Question

The locus of z satisfying the inequality $|\frac{z+2i}{2z+i}|<1$, where $z=x+iy$, is

• A. $x^2+y^2<1$
• B. $x^2-y^2<1$
• C. $x^2+y^2>1$
• D. $2x^2+3y^2<1$

Answer 1

Answer: (C)

$x^2+y^2>1$

Given inequality:

$|\frac{z+2i}{2z+i}|<1$ and $z=x+iy$

Hence,

$⟹\frac{|x+iy+2i|}{|2(x+iy)+i|}<1$

$⟹\frac{|x+(2+y)i|}{|2x+(2y+1)i|}<1$

$⟹|x+(2+y)i|<|2x+(2y+1)i|$

$⟹\sqrt{x^2+(2+y{)}^2}<\sqrt{(2x{)}^2+(2y+1{)}^2}$

Squaring both sides:

$x^2+4+y^2+4y<4x^2+4y^2+4y+1$

$⟹3x^2+3y^2>3$

$⟹x^2+y^2>1$ is required locus.