Question:
The locus of z satisfying the inequality \( \left| \frac{z+2i}{2z+i} \right| < 1 \), where \( z = x + iy \), is
The locus of z satisfying the inequality \( \left| \frac{z+2i}{2z+i} \right| < 1 \), where \( z = x + iy \), is
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$|z-a|<|z-b|$ represents a half-plane or circle depending on the coefficients of $z$.
Updated On: Apr 10, 2026
- $x^{2}+y^{2}<1$
- $x^{2}-y^{2}<1$
- $x^{2}+y^{2}>1$
- $2x^{2}+3y^{2}<1$
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The Correct Option is C
Solution and Explanation
Step 1: Modulus Expression
$|z+2i|<|2z+i|$.
Step 2: Real and Imaginary Parts
$|x + i(y+2)|<|2x + i(2y+1)|$.
Step 3: Square Both Sides
$x^2 + (y+2)^2<(2x)^2 + (2y+1)^2 \Rightarrow x^2 + y^2 + 4y + 4<4x^2 + 4y^2 + 4y + 1$.
Step 4: Simplify
$3x^2 + 3y^2>3 \Rightarrow x^2 + y^2>1$.
Final Answer: (c)
$|z+2i|<|2z+i|$.
Step 2: Real and Imaginary Parts
$|x + i(y+2)|<|2x + i(2y+1)|$.
Step 3: Square Both Sides
$x^2 + (y+2)^2<(2x)^2 + (2y+1)^2 \Rightarrow x^2 + y^2 + 4y + 4<4x^2 + 4y^2 + 4y + 1$.
Step 4: Simplify
$3x^2 + 3y^2>3 \Rightarrow x^2 + y^2>1$.
Final Answer: (c)
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