open sets, connectedness and domains
Before you can talk about a complex function being continuous, differentiable, or analytic, you have to say WHERE it lives — on what part of the plane. A few pieces of vocabulary set that stage. A neighborhood of a point z_0 is a small open disk around it, all the points z with |z - z_0| < r for some radius r > 0. A set is open if it is 'all interior, no edge': around every point of the set there is a whole little neighborhood still inside the set. The open disk |z| < 1 is open; the closed disk |z| <= 1 is not, because points right on the rim have neighbors poking outside.
A set is connected, loosely, if it is in one piece — you can travel between any two of its points without leaving the set (for open sets in the plane, you can even do it along a path of straight segments). The word domain has a precise technical meaning in complex analysis: a domain is a set that is BOTH open AND connected. (Beware — in everyday math 'domain' also just means the input set of a function, so context matters; here it is this open-and-connected region.) An open disk, the upper half-plane, and an annulus are all domains; two separate disks together are open but not connected, so not a domain.
Why fuss over this? Because the central theorems of complex analysis are statements about functions on a domain, and they genuinely need both properties. Openness ensures every point has room around it for a derivative or a limit to make sense (no awkward boundary points). Connectedness is what makes uniqueness work: the identity theorem says a function determined on a tiny piece of a domain is forced everywhere on it — but only because the domain is all one connected piece. Get these foundations right and the powerful results later stand on solid ground.
The set {z : 1 < |z| < 2} (an annulus) is a domain: it is open (no point sits on either rim) and connected (you can walk around the ring from any point to any other). But {z : |z| < 1 or |z - 5| < 1} — two separate disks — is open yet not connected, so it is not a domain.
A domain = open AND connected. Two disconnected open pieces do not qualify.
Watch the word 'domain': in complex analysis it specifically means an open connected set, but in general usage it just means a function's input set. The two readings usually agree because we define complex functions on such regions, but the precise sense is the open-and-connected one.