The point z0 is called an accumulation point of the set S if, for each ϵ, the punctured disk Dϵ*(z0) contains at least one point of S. We ask you to show in the exercises that the set of accumu-lation points of D1(0) is D1(0), and that there is only one accumulation point of
S = ⅈ
n : n = 1,2, ..., namely, the point 0.
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 open set if every point of S is an interior point of S. Thus, Example 1.25 shows that D1(0) is open. A set S s 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 z1andz2 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
Chapter01Section06.nb 5
D1(0) = {z : z || < 1} is a connected open set. We ask you to verify in the exercises that, if z1andz2 lie in D1(0), then the straight-line segment joining them lies entirely in D1(0). The annu-lus A = {z : 1 < z <2} is a connected open set because any two points in A can be joined by a curve C that lies entirely in A, as shown in Figure 1.26. The set
B = {z : z + 2 <1 or z - 2 <1} consists of two disjoint disks. We leave it as an exercise for you to show that the set is not connected, as shown in Figure 1.27.
Figure 1.26 The annulus A = {z : 1 < z <2} is a connected set.
Figure 1.27 The set B = {z : z + 2 <1 or z - 2 <1} is not a connected set.
We call a connected open set a domain. In the exercises we ask you to show that the open unit disk D1(0) = {z : z || < 1} is a domain and that the closed unit disk
D1(0) = {z : z || ≤ 1} is not a domain. The term domain is a noun and is a type of set. In Chapter 2 we note that it also refers to the set of points on which a function is defined. In the latter context, it does not necessarily mean a connected open set.
Example 1.25. Show that the right half-plane H = {z : Re (z) > 0} is a domain.
Solution. First we show that H is connected. Let z0andz1 be any two points in H. We claim the obvious, that the straight-line segment C : z (t) = z0 + (z1-z0)t, for 0 ≤ t ≤ 1, lies entirely within H. To prove this claim, we let z (t*) = z0 + (z1-z0)t*, for some t* ∈ [0, 1], be an arbitrary point on C. We must show that Re[z (t*)] >0. Now,
(1 - 53) Re[z (t*)] = Re[z0 + (z1-z0)t*]
= Re[ (1 - t*)z0 + t*z1]
= Re[ (1 - t*)z0] + Re[t*z1]
= (1 - t*)Re[z0] + t*Re[z1]
If t = 0, the last expression becomes Re[z0], which is greater than zero because z0∈H. Likewise, if t = 1, then Equation (1-53) becomes Re[z1], which also is positive. Finally, if 0 ≤ t ≤ 1, then each term in Equation (1-53) is positive, so in this case we also have Re[z (t*)] >0.
To show that H is open, we suppose without loss of generality that Re[z0] <Re[z1]. We claim that Dϵ(z0) ⊂H, where ∈ = Re[z0]. We leave the proof of this claim as an exercise.
A domain, together with some, none, or all its boundary points, is called a region. For example, the horizontal strip {z : 1 < Im (z)} is a region. A set formed by taking the union of a domain and its boundary is called a closed region; thus {z : 1 ≤ Im (z) ≤ 2} is a closed region. A set S is said to be a bounded set if it can be completely contained in some closed disk, that is, if there exists an R > 0 such that for each z in S we have z ≤ R. The rectangle given by {z : x ≤4 and y ≤3} is bounded because it is contained inside the disk
D5(0) = {z : z ≤ 5}. A set that cannot be enclosed by any closed disk is called an unbounded set.
We mentioned earlier that a simple closed curve is positively oriented if its interior is on the left when the curve is traversed. How do we know, though, that any given simple closed curve will have an interior and exterior? Theorem 1.6 guarantees that this is indeed the case. It is due in part to the work of the French mathematician Marie Ennemond Camille Jordan (1838--1922).
Theorem 1.5 (Jordan Curve Theorem). The compliment of any simple closed curve C can be partitioned into two mutually exclusive domains I and E in such a way that I is bounded, E is unbounded, and C is the boundary for both I and E. In addition I ⋃ E ⋃ C is the entire complex plane. The domain I is called the interior of C, and the domain E is called the exterior of C.
The Jordan curve theorem is a classic example of a result in mathematics that seems obvious but is very hard to demonstrate, and its proof is beyond the scope of this book. Jordan's original argument, in fact, was inadequate, and not until 1905 was a correct version finally given by the American topologist Oswald Veblen (1880-1960). The difficulty lies in describing the interior and exterior of a simple closed curve analytically, and in showing that they are connected sets. For example, in which domain (interior or exterior) do the two points depicted in Figure 1.28 lie? If they are in the same domain, how, specifically, can they be connected with a curve? If you appreciated the subtleties involved in showing that the right half-plane of Example 1.26 is connected, you can begin to appreciate the obstacles that Veblen had to navigate.
Chapter01Section06.nb 7
Figure 1.28 Are the points z1(blue),and z2(red) in the interior or exterior of this simple closed curve?
Exploration
Although an introductory treatment of complex analysis can be given without using this theorem, we think it is important for the well-informed student at least to be aware of it.
Complex Analysis for Mathematics and Engineering by
John H. Mathews and Russell W. Howell Jones and Bartlett Learning
The last chapter developed a basic theory of complex numbers. For the next few chapters we turn our attention to functions of complex numbers. They are defined in a similar way to func-tions of real numbers that you studied in calculus; the only difference is that they operate on com-plex numbers rather than real numbers. This chapter focuses primarily on very basic functions, their representations, and properties associated with functions such as limits and continuity. You will learn some interesting applications as well as some exciting new ideas.
2.1 Functions and Linear Mappings
A complex-valued function f of the complex variable z is a rule that assigns to each complex number z in a set D one and only one complex number w. We write w = f (z) and call w the image of z under f. A simple example of a complex-valued function is given by the formula
w = f (z) = z4. The set D is called the domain of f, and the set of all images {w = f (z) : z ∈ D}
is called the range of f. When the context is obvious, we omit the phrase complex-valued, and simply refer to a function f, or to a complex function f.
We can define the domain to be any set that makes sense for a given rule, so for w = f (z) = z4, we could have the entire complex plane for the domain D, or we might artificially restrict the domain to some set such as D = D1(0) = {z : z <1}. Determining the range for a function defined by a formula is not always easy, but we will see plenty of examples later on. In some contexts functions are referred to as mappings or transformations.
In Section 1.6, we used the term domain to indicate a connected open set. When speaking about the domain of a function, however, we mean only the set of points on which the function is defined. This distinction is worth noting, and context will make clear the use intended.
Just as z can be expressed by its real and imaginary parts, z = x + ⅈ y, we write
f (z) = w = v + ⅈ v, where u and v are the real and imaginary parts of w, respectively. Doing so gives us the representation
w = f (z) = f (x, y) = f (x + ⅈ y) = u + ⅈ v.
Because u and v depend on x and y, they can be considered to be real-valued functions of the real variables x and y; that is,
u = u (x, y) and v = v (x, y).
Combining these ideas, we often write a complex function f in the form
f (z) = f (x + ⅈ y) = u (x, y) + ⅈ v (x, y).
Figure 2.1 illustrates the notion of a function (mapping) using these symbols.
Figure 2.1 The mapping w = f (z) = u (x, y) + ⅈ v (x, y).
There are two methods for defining a complex function in Mathematica.
Exploration.
We now give several examples that illustrate how to express a complex function.
Example 2.1. Write f (z) = z4 in the for f (z) = u (x, y) + ⅈ v (x, y).
Solution. Using the binomial formula, we obtain
f (z) = f (x + ⅈ y) = (x + ⅈ y)4
= x4+4 x3(ⅈy) + 6 x2(ⅈy)2+4 x (ⅈ y)3+ (ⅈy)4
= x4+4 ⅈ x3y - 6 x2y2-4 ⅈ x y3+y4
= x4-6 x2y2+y4 + ⅈ (4 x3y - 4 x y3)
= u (x, y) + ⅈ v (x, y)
so that u (x, y) = x4-6 x2y2+y4 and v (x, y) = 4 x3y - 4 x y3.