JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 1)
Let $$S = \left\{ {z = x + iy:{{2z - 3i} \over {4z + 2i}}\,\mathrm{is\,a\,real\,number}} \right\}$$. Then which of the following is NOT correct?
$$y + {x^2} + {y^2} \ne - {1 \over 4}$$
$$(x,y) = \left( {0, - {1 \over 2}} \right)$$
$$x = 0$$
$$y \in \left( { - \infty , - {1 \over 2}} \right) \cup \left( { - {1 \over 2},\infty } \right)$$
Explanation
Given that $z=x+i y$
$$ \begin{aligned} & \text { then } \frac{2 z-3 i}{4 z+2 i}=\frac{2(x+i y)-3 i}{4(x+i y)+2 i} \\\\ & =\frac{2 x+i(2 y-3)}{4 x+i(4 y+2)} \times \frac{4 x-i(4 y+2)}{4 x-i(4 y+2)} \\\\ & =\frac{8 x^2+(2 y-3)(4 y+2)}{(4 x)^2+(4 y+2)^2}+i\left(\frac{4 x(2 y-3)-2 x(4 y+2)}{(4 x)^2+(4 y+2)^2}\right) \end{aligned} $$
Since, $\frac{2 z-3 i}{4 z+2 i}$ is Real $\Rightarrow \operatorname{Img}\left(\frac{2 z-3 i}{4 z+2 i}\right)=0$
$$ \begin{aligned} & \Rightarrow 4 x(2 y-3)-2 x(4 y+2)=0 \\\\ & \Rightarrow 2 x(4 y-6-4 y-2)=0 \\\\ & \Rightarrow 2 x(-8)=0 \Rightarrow x=0 \end{aligned} $$
Also, $(4 x)^2+(4 y+2)^2 \neq 0 \Rightarrow y+x^2+y^2 \neq \frac{-1}{4}$
If $x=0$, then $y \neq-\frac{1}{2}$
$$ \begin{aligned} & \text { then } \frac{2 z-3 i}{4 z+2 i}=\frac{2(x+i y)-3 i}{4(x+i y)+2 i} \\\\ & =\frac{2 x+i(2 y-3)}{4 x+i(4 y+2)} \times \frac{4 x-i(4 y+2)}{4 x-i(4 y+2)} \\\\ & =\frac{8 x^2+(2 y-3)(4 y+2)}{(4 x)^2+(4 y+2)^2}+i\left(\frac{4 x(2 y-3)-2 x(4 y+2)}{(4 x)^2+(4 y+2)^2}\right) \end{aligned} $$
Since, $\frac{2 z-3 i}{4 z+2 i}$ is Real $\Rightarrow \operatorname{Img}\left(\frac{2 z-3 i}{4 z+2 i}\right)=0$
$$ \begin{aligned} & \Rightarrow 4 x(2 y-3)-2 x(4 y+2)=0 \\\\ & \Rightarrow 2 x(4 y-6-4 y-2)=0 \\\\ & \Rightarrow 2 x(-8)=0 \Rightarrow x=0 \end{aligned} $$
Also, $(4 x)^2+(4 y+2)^2 \neq 0 \Rightarrow y+x^2+y^2 \neq \frac{-1}{4}$
If $x=0$, then $y \neq-\frac{1}{2}$
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