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

$\left|\frac{z+2i}{2z+i}\right|<1\Rightarrow \left|z+2i\right|<\left|2z+i\right|\Rightarrow \left|x+iy+2i\right|<\left|2\left(x+iy\right)+i\right|$ 

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

$\Rightarrow x^2+y^2+4y+4<4x^2+4y^{2}4y+1\Rightarrow 3x^2+3y^2>3\Rightarrow x^2+y^2>1.$