`Re((z+4)/(2z-1)) = 1/2`, then z is represented by a point lying on

Options

A. A circle
B. An ellipse
C. A straight line
D. No real locus

Answer

To solve the equation \( \text{Re}\left(\frac{z+4}{2z-1}\right) = \frac{1}{2} \), we will follow these steps:

### Step 1: Substitute \( z \) with \( x + iy \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Thus, we can rewrite the expression as:
\[ \frac{z+4}{2z-1} = \frac{(x + iy) + 4}{2(x + iy) - 1} = \frac{x + 4 + iy}{(2x - 1) + 2iy} \]

### Step 2: Separate the real and imaginary parts
Now, we separate the real and imaginary parts of the fraction:
\[ \frac{x + 4 + iy}{(2x - 1) + 2iy} \]
To do this, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(x + 4 + iy)((2x - 1) - 2iy)}{((2x - 1) + 2iy)((2x - 1) - 2iy)} \]

### Step 3: Simplify the denominator
The denominator simplifies to:
\[ (2x - 1)^2 + (2y)^2 = 4x^2 - 4x + 1 + 4y^2 \]

### Step 4: Expand the numerator
The numerator expands to:
\[ (x + 4)(2x - 1) - 2y^2 + i[(x + 4)(-2y) + 2y(2x - 1)] \]
This gives:
\[ (2x^2 + 8x - x - 4 - 2y^2) + i[-2xy - 8y + 4y] \]
Thus, the numerator becomes:
\[ (2x^2 + 7x - 4 - 2y^2) + i[-2xy - 4y] \]

### Step 5: Write the real part
The real part of the fraction is:
\[ \frac{2x^2 + 7x - 4 - 2y^2}{4x^2 - 4x + 1 + 4y^2} \]

### Step 6: Set the real part equal to \( \frac{1}{2} \)
Now, we set this equal to \( \frac{1}{2} \):
\[ \frac{2x^2 + 7x - 4 - 2y^2}{4x^2 - 4x + 1 + 4y^2} = \frac{1}{2} \]

### Step 7: Cross-multiply
Cross-multiplying gives:
\[ 2(2x^2 + 7x - 4 - 2y^2) = 4x^2 - 4x + 1 + 4y^2 \]
This simplifies to:
\[ 4x^2 + 14x - 8 - 4y^2 = 4x^2 - 4x + 1 + 4y^2 \]

### Step 8: Simplify the equation
Cancel \( 4x^2 \) and \( 4y^2 \) from both sides:
\[ 14x + 4x - 8 - 1 = 0 \]
This simplifies to:
\[ 18x - 9 = 0 \]

### Step 9: Solve for \( x \)
Thus, we find:
\[ 18x = 9 \implies x = \frac{1}{2} \]

### Step 10: Conclusion
Since \( x \) is constant, \( z \) can take any value of \( y \). Therefore, the points represented by \( z \) lie on the vertical line \( x = \frac{1}{2} \).