Question

The locus of $Z$ satisfying the inequality $\left|\frac{z+2i}{2z+i}\right|<1$, 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

The correct option is C

$x^2+y^2>1$

Explanation of the correct option.

Compute the locus:

Given : $\left|\frac{z+2i}{2z+i}\right|<1$

It can be written as, $\left|z+2i\right|<\left|2z+i\right|$

Substitute $z=x+iy$

$\left|x+iy+2i\right|<\left|2(x+iy)+i\right|$

$\Rightarrow$ $\left|x+i(y+2)\right|<\left|2x+i(2y+1)\right|$

$\Rightarrow$ $\sqrt{x^2+{(y+2)}^2}<\sqrt{4x^2+{(2y+1)}^2}$

$\Rightarrow$ $x^2+{(y+2)}^2<4x^2+{(2y+1)}^2$

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

$\Rightarrow$ $3x^2+3y^2>3$

$\Rightarrow$ $x^2+y^2>1$

Hence option $\left(C\right)$ is the correct option.