Complex Number Inequality $|2+z|\leq 2$ If I have a complex number $z = a + ib$, how do you interpret the inequality 
$|2+z|\leq 2$?
I believe the answer is a circle in the complex plane, but where I am getting confused is understanding the inequality due to the use of imaginary numbers on the left hand side and no imaginary numbers on the right hand side.
Thanks in advance.
 A: It is a disc, centered at point $-2$ on the complex plane and with radius $2$. The non existence of imaginary numbers in the inequality is only natural: there is no "order" in $\mathbb{C}$, you cannot compare two complex numbers. No such relation is defined. If you want to get some more intuition on that stuff, try expressing complex numbers as numbers of $\mathbb{R}^2$. How would you write this equation for $z=x+iy$? simply as $\sqrt{(x+2)^2+y^2}\leq2$, which is precisely the object we described: a disc center at point $(-2,0)$ and of radius $2$.
A: Write this as $|z-(-2)| \leq 2$. In that case, this would mean that the distance of your point z from the point (-2,0) is always less than or equal to 2. Or, make the circle centered at (-2,0) radius 2, this gives you all points on the boundary and inside the circle.
A: $$|2+z|\leq 2 \iff |z-(-2)|\leq 2 \iff d(z, -2)\leq 2$$
That is the closed ball with center at $-2$ and radius $2$ 
A: I will ake for granted that you know the equation of the circle.
We know that $z=x+iy$, hence collecting real and imaginary parts together we have
$$\left|[x-(-2)]+iy\right|\leq2$$
or
$$(x+2)^2+y^2\leq4$$
Clearly this represents a closed disk of radius $2$ with a centre $(-2,0)$.
