JEE Mathematics 2023 Question with Solution

Given: S = \left\{ z = x + iy : \frac{2z - 3i}{4z + 2i} is a real number \right\}

Find: Which statement is not correct.

Let

$$z = x + iy$$

Then

$$\frac{2z-3i}{4z+2i} = \frac{2x + (2y-3)i}{4x + (4y+2)i}$$

For a quotient $\frac{a+ib}{c+id}$ to be real, we need

$$bc - ad = 0$$

Here,

$$a = 2x, \quad b = 2y-3, \quad c = 4x, \quad d = 4y+2$$

So,

$$(2y-3)(4x) - (2x)(4y+2) = 0$$ $$8xy - 12x - 8xy - 4x = 0$$ $$-16x = 0$$

Hence,

$$x = 0$$

Also, the denominator must be non-zero:

$$4z + 2i \neq 0$$

With $x=0$ and $z = iy$,

$$4iy + 2i = 2i(2y+1) \neq 0$$

So,

$$y \neq -\frac{1}{2}$$

Thus the set $S$ is all points of the form

$$z = iy \quad \text{with} \quad y \neq -\frac{1}{2}$$

That means:

• $x = 0$ is correct.
• $y \in \left( -\infty, -\frac{1}{2} \right) \cup \left( -\frac{1}{2}, \infty \right)$ is correct.
• $(x,y) = \left(0,-\frac{1}{2}\right)$ is not allowed because it makes the denominator zero.
• If $x=0$, then $y + x^2 + y^2 = y + y^2$. The value $-\frac{1}{4}$ occurs only at $y=-\frac{1}{2}$, which is excluded, so $y + x^2 + y^2 \neq -\frac{1}{4}$ is correct.

Therefore, the incorrect statement is option B.