Inequalities have always gotten me for whatever reason
$|z+i| \le 3$
well, $z=x+iy$
so it becomes $|x+iy+i| \le 3 $ which is $|x+i(y+1)| \le 3$
$\sqrt{x^2+(y+1)^2} \le 3$
$-9 \le x^2 + y^2 + 2y +1 \le 9$
$-9 - 2y -1 \le x^2 +y^2 \le 9 - 2y -1$ I think Im just complicating it
$|z+i| \le 3$
well, $z=x+iy$
so it becomes $|x+iy+i| \le 3 $ which is $|x+i(y+1)| \le 3$
$\sqrt{x^2+(y+1)^2} \le 3$
$-9 \le x^2 + y^2 + 2y +1 \le 9$
$-9 - 2y -1 \le x^2 +y^2 \le 9 - 2y -1$ I think Im just complicating it