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 find the locus of the complex number \( z \) given the equation \( \text{Im}\left(\frac{2z + 1}{iz + 1}\right) = -2 \), we will follow these steps:

### Step 1: Substitute \( z \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can express \( 2z + 1 \) and \( iz + 1 \) as follows:
\[ 2z + 1 = 2(x + iy) + 1 = 2x + 1 + 2iy \]
\[ iz + 1 = i(x + iy) + 1 = ix - y + 1 = 1 - y + ix \]

### Step 2: Write the expression
Now we can write the expression:
\[ \frac{2z + 1}{iz + 1} = \frac{2x + 1 + 2iy}{1 - y + ix} \]

### Step 3: Multiply by the conjugate
To simplify this expression, multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(2x + 1 + 2iy)(1 - y - ix)}{(1 - y + ix)(1 - y - ix)} \]

### Step 4: Simplify the denominator
The denominator simplifies as follows:
\[ (1 - y)^2 + x^2 \]

### Step 5: Expand the numerator
Now, expand the numerator:
\[ (2x + 1)(1 - y) + (2iy)(1 - y) - (2x + 1)(ix) - (2iy)(ix) \]
This gives:
\[ (2x + 1 - 2xy - y) + i(2y - 2x - x(2y + 1)) \]

### Step 6: Collect real and imaginary parts
The real part is:
\[ 2x + 1 - 2xy - y \]
The imaginary part is:
\[ 2y - 2x - x(2y + 1) \]

### Step 7: Set the imaginary part equal to -2
From the problem statement, we have:
\[ \text{Im}\left(\frac{2z + 1}{iz + 1}\right) = -2 \]
Thus, we set:
\[ 2y - 2x - x(2y + 1) = -2 \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 2y - 2x - 2xy - x + 2 = 0 \]
This simplifies to:
\[ -2xy - 3x + 2y + 2 = 0 \]

### Step 9: Rearranging to find the locus
Rearranging further gives:
\[ 2y - 2xy - 3x + 2 = 0 \]
This is a linear equation in \( x \) and \( y \).

### Step 10: Identify the locus
The equation represents a straight line in the \( xy \)-plane.

### Final Result
The locus of \( z \) is a straight line.

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