If the imaginary part of `(2z + 1)/(iz + 1)` is -4, then the locus of the point representing z in the complex plane is

Options

A. a straight line
B. a parabola
C. a circle
D. an ellipse

Answer

To solve the problem, we need to find the locus of the point representing \( z \) in the complex plane given that the imaginary part of \( \frac{2z + 1}{iz + 1} \) is equal to -4. Let's denote \( z \) as \( x + iy \), where \( x \) is the real part and \( y \) is the imaginary part.

### Step-by-Step Solution:

1. **Substituting \( z \)**:
\[ z = x + iy \]
Substitute \( z \) into the expression:
\[ \frac{2z + 1}{iz + 1} = \frac{2(x + iy) + 1}{i(x + iy) + 1} \]
Simplifying the numerator:
\[ 2z + 1 = 2x + 2iy + 1 = (2x + 1) + 2iy \]
Simplifying the denominator:
\[ iz + 1 = i(x + iy) + 1 = ix - y + 1 = (1 - y) + ix \]

2. **Writing the expression**:
Now we have:
\[ \frac{(2x + 1) + 2iy}{(1 - y) + ix} \]

3. **Rationalizing the denominator**:
Multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((2x + 1) + 2iy)((1 - y) - ix)}{((1 - y) + ix)((1 - y) - ix)} \]
The denominator simplifies to:
\[ (1 - y)^2 + x^2 \]

4. **Expand the numerator**:
Expanding the numerator:
\[ (2x + 1)(1 - y) - (2iy)(ix) + 2iy(1 - y) - (2x + 1)ix \]
This results in:
\[ (2x + 1)(1 - y) + 2y - 2x^2 - 2y^2 - x \]
Combine the terms to separate real and imaginary parts.

5. **Finding the imaginary part**:
The imaginary part of the fraction is given as:
\[ \frac{2y - 2x^2 - 2y^2 - x}{(1 - y)^2 + x^2} = -4 \]
Cross-multiplying gives:
\[ 2y - 2x^2 - 2y^2 - x = -4((1 - y)^2 + x^2) \]

6. **Expanding and rearranging**:
Expanding the right side:
\[ -4(1 - 2y + y^2 + x^2) = -4 + 8y - 4y^2 - 4x^2 \]
Rearranging gives:
\[ 2y - 2x^2 - 2y^2 - x + 4 - 8y + 4y^2 + 4x^2 = 0 \]
Combine like terms:
\[ 2y^2 + 2x^2 - 6y - x + 4 = 0 \]

7. **Final equation**:
Dividing through by 2:
\[ y^2 + x^2 - 3y - \frac{x}{2} + 2 = 0 \]
Rearranging gives:
\[ x^2 + y^2 - \frac{x}{2} - 3y + 2 = 0 \]

8. **Identifying the locus**:
This is the equation of a circle. To see this, we can complete the square for both \( x \) and \( y \).

### Conclusion:
The locus of the point representing \( z \) in the complex plane is a circle.

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