QJEEM23Maa8ce: JEE Main 2023 Mathematics - Class 11 JEEM23Maa8ce Solution: Let . Then which of the following is NOT correct?

Question

Verified Solution • 2025–26 Session moderate

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?

A $y + {x^2} + {y^2} \ne - {1 \over 4}$
B $(x,y) = \left( {0, - {1 \over 2}} \right)$
C $x = 0$
D $y \in \left( { - \infty , - {1 \over 2}} \right) \cup \left( { - {1 \over 2},\infty } \right)$

Accepted Answer

Given that

z=x+i y

\begin{aligned}&amp; \text { then } \frac{2 z-3 i}{4 z+2 i}=\frac{2(x+i y)-3 i}{4(x+i y)+2 i} \\\\&amp; =\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)} \\\\&amp; =\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}&amp; \Rightarrow 4 x(2 y-3)-2 x(4 y+2)=0 \\\\&amp; \Rightarrow 2 x(4 y-6-4 y-2)=0 \\\\&amp; \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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