the imaginary part equal to -2**:
According to the problem, the imaginary part equals -2:
\[ 2y - 2x - 2y^2 = -2 \]

8. **Rearrange the equation**:
Rearranging gives:
\[ 2y - 2x - 2y^2 + 2 = 0 \]
Dividing through by 2:
\[ y - x - y^2 + 1 = 0 \]
Rearranging gives:
\[ y^2 - y + x - 1 = 0 \]

9. **Identify the locus**:
This is a quadratic equation in \( y \). The locus of \( z \) is a parabola.

### Final Answer:
The locus of \( z \) is given by the equation:
\[ y^2 - y + x - 1 = 0 \]