JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let the complex number \(z=x+\) iy be such that \(\frac{2 z-3 i}{2 z+i}\) is purely imaginary. If \(x + y ^2=0\), then \(y^4+y^2-y\) is equal to :
- A \(\frac{3}{2}\)
- B \(\frac{4}{3}\)
- C \(\frac{2}{3}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
\(\frac{2 z-3 i}{2 z+i}\) is purely imaginary \(\therefore \frac{2 z-3 i}{2 z+i}+\frac{2 \bar{z}+3 i}{2 \bar{z}-i}=0\) \(z=x+i y\) \(\Rightarrow 4 x^2+4 y^2-4 y-3=0\) Given that \(x+y^2=0\) \(y^4+y^2-y=3 / 4\)
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