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

To determine which option is not correct, we need to evaluate the given expression and check for conditions under which it is a real number. We are given:

Let \( z = x + ry \). We need to find when \(\frac{2z - 3i}{4z + 2i}\) is real.

1. First, express \( z \) as \( z = x + yi \) since \( r \) denotes the imaginary unit.
2. Substitute \( z = x + yi \) into the expression: \[ \frac{2(x + yi) - 3i}{4(x + yi) + 2i} = \frac{2x + 2yi - 3i}{4x + 4yi + 2i} \]
3. Simplify the numerator and the denominator: \[ \text{Numerator: } 2x + (2y - 3)i \] \[ \text{Denominator: } 4x + (4y + 2)i \]
4. The expression is real when its imaginary part is zero. Set the imaginary part to zero: \[ \frac{(2y - 3)(4x) - (2x)(4y + 2)}{(4x)^2 + (4y + 2)^2} = 0 \]
5. Solving the condition \((2y - 3)(4x) = (2x)(4y + 2)\) gives: \[ 8xy - 12x = 8xy + 4x \] \] \] \(-16x = 0\)
6. This implies \( x = 0 \). Substitute \( x = 0 \) into the original expression: \[ \frac{2(0) + (2y - 3)i}{(4y + 2)i} = \frac{(2y - 3)i}{(4y + 2)i} \] \] \] \] \] \] \] \]
7. For the above to be real, the coefficients of \( i \) should balance, providing: \[ 2y - 3 = -(4y + 2) \] \] \text{Solving gives } \( 6y = -1 \), so \( y = -\frac{1}{2} \).
8. Thus, the option \((x, y) = (0, -\frac{1}{2})\) is not a valid case because it contradicts with set conditions for \( z \) leading to a real number.
9. Therefore, the incorrect option is: \((x, y) = (0, -\frac{1}{2})\).

The correct answer is the second option, \((x, y) = (0, -\frac{1}{2})\), as it leads the expression to an undefined imaginary division condition contrary to it being real.