TS EAMCET · Maths · Complex Number
The locus of \(z\) satisfying the inequality \(\left|\frac{z+2 i}{2 z+i}\right| < 1\), where \(z=x+i y\), is
- A \(x^2+y^2 < 1\)
- B \(x^2-y^2 < 1\)
- C \(x^2+y^2>1\)
- D \(2 x^2+3 y^2 < 1\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2>1\)
Step-by-step Solution
Detailed explanation
Let \(z=x+i y\) Given, \(\quad\left|\frac{z+2 i}{2 z+i}\right| 3 \\ \Rightarrow & x^2+y^2>1\end{array}\)
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