Question

Let s={\(z=x+ry:\frac{2z-3i}{4z+2i}\) is a real number}. Then which of the following is not correct?

• A. x=0
• B. \((x,y)=(0,-\frac{1}{2})\)
• C. \(y+x^2+y^2 ≠-\frac{1}{4}\)
• D. \(y∈(-∞,-\frac{1}{2})∪(-\frac{1}{2},-∞)\)

Answer 1

The Correct Option is B

We are given the equation: \[ \frac{2z - 3i}{4z + 2i} \text{ is a real number}. \] Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Substituting this into the given equation: \[ \frac{2(x + iy) - 3i}{4(x + iy) + 2i} = \frac{2x + 2iy - 3i}{4x + 4iy + 2i}. \] Now simplify the numerator and denominator: \[ \frac{2x + (2y - 3)i}{4x + (4y + 2)i}. \] For the expression to be a real number, the imaginary part must be zero. So, we equate the imaginary part to zero: \[ \text{Imaginary part: } (2y - 3)(4x - (4y + 2)) = 0. \] This gives us two cases: 1. \( 2y - 3 = 0 \Rightarrow y = \frac{3}{2} \) 2. \( 4x - (4y + 2) = 0 \Rightarrow x = \frac{y + \frac{1}{2}}{2} \) Since \( x = 0 \) from the real part, we substitute into the second equation: \[ 4(0) - (4y + 2) = 0 \Rightarrow y = -\frac{1}{2}. \] Thus, \( x = 0 \) and \( y = -\frac{1}{2} \). 
Step 2: Now, let's check which of the given options is not correct. - Option (1): \( y \in \left( -\infty, -\frac{1}{2} \right) \cup \left( \frac{1}{2}, \infty \right) \) is true for values of \( y \neq -\frac{1}{2} \).
- Option (2): \( (x, y) = (0, -\frac{1}{2}) \) is true.
- Option (3): \( x = 0 \) is correct.
- Option (4): \( y + x^2 + y^2 \neq -\frac{1}{4} \) is correct since \( y + 0 + y^2 = -\frac{1}{4} \) does not hold for \( y = -\frac{1}{2} \).
Thus, the incorrect statement is option (2).