Math 461, Fall 2010
Chapter 2
Topology of the Complex Plane
This chapter concerns the basic topological ideas we require for our study of complex
analysis. Differentiation is naturally set against a background of limits and continuity, and these
are best dealt with on open sets (analogous to open intervals on the real line). On the other hand,
an integral computed from one complex number to another is computed along a specific,
possibly convoluted, path between the two numbers, rather than on a closed interval [a, b] as
when integrating on the real line. A set within which any two points can be joined by a path is
said to be connected. To be able to cope with both integration and differentiation in the simplest
possible manner later on, we shall restrict our complex functions to those defined on open
connected sets. Such a set is called a domain.
Domains can have exotic shapes and paths can wiggle around a great deal. To be able to
appeal to geometric intuition without our imagination having to work overtime, we will use a
carefully-conceived technical device called the Paving Lemma. We show in this lemma that a
path in a domain can always be subdivided into a finite number of smaller pieces in such a way
that each piece is contained in a disc which lies wholly within the domain (thus 'paving' the path
with discs). With the Paving Lemma in mind, we will then see that for any path in a domain
there is an alternative path between the same two points which is made up of a finite number of
straight segments. We can even insist that each segment is either horizontal or vertical, giving a
step-path in the domain.
By techniques such as this, we can use the Paving Lemma to illuminate complex analysis,
yielding fully rigorous analytic proofs linked firmly to geometric intuition.
Section 1: The standard topology (open and closed sets)
Definition. For a complex number z0 and a positive real number ε, we define the
ε-neighborhood of z0 to be the open disc in the complex plane which is centered at z0 and has
radius ε. That is, N ( z0 )  {z  C : | z  z0 |   } .
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Exercise 2.1. Draw the following neighborhoods in the complex plane. Shade the interior of
each disc, and (since the boundary is not included in the neighborhood) indicate each boundary
with a dashed line.
1) N1 (0)
2) N1/ 2 (i)
3) N 2 (1  i )
4) N1/ 5 (1)
Definition. A subset S  C is said to be open (specifically, open in C) provided that for every
z 0  S there exists a corresponding real number   0 such that N  ( z0 )  S .
Exercise 2.2. Let a, b be real numbers with a < b. Let S be the open interval (a, b) in R. Is S
open in C? Explain.
Theorem 2.3. The empty set is open in C.
Theorem 2.4. C is open in C.
Theorem 2.5. If Λ is any collection of open subsets in C, then
 S is open in C.
S
Theorem 2.6. If F is any finite collection of open subsets in C, then
 S is open in C.
SF
A topological space is a pair (X, Λ) where X is any set and Λ is a collection of subsets of X,
called the open sets in X. In any topological space we require that  and X are both open, and
that arbitrary unions and finite intersections of open sets are open. Thus, theorems 2.3 through
2.6 show that the definition given above for open in C makes the complex plane C into a
topological space. Since neighborhoods in C are analogous to open intervals in R, we call this
the standard topology for C. The ε-neighborhoods (open discs) obviously have a starring role in
the definition of this topology; they are called a basis for the standard topology on C.
Theorem 2.7. For every z 0  C and every   0 , N ( z0 ) is open in C.
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Exercise 2.8. Draw each of the following regions in the complex plane. Indicate any "missing"
boundaries with dashed lines. Explain why each region is, or is not, open in C.
1) R1  {z  C : 1  | z |  2}
2) R2  {z  C : re( z)  0 and 1  | z |  2}
3) R3  {z  C : re( z 2 )  0 }
4) R4  {z  C : re( z 2 )  0 }
5) R5  {z  C : z  0 and z  i }
6) R6  {z  C : re( z )  Z and im( z )  0 }
Definition. Let S  C . The complement of S is C \ S = {z  C | z  S} . S is said to be closed
(specifically, closed in C) provided C \ S is open. In other words, the complement of a closed set
is open.
Definition. A set S is called finite provided S is either empty or |S| is a positive integer. A set S
is called co-finite (relative to some superset T) provided its complement T \ S is finite.
Exercise 2.9. For each region R in exercise 2.8, draw the complement C \ R of the region, then
explain why the original region is, or is not, closed in C. Are any of the regions in exercise 2.8
co-finite?
Theorem 2.10. Let S  C . If S is open then C \ S is closed. (The complement of an open set is
closed.)
Theorem 2.11. Every finite subset of C is closed.
Corollary 2.12. Every co-finite subset of C is open.
Exercise 2.13. Give an example of a subset S  C which is neither open nor closed.
Exercise 2.14. Give two examples of subsets of C which are clopen (both open and closed).
Theorem 2.15. Let S  C . S is open iff there is a set Λ of ε-neighborhoods in C such that
S=
X .
X 
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Theorem 2.16. Arbitrary intersections of closed sets are closed, and finite unions of closed sets
are closed. That is:
(a) If Λ is any collection of closed subsets in C, then  S is closed in C.
S
(b) If F is any finite collection of closed subsets in C, then
 S is closed in C.
S
There is another way of characterizing closed sets using the notion of a limit point of a subset
S.
Definition. Let z 0  C and let S  C . We say z 0 is a limit point of S provided every
ε-neighborhood of z 0 contains a point w in S with w ≠ z 0 ; that is, z 0 is a limit point of S
provided that for all ε > 0, there exists w  S  N  ( z 0 ) such that w ≠ z 0 .
Exercise 2.17. Consider the region R2  {z  C : re( z)  0 and 1  | z |  2} from exercise 2.8.
(a) Prove that 2 is a limit point of R2 and 2 is not an element of R2.
(b) Prove that 3i/2 is a limit point of R2 and 3i/2 is an element of R2.
Thus, a limit point of a set may, or may not, be an element of the set.
Exercise 2.18. For each region in exercise 2.8, draw the set of limit points of the region.
The essential feature of a limit point z 0 is that there are points of S arbitrarily close (and
unequal) to z 0 . In fact, each ε-neighborhood of a limit point of S must contain infinitely many
points of S (do you see why?). As an alternative characterization of a closed set, we have the
following.
Theorem 2.19. Let S  C . S is closed iff S contains all its limit points.
Definition. Let S  C . Points which are in S but are not limit points of S are called isolated
points of S.
Exercise 2.20. For each region in exercise 2.8, find all the isolated points of the region.
Theorem 2.21. If S is an ε-neighborhood in C then S has no isolated points.
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Section 2: Limits of functions
Suppose f is a function which maps complex numbers to complex numbers. The notion of the
limit of such a function is analogous to the limit in the real case, and its properties follow by
similar arguments. We begin with the familiar ε-δ definition of limit, adapted to the complex
setting.
Definition. Let f be a complex-valued function, f : S → C, where S is a subset of C. Let z 0 be a
limit point of S. Let L be a complex number. Then the limit of f (z) as z approaches z 0 is equal
to L (that is, lim f ( z )  L ) provided that for all ε > 0 there exists a corresponding δ > 0 such that
z  z0
for all z in S, if 0  | z  z 0 |   then | f ( z )  L |   .
Exercise 2.22. Write a formal boolean expression corresponding to the preceding definition,
including all the necessary quantifier symbols; then, form the negation of your expression and
simplify it to learn what it means for the limit of f (z) not to equal L.
Three points should be emphasized about the preceding definition. First, since a limit point of
S is not necessarily an element of S, f ( z 0 ) need not exist for the limit to exist. Second, when the
limit point z 0 is an element of S, the actual value of f ( z 0 ) is not necessarily equal to the limit;
it's possible that f ( z 0 ) ≠ lim f ( z) . Third, it's an essential part of the definition that z 0 be a
z z0
limit point of S, that is, it's necessary for z 0 to have points of S arbitrarily close to itself so that
there are points z in S with 0  | z  z 0 |   ; if the implication (for all z in S, if 0  | z  z 0 |  
then | f ( z )  L |   ) were vacuously true, then the value L would be void of meaning. We will
only concern ourselves with the limit of a function at a limit point of the function's domain.
Exercise 2.23. Give an example of a function f : C → C such that f (0) exists, and lim f ( z )
z 0
exists, but they are not equal.
Exercise 2.24. Show the outline for a direct proof of a statement with the following form: For
all ε > 0 there exists a corresponding δ > 0 such that for all z in S, if P(z) then Q(z).
Theorem 2.25. (Uniqueness of limits.) Let f : S → C, where S is a subset of C. Let L and K be
complex numbers. If lim f ( z )  L and lim f ( z )  K , then L = K.
z  z0
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z  z0
Recall the Triangle Inequality, which we proved in Chapter 1: For all complex numbers z and
w, | z + w | ≤ | z | + | w |. Here is the generalized Triangle Inequality; we just need to prove part
(b).
Theorem 2.26 (Triangle Inequality). For all complex numbers z and w,
(a) | z + w | ≤ | z | + | w |
(b) | z - w | ≥ | | z | - | w | |
Exercise 2.27. Let f : S → C, where S is a subset of C. Prove (from the definition of limits): If
lim f ( z )  L then lim ( f ( z ))   L .
z  z0
z  z0
Exercise 2.28. Let f : S → C, where S is a subset of C. Let c be a complex number. Prove (from
the definition of limits): If lim f ( z )  L then lim (c  f ( z ))  c  L .
z  z0
z  z0
Theorem 2.29. Let f : S → C, where S is a subset of C. If lim f ( z )  L then lim | f ( z ) |  | L | .
z  z0
z  z0
Theorem 2.30. Let f : S → C, where S is a subset of C. If lim f ( z )  L and L ≠ 0, then there
z  z0
exists a real number δ > 0 such that for all z  S , if | z  z 0 |   then | f ( z ) | 
|L|
.
2
Theorem 2.31. Let f : S → C, where S is a subset of C. If lim f ( z )  L and L ≠ 0, then
z  z0
lim
z  z0
1
1
 .
f ( z) L
Theorem 2.32. (Algebraic properties of limits -- to save time, we will skip the proofs.)
Let f and g be functions from C to C (that is, their domains and ranges are subsets of C). Let L,
K, and z 0 be complex numbers. If lim f ( z )  L and lim g ( z)  K , then
z  z0
(a) lim ( f ( z)  g ( z))  L  K
z  z0
(b) lim ( f ( z)  g ( z ))  L  K
z  z0
(c) lim ( f ( z ) g ( z ))  LK
z  z0
(d) for K ≠ 0, lim ( f ( z) / g ( z))  L / K .
z  z0
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z  z0
The real and imaginary parts of a function may be considered separately as functions from C
to R; by definition, these functions satisfy the following condition:
For all z, f ( z )  re( f ( z ))  i  im( f ( z )) .
For example, if f (z) = iz + 5, then f (x + iy) = i(x + iy) + 5 = 5 - y + ix. Thus
re( f ( z ))  5  y  5  im ( z )  R and im ( f ( z ))  x  re( z )  R .
Exercise 2.33. If f ( z )  1  i  z 2 , find expressions for re( f ( z )) and im ( f ( z )) . Then evaluate
the following limits:
(a) lim re( f ( z ))
z i
(b) lim im ( f ( z ))
z i
(c) lim f ( z )
z i
Lemma. Let f : S → C, where S  C . Let L be any complex number. If lim f ( z )  L then
z  z0
lim f ( z )  L .
z  z0
Theorem 2.34. Let f : S → C, where S  C . If lim f ( z )  a  bi , where a, b are real numbers,
z  z0
then lim re( f ( z))  a and lim im( f ( z))  b .
z  z0
z  z0
Theorem 2.35. (Converse to Theorem 2.30.) Let f : S → C, where S  C . Let a, b be real
numbers. If lim re( f ( z))  a and lim im( f ( z))  b , then lim f ( z )  a  bi .
z  z0
z  z0
z  z0
Exercise 2.36. Find each limit, if it exists. Justify your answers.
z
(a) lim
z 0 | z |
|z|
z 0 z
(b) lim
| z |2
z 0
z
z
(d) lim
z 0 | z | 2
(c) lim
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Exercise 2.37. Let c  C . Prove: If f is a constant function on C (f (z) = c), then lim f ( z)  c
z  z0
for all z 0  C .
Lemma. For all z 0  C and for all n  N , lim z n  z 0n .
z  z0
Exercise 2.38. Let n be a non-negative integer. Let f be a polynomial function of degree n on C,
f (z) = a0  a1 z  a2 z 2    an z n (so, an ≠ 0). Prove that lim f ( z )  f ( z 0 ) for all z 0  C .
z  z0
Section 3: Continuity
A continuous function has the property that, for every limit point z0 in its domain, the limit of
the function at z0 is equal to the value of the function at z0. For example, the last two exercises in
Section 2 show that constant functions and polynomial functions are continuous on C.
Functions are considered automatically to be continuous at any isolated points of their
domains. Here is a definition of continuity which covers both cases (z0 is a limit point of S vs. z0
is an isolated point of S) at once.
Definition. Let f : S → C, where S is a subset of C. Let z0 be in S. f is continuous at z0
provided that for all ε > 0 there exists a corresponding δ > 0 such that for all z in S, if
| z  z 0 |   then | f ( z )  f ( z 0 ) |   . Moreover, if f is continuous at every point in S, we say
that f is continuous on its domain or, simply, continuous.
We need to verify that the definition works as claimed.
Theorem 2.39. Let f : S → C, where S is a subset of C. If z0 is an isolated point of S, then f is
continuous at z0.
Theorem 2.40. Let f : S → C, where S is a subset of C. Let z0 in S be a limit point of S. f is
continuous at z0 iff lim f ( z )  f ( z 0 ) .
z  z0
Exercise 2.41. Find the domain of each function. Then, use the definition of continuity, or
Theorem 2.40, to prove that each of the following functions is continuous on its domain.
(a) re z
(b) im z
(c) z + | z |
(d) 1/z
(e) | z |2
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Continuity of functions is preserved under the operations of addition, subtraction,
multiplication, division (as long as we avoid division by zero), and composition (as long as the
functions are composable).
Theorem 2.42. Let f : S → C and g : S  C , where S is a subset of C. If f and g are continuous
on S, then so are f  g , f  g , and f  g . Moreover, f / g is continuous on its domain,
{z  S : g ( z )  0} .
Theorem 2.43. Let f : S → C and g : T → C. Let z 0  S such that f ( z 0 )  T . If f is continuous
at z0 and g is continuous at f (z0), then g  f is continuous at z0.
Theorem 2.44. Let f : S → C, where S is a subset of C. Suppose f ( z )  u ( x, y )  iv ( x, y ) ,
where u and v are real-valued functions on R2 (that is, u(x, y) = re f (x + iy) and
v(x, y) = im f (x + iy)). Then f is continuous at z 0  x0  iy 0 iff u and v are continuous at ( x0 , y0 ) .
[To keep things moving, we will omit this proof.]
Our next order of business is to give a topological characterization of continuity, that is, we
need to see how continuity relates to open sets in C. To begin, we generalize the concept of
"open" to consider sets which are "relatively open."
Definition. Let V  S  C . We say V is relatively open in S, or just open in S, if for every z0 in
V there exists a corresponding σ > 0 such that N  ( z 0 )  S  V .
Exercise 2.45. Let S = {z  C : 1  | z |  2}. Find a subset V of S so that V is relatively open in
S, but V is not open in C.
Exercise 2.46. Find a subset V of the real numbers so that V is open in R but V is not open in C.
Theorem 2.47. Let V  S  C . If V is open in S and S is open in C, then V is open in C.
(Relatively open subsets of an open set are open.)
Theorem 2.48. Let V  S  C . If V is open in C, then V is open in S. (Open sets are relatively
open.)
A function f from S to C maps complex numbers to complex numbers; z corresponds with
f (z). In a similar way, we can use a function f : S → C to create correspondences between
subsets of C. A subset V of the domain determines a corresponding subset f (V), the set of all
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f (z) values where z is in V. And an arbitrary subset U of C determines a corresponding subset
f 1 (U ) , the set of all z in the domain which are mapped into U by f.
Definition. Let f : S → C, where S is a subset of C. Let V be any subset of S, and let U be any
subset of C.
 The image of V is f (V )  { f ( z ) : z  V } .

The inverse image of U is f 1 (U )  {z  S : f ( z )  U } .
Exercise 2.49. For each function f and each pair of sets U and V, find f (V ) and f 1 (U ) .
Express your answers as simply as possible. Also, draw the regions f (V ) and f 1 (U ) in a
sketch of the complex plane.
a) f ( z)  iz , V  N1 (1), U  {z  C | 0  arg( z)   / 2}
b)
f ( z)  2z  i, V  N1 (0), U  N1 (0)
c)
f ( z )  z 2 , V  N1 (1), U  {z  C : re( z )  0}
d)
f ( z )  z / | z |, V  R \ {0}, U  {i}
e)
f ( z )  z / | z |, V  C \ {0}, U  {z  C : | z |  1 and 0  arg( z )   / 2}
We are now ready to state our topological characterization of continuity for complex-valued
functions.
Theorem 2.50 (topological characterization of continuity). Let f : S → C, where S is an arbitrary
subset of C. f is continuous on S iff for every open set U  C , f 1 (U ) is relatively open in S.
Corollary 2.51. Let f : S → C, where S is an open subset of C. f is continuous on S iff for every
open set U  C , f 1 (U ) is open.
Exercise 2.52. Let f be a function from [-1, 1] to C, defined by f ( x)  x  ix .
a) Draw the range of f (that is, the image of [-1, 1]).
b) Find the inverse image of U = {z  C : | z |  1} .
c) Prove or disprove: f is continuous on [-1, 1].
Exercise 2.53. Let f be a function from [0, 2π] to C, defined by f ( )  cos( )  i  sin(  ) .
a) Draw the range of f (that is, the image of [0, 2π]).
b) Find the inverse image of U = {z  C : re( z )  0 and im ( z )  0} .
c) Prove or disprove: f is continuous on [0, 2π].
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Exercise. Consider two rays in the complex plane. R1 starts from 0 (with an open endpoint; z is
not equal to 0) and proceeds at a 45 degree angle relative to the positive x axis. R2 is parallel to
R1 but starts from -5i (with an open endpoint).
Consider the function f ( z )  arg( z ) , the principal argument of z, on its natural domain. Justify
each answer.
a) What is the limit of f (z ) as z  0 along R1?
b) What is the limit of f (z ) as z  5i along R2?
c) What is the limit of f (z ) as z   along R1?
d) What is the limit of f (z ) as z   along R2?
e) What is f ( R1 ) ?
f) What is f ( R2 ) ?
g) What is the domain of f ? Range of f ?
h) Is f continuous on its domain?
Section 4: Paths
Definition. A path in the complex plane is a continuous function  : [a, b]  C , where a and b
are real numbers with a ≤ b. The initial point of the path is  (a) , and the final (or terminal)
point of the path is  (b) .
Sometimes we speak of a "path in C from z1 to z2" if z1 is the initial point of the path and z2 is
the final point. We also refer to the point z   (t ) as a "point on the path  ," although, strictly
speaking, z is on the image of  ( z   ([ a, b]) ). If we think of t as "time" and imagine t
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increasing from a to b, then  (t ) traces out a curve in the plane from  (a) to  (b) . In drawing
a diagram of the curve, we often indicate the direction of this motion by an arrow, though it
needs to be emphasized that this is a "makeshift" device; the curve may cross itself and the
picture may get quite complicated.
In theory, the path could be so complicated it's undrawable; for example, it could be a "spacefilling" curve. We should therefore beware of representing the general case of a path in the
plane by a simplistic geometric picture.
Even in simple cases, it is possible for two different paths to have the same image curve.
(The paths are not literally equal unless the underlying functions are equal.)
Exercise 2.54. Show that the following paths traverse the same curve in C, and draw the curve.
 1 (t )  2(t  i  t ), 0  t  1 / 2
 2 (t )  t 2  i  t 2 , 0  t  1
To simplify matters, when we speak of paths which traverse line segments and circles, we
will usually assume them to be given by the following standard functions.
 The line segment from z1 to z2:  (t )  z1 (1  t )  z 2 (t ), 0  t  1

The unit circle:  (t )  cos t  i  sin t , 0  t  2

The circle centered at z0 with radius r:  (t )  z 0  r (cos t  i  sin t ), 0  t  2
Exercise 2.55. Write the standard function for the straight path from 0 to 1+i. Remember to
include the function's domain.
Definition. Let  : [a, b]  C be an arbitrary path in C. A subpath of  is specified by
restricting the domain to [c, d ] , where a  c  d  b . If a  x0  x1  x2    xn  b , and if
subpath  r is the restriction of  to [ xr 1 , xr ] , we write    1   2     n . In doing so, we
think of  being made up by tracing along the subpaths  1 ,  2 ,,  n taken in the order in which
they appear in the sum (left to right).
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Exercise 2.56. Draw the equilateral triangle with vertices at z1  cos 0  i  sin 0 ,
2
2
4
4
 i  sin
 i  sin
, and z 3  cos
.
3
3
3
3
a)  1 (t ) traverses the segment from z1 to z2 for 0 ≤ t ≤ 1. Give a possible definition of  1 (t ) .
z 2  cos
b)  2 (t ) traverses the segment from z2 to z3 for 1 ≤ t ≤ 2. Give a possible definition of  2 (t ) .
c)  3 (t ) traverses the segment from z2 to z0 for 2 ≤ t ≤ 3. Give a possible definition of  3 (t ) .
d)    1   2   3 . Find the domain of  . On your drawing of the equilateral triangle, use
arrows to show the direction in which the triangle is traversed by  . What is the initial
point of  ? Terminal point of  ?
Exercise 2.57.  1 (t )  t (0  t  1) and  2 (t )  cos t  i  sin t (0  t   ) .
a) Draw the images of  1 and  2 in the same plane. Use arrows to show the direction of
traversal.
b) The terminal point of  1 is the same as the initial point of  2 , so  1   2 makes sense -sort of. The problem is, the domain of  2 does not begin where the domain of  1 ends.
Resolve this problem by shifting the domain of  2 to begin at 1. That is, give a revised
definition for the path  2 so that it still traces a counter-clockwise semi-circular path from
1 to -1 in the plane, but adjust the formula to work with the domain [1, 1+π].
c) Now it is "legal" to write    1   2 . Write a "piecewise" expression for    1   2 .
When we write  1   2 , we require that the terminal point of  1 is the same as the initial
point of  2 . If the domains do not match up (as in the previous exercise), we assume the domain
of one of the paths will be adjusted so that the maximum of the domain of  1 is the same as the
minimum of the domain of  2 .
Definition. Let  : [a, b]  C be a path in C. Then the opposite path,   : [a, b]  C , is
defined as follows:   (t )   (a  b  t ), a  t  b .
Exercise 2.58. Let  : [a, b]  C be a path in C. Is   , the opposite path of  , the same as
(1)   ? Explain.
We will write  1   2 as an abbreviation for  1  ( 2 ) . This is simply the path which traces
first along  1 , then back along the opposite path of  2 (assuming that the appropriate endpoints
of the paths match up).
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Section 5: The Paving Lemma
Definition. Let S  C . A path  : [a, b]  C is said to be a path in S provided the image of 
is a subset of S:  ([ a, b])  S . We can also signify this same concept by the notation
 : [ a, b]  S .
Exercise 2.59. Let V  C \ {1 / n | n  Z and n  0} . Let  : [1,1]  V where
t  i  t  cos( / t ), t  0
.
t 0
 0,
 (t )  
a) Show that lim t  cos( / t ) exists, and find its value.
t 0
b) Show that  is a path.
c) Show that  is a path in V.
d) Make a sketch of the image of  in the plane; in the same sketch, show the points of the
plane which are not in V.
A path in S can be very intricate, and can be strangely entwined about the points of C which
are not in S. Rather than trying to imagine all the possible intricacies, we side-step the issue, as
follows. If S is an open set, we can cover any path  in S with a finite number of open discs
within S, in the following fashion.
The Paving Lemma. Let  : [a, b]  S be a path in an open set S. Then there exists a
subdivision a  t 0  t1  t 2    t n  b of [a, b] and a sequence of open discs (neighborhoods)
D1 , D2 ,, Dn in S such that each subpath  j (given by restricting  to [t j 1 , t j ] ) lies inside the
open disc D j  S . [Proof omitted.]
Exercise 2.60. Draw an oddly-shaped open set S in C and the image of a complicated path
 : [0,1]  S . (Do not worry about an explicit function which defines  .) Add to your drawing
a sequence of points z j   (t j ) along the path, and a sequence of discs D1 , D2 ,, Dn in S such
that each subpath  j (from z j 1 to z j ) lies inside the corresponding disc D j  S .
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Section 6: Connectedness
Definition. A subset S  C is said to be path connected if for all z1 , z 2  S there exists a path
 in S from z1 to z2.
Theorem 2.61. Every open disc N  ( z 0 ) in C is path connected.
Exercise 2.62. Prove, by constructing an appropriate path given by an explicit function, that
S  {z  C | z  t  it 2 for some t  R} is path connected.
Sometimes it is more convenient to use a simpler type of path.
Definition. A step path in S is a path  : [a, b]  S together with a subdivision
a  t 0  t1  t 2    t n  b of [a, b] such that in each subinterval [t j 1 , t j ] either re ( ) or im ( )
is constant.
This means that the image of  consists of a finite number of straight line segments, each
parallel to the real or imaginary axis.
Exercise 2.63. (a) Draw a complicated step path from 3 to 7i. (b) Draw a simple step path
from 3 to 7i.
Definition. A subset S  C is said to be step connected if for all z1 , z 2  S there exists a step
path  in S from z1 to z2.
Apparently every step connected set is also path connected. The converse, however, is not
true.
Exercise 2.64. Give an example of a subset S  C which is path connected but is not step
connected.
Theorem 2.65. Every open disc N  ( z 0 ) in C is step connected.
Theorem 2.66. Let S be an open subset of C. If S is path connected, then S is step connected.
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An arbitrary subset S  C may not be path connected. However, for any S  C we may
define the "is-connected-to" relation on S as follows: z1 ~ z2 iff there exists a path  in S from z1
to z2.
Theorem 2.67. Let S  C . The "is-connected-to" relation on S (defined above) is an
equivalence relation on S. Moreover, each equivalence class is path connected.
Definition. The equivalence classes of the "is-connected-to" relation on S  C are called the
connected components of S.
Exercise 2.68. Find the connected components of S  {z  C : | z |  1 and | z |  2} .
Theorem 2.69. If S is open in C then all the connected components of S are open in C.
One interesting way to generate open sets in C is to consider the complement of a path.
Definition. Let  : [a, b]  C be a path. The complement of  is defined to be C \  ([ a, b]) .
Exercise 2.70. Draw a path which satisfies the given description.
a) The complement of  has one connected component.
b) The complement of  has two connected components.
c) The complement of  has three connected components.
Definition. Let S  C . If there exists a positive real number B such that S  N B (0) then we
say S is a bounded subset of C. That is, S is bounded provided there exists B > 0 such that for all
z in S, | z | < B. If no such bound B exists, we say S is unbounded.
Exercise 2.71. For each path you drew in Exercise 2.70, how many of the complement's
components are bounded? How many are unbounded?
Exercise 2.72. Is there a path whose complement has infinitely many connected components?
Explain.
Theorem 2.73. Let  : [a, b]  C be a path. Then the image of  is closed and bounded.
[Proof omitted.]
Theorem 2.74. Let  : [a, b]  C be a path. Then the complement of  is open and has exactly
one unbounded component. [Proof omitted.]
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We now define the particular type of set which is fundamental to the theory of complex
analysis.
Definition. A domain is a non-empty, path connected, open set in the complex plane.
According to Theorem 2.66, a domain is also step connected. This means we can refer to a
domain as being "connected," meaning either path or step connected as appropriate.
As the theory of complex analysis unfolds, the reader will see the immense importance of this
definition. Complex functions f will be restricted to those of the form f : D  C where D is a
domain. The fact that D is open will allow us to deal neatly with limits, continuity, and
differentiability because z 0  D implies the existence of a positive  such that N  ( z 0 )  D and
so f (z) is defined for all z near z0. The connectedness of D guarantees a (step) path between any
two points in D and this in turn will allow us to define the integral from one point to another
along such a path.
However, restricting complex functions to those defined on domains has far more subtle
consequences than merely providing a platform for the appropriate definitions. Those who have
studied real analysis will find the complex case to be much richer in results. For example, if two
differentiable complex functions are defined on the same domain D and they happen to be equal
in a small disc in D, then they are equal throughout D! No analogous result holds for general
differentiable functions in the real case, and this result is just one of many that illustrate the
beauty and simplicity of complex analysis. Unfortunately, we shall not reach this result until
Chapter 10; but we mention it here to underline the fundamental importance of establishing the
correct topological foundations for the subject. We are ready to move on from topology!
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