1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

4 2 CHAPTER l • COMPLEX NUMBERS

The boundary of DR (zo) is the circle depicted in Figure 1.23. We denote
this circle CR. (zo) and refer to it as the circle of radius R centered a t zo.
Thus,

CR. (zo) = {z: iz -zol = R}. (1-52)

We use the notation C~ (zo) to indicate that the parametrization we chose for
this simple closed curve resulted in a positive orientation; Cfi (zo) denotes the
same circle, but with a negative orientation. (In both cases, counterclockwise
denotes the positive direction.) Using notation that we have already introduced,
we get Cfi (zo) = -C~ (zo).

• EXAMPLE 1.25 Let S = D 1 (0) = {z: izl < 1}. Find the interior, exterior,

and boundary of S.

Solution We show that every point of S is an interior point of S. Let zo be a

point of S. Then lzol < 1, and we can choose e = 1 - lzol > 0. We claim that

De (zo) ~ S. If z ED, (zo), then

lzl = lz - zo + zol :S lz - zol + izol < € + 12-0 1 = 1 - lzol + lzol = 1.

Hence the c neighborhood of z 0 is contained in S, which shows that zo is an

interior point of S. It follows that the interior of S is the set S itself.

Similarly, it can be shown that the exterior of S is the set {z: izl > 1}. The

boundary of Sis the unit circle C1 (0) = {z: lzl = 1}. This condition is true

because if zo = ei^8 o is any point on the circle, then any t: neighborhood of zo will

contain the point (1 - ~) ei^8 o, which belongs to S, and (1 + ~) ei^8 o, which does

not belong to S. We leave the details as an exercise.

The point zo is called an accumulation point of the set S if, for each t:,

the punctured disk D; (zo) contains at least one point of S. We ask you to show

in the exercises that the set of accumulation points of D 1 (O) is D 1 (O), and that

there is only one accumulation point of S = { ~ : n = 1, 2, ... } , namely, the

point O. We also ask you to prove that a set is closed if and only if it contains

all of its accumulation points.

A set S is called an o pen set if every point of S is an interior point of S.

Thus, Example (1.25) shows that D 1 (O) is open. A set Sis called a closed set

if it contains all its boundary points. A set S is said to be a connected set if

every pair of points z 1 and z 2 contained in S can be joined by a curve that lies

entirely in S. Roughly speaking, a connected set consists of a "single piece." The

unit disk D1(0) = {z: lzl < 1} is a connected open set. We ask you to verify in

the exercises that, if z 1 and z2 lie in D 1 (0), then the straight-line segment joining

them lies entirely in D1(0). The annulus A = {z: 1 < lzl < 2} is a connected

open set because any two points in A can be joined by a curve C that lies entirely