JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\quad S=\left\{z=x+i y: \frac{2 z-3 i}{4 z+2 i}\right.\) is a real number \(\}\). Then which of the following is NOT correct?
- A \(y + x ^2+ y ^2 \neq-\frac{1}{4}\)
- B \(x=0\)
- C \(( x , y )=\left(0,-\frac{1}{2}\right)\)
- D \(y \in\left(-\infty,-\frac{1}{2}\right) \cup\left(-\frac{1}{2}, \infty\right)\)
Answer & Solution
Correct Answer
(C) \(( x , y )=\left(0,-\frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{2 z-3 i}{q z+2 i} \in R\) \(\frac{2(x+i y)-3 i}{4(x+i t)+2 i}=\frac{2 x+(2 y-3) i}{4 x+(4 y+2) i} \times \frac{4 x-(4 y+2) i}{4 x-(4 y+2) i}\) \(4 x(2 y-3)-2 x(4 y+2)=0\) \(x=0 \quad y \neq-\frac{1}{2}\) \(\text { Ans. }=3\)
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