In Section 1.2 we saw that the complex numbers \(\C\text{,}\) which were initially defined algebraically, can be identified with the points in the Euclidean plane \(\R^2\text{.}\) In this section we collect some definitions and results concerning the topology of the plane.
In Proposition 1.2.3, we interpreted \(|z - w|\) as the distance between the complex numbers \(z\) and \(w\text{,}\) viewed as points in the plane. So if we fix a complex number \(a\) and a positive real number \(r\text{,}\) then all \(z \in
\C\) satisfying \(\abs{z-a}=r\) form the set of points at distance \(r\) from \(a\text{;}\) that is, this set is the circle with center \(a\) and radius \(r\text{,}\) which we denote by
\begin{equation*}
C[a,r] \ := \ \left\{ z \in \C : \, \abs{ z-a } = r \right\} \, .
\end{equation*}
A point \(b \in \C\) is a boundary point of \(G\) if every open disk centered at \(b\) contains a point in \(G\) and also a point that is not in \(G\text{.}\)
The idea is that if you don’t move too far from an interior point of \(G\) then you remain in \(G\text{;}\) but at a boundary point you can make an arbitrarily small move and get to a point inside \(G\) and you can also make an arbitrarily small move and get to a point outside \(G\text{.}\)
For \(r>0\) and \(a \in \C\text{,}\) the sets \(\left\{ z \in \C : \, |z-a| \lt r \right\} =
D[a,r]\) and \(\left\{ z \in \C : \, |z-a| > r \right\}\) are open. The closed disk
\begin{equation*}
\overline D[a,r] \ := \ \left\{ z \in \C : \, |z-a| \le r \right\}
\end{equation*}
A given set might be neither open nor closed. The complex plane \(\C\) and the empty set\(\emptyset\) are (the only sets that are) both open and closed.
The boundary\(\partial G\) of a set \(G\) is the set of all boundary points of \(G\text{.}\) The interior of \(G\) is the set of all interior points of \(G\text{.}\) The closure of \(G\) is the set \(G \cup \partial G\text{.}\)
One notion that is somewhat subtle in the complex domain is the idea of connectedness. Intuitively, a set is connected if it is “in one piece.” In \(\R\) a set is connected if and only if it is an interval, so there is little reason to discuss the matter. However, in the plane there is a vast variety of connected subsets.
Two sets \(X, Y \subseteq \C\) are separated if there are disjoint open sets \(A,B \subset \C\) so that \(X\subseteq A\) and \(Y\subseteq B\text{.}\) A set \(G \subseteq \C\) is connected if it is impossible to find two separated nonempty sets whose union is \(G\text{.}\) A region is a connected open set.
The idea of separation is that the two open sets \(A\) and \(B\) ensure that \(X\) and \(Y\) cannot just “stick together.” It is usually easy to check that a set is not connected. On the other hand, it is hard to use the above definition to show that a set is connected, since we have to rule out any possible separation.
The intervals \(X=[0,1)\) and \(Y=(1,2]\) on the real axis are separated: There are infinitely many choices for \(A\) and \(B\) that work; one choice is \(A=D[0,1]\) and \(B=D[2,1]\text{,}\) depicted in Figure 1.4.10. Hence \(X \cup Y =
[0,2]\setminus\listset1\) is not connected.
A path (or curve) in \(\C\) is a continuous function \(\gg \colon [a,b] \to \C\text{,}\) where \([a,b]\) is a closed interval in \(\R\text{.}\) We may think of \(\gg\) as a parametrization of the image that is painted by the path and will often write this parametrization as \(\gamma(t), \ a \le t \le b\text{.}\) The path is smooth if \(\gg\) is differentiable and the derivative \(\gg'\) is continuous and nonzero. 1
There is a subtlety here, because \(\gg\) is defined on a closed interval. For \(\gg: [a,b] \to \C\) to be smooth, we demand both that \(\gg'(t)\) exists for all \(a \lt t
\lt b\text{,}\) and that \(\lim_{ t \to a^+ } \gg'(t)\) and \(\lim_{ t \to b^- } \gg'(t)\) exist.
This definition uses the calculus notions of continuity and differentiability; that is, \(\gg \colon [a,b] \to \C\) being continuous means that for all \(t_0 \in [a,b]\)
Figure1.4.13.Two paths and their parametrizations; \(\gg_1\) is smooth and \(\gg_2\) is continuous and piecewise smooth (a term which we will define in Section 4.1).
even though both \(\gg_1\) and \(\gg_3\) yield the same picture: \(\gg_1\) features a counter-clockwise orientation, where as that of \(\gg_3\) is clockwise.
It is a customary and practical abuse of notation to use the same letter for the path and its parametrization. We emphasize that a path must have a parametrization, and that the parametrization must be defined and continuous on a closed and bounded interval \([a,b]\text{.}\) Since topologically we may identify \(\C\) with \(\R^2\text{,}\) a path can be specified by giving two continuous real-valued functions of a real variable, \(x(t)\) and \(y(t)\text{,}\) and setting \(\gg(t) = x(t) + i \, y(t)\text{.}\)
The path \(\gg: [a,b] \to \C\) is simple if \(\gg(t)\) is one-to-one, with the possible exception that \(\gg(a) = \gg(b)\) (in plain English: the path does not cross itself). A path \(\gg: [a,b] \to \C\) is closed if \(\gg(a) = \gg(b)\text{.}\)
As seems intuitively clear, any path is connected; however, a proof of this fact requires a bit more preparation in topology. The same goes for the following result, which gives a useful property of open connected sets.
If any two points in \(G \subseteq \C\) can be connected by a path in \(G\text{,}\) then \(G\) is connected. Conversely, if \(G \subseteq \C\) is open and connected, then any two points of \(G\) can be connected by a path in \(G\text{;}\) in fact, we can connect any two points of \(G\) by a chain of horizontal and vertical segments lying in \(G\text{.}\)
Here a chain of segments in \(G\) means the following: there are points \(z_0, z_1, \ldots, z_n\) so that \(z_k\) and \(z_{k+1}\) are the endpoints of a horizontal or vertical segment in \(G\text{,}\) for all \(k = 0, 1, \dots, n-1\text{.}\) (It is not hard to parametrize such a chain, so it determines a path.)
Consider the open unit disk\(D[0,1]\text{.}\) Any two points in \(D[0,1]\) can be connected by a chain of at most two segments in \(D[0,1]\text{,}\) and so \(D[0,1]\) is connected. Now let \(G = D[0,1] \setminus\listset0\text{;}\) this is the punctured disk obtained by removing the center from \(D[0,1]\text{.}\) Then \(G\) is open and it is connected, but now you may need more than two segments to connect points. For example, you need three segments to connect \(- \frac 1 2\) to \(\frac 1 2\) since we cannot go through \(0\text{.}\)
We remark that the second part of Theorem 1.4.16 is not generally true if \(G\) is not open. For example, circles are connected but there is no way to connect two distinct points of a circle by a chain of segments that are subsets of the circle. A more extreme example, discussed in topology texts, is the “topologist’s sine curve,” which is a connected set \(S \subset \C\) that contains points that cannot be connected by a path of any sort within \(S\text{.}\)
