If `lm ((z-1)/(2z+1)) = -4` then locus of z is

Options

A. An ellipse
B. A parabola
C. A straight line
D. A circle

Answer

To solve the problem, we need to determine the locus of the complex number \( z \) such that the imaginary part of \( \frac{z-1}{2z+1} = -4 \).

### Step 1: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we have:
\[ z - 1 = (x - 1) + iy \]
\[ 2z + 1 = 2(x + iy) + 1 = (2x + 1) + 2iy \]

### Step 2: Substitute into the expression
Now substituting these into the expression:
\[ \frac{z - 1}{2z + 1} = \frac{(x - 1) + iy}{(2x + 1) + 2iy} \]

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

### Step 4: Calculate the denominator
The denominator simplifies to:
\[ (2x + 1)^2 + (2y)^2 = 4x^2 + 4x + 1 + 4y^2 \]

### Step 5: Calculate the numerator
The numerator expands as follows:
\[ (x - 1)(2x + 1) - 2y^2 + i[(x - 1)(-2y) + 2y(2x + 1)] \]
This results in:
\[ (2x^2 + x - 2x - 1 - 2y^2) + i[-2y(x - 1) + 4xy + 2y] \]
Simplifying gives:
\[ (2x^2 - x - 1 - 2y^2) + i[2y(2x - 1)] \]

### Step 6: Find the imaginary part
The imaginary part of \( \frac{z - 1}{2z + 1} \) is:
\[ \frac{2y(2x - 1)}{4x^2 + 4y^2 + 4x + 1} \]

### Step 7: Set the imaginary part equal to -4
Setting the imaginary part equal to -4 gives:
\[ \frac{2y(2x - 1)}{4x^2 + 4y^2 + 4x + 1} = -4 \]

### Step 8: Cross-multiply and simplify
Cross-multiplying yields:
\[ 2y(2x - 1) = -4(4x^2 + 4y^2 + 4x + 1) \]
This simplifies to:
\[ 2y(2x - 1) + 16x^2 + 16y^2 + 16x + 4 = 0 \]

### Step 9: Rearranging the equation
Rearranging gives:
\[ 16x^2 + 16y^2 + 2y(2x - 1) + 16x + 4 = 0 \]

### Step 10: Identify the locus
This equation represents a circle. Therefore, the locus of \( z \) is a circle.

### Conclusion
Thus, the locus of \( z \) is a circle.

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