Understanding the definition of domain in Complex Analysis I have a definition in my book which states, "a nonempty open set that is connected is called a domain." I understand what an open set is (a set containing none of its boundary points and I know what a boundary point is). I am a bit confused with the definition of connected.
Ex. |z-3+2i| $\ge$ 1. We can translate this to (x-3)^2+(y+2)^2 $\ge$ 1. Is this not a domain because this set contains the boundary points of the circle centered at (3,-2)?
 A: Formally, connected means that we cannot break the domain up into two disjoint non-empty open sets. The picture you should have in mind is a region that is "all one piece."
So the example you gave is connected (it's the entire plane except for an open disk of radius $1$ around the point $3 - 2i$) but it's not a domain, since it's not open. 
A: In this situation a domain is also path-connected (this means that given any two points in the domain you can connect them by a path that stays in the domain). So the intuitive picture that you should is to draw a curve that a.) does not intersect itself b.) cuts the plane into 2 pieces.
If you think about it there are two ways to do this.
1.) Draw a closed loop.
2.) Draw a curve such that both ends go off to infinity at some point
(Note: there are some other issues like drawing a curve that looks like a number 6 but that actually work for what I am saying, and to get rid of this case you can only consider curves that extend to embeddings of $S^1$ in the Riemann sphere)
Then if we look at the two parts of the plane and DON'T include the curve these will be domains. This is the type of picture you should have in your head. Though not all domains look exactly like this, they are pretty close.
A: I think the deep meaning of "Domain $D\subset \mathbb{C}$" is that you cannot write $D=D_1\sqcup D_2, D_i\not=\emptyset$ (disjoint union). Indeed, if $D\subset \mathbb{C}$ is only open, consider the algebra $\mathcal{H}(D)$ of analytic functions on it. This algebra is a domain (in the algebraic sense, i.e. without zero divisor) iff $D$ is connected as if $D=D_1\sqcup D_2, D_i\not=\emptyset$, we have $0_D=1_{D_1}.1_{D_2}$ and $1_{D_i}\not=0$. Hope it helps. 
