In beginning calculus, the notion of an antiderivative was very important. It is similar for functions of
complex variables. The role of a primitive is also a lot like a potential in computing line integrals, which
should be familiar from calculus.
Definition 15.4.1A function F such that F^{′} = f is called a primitive of f.
See how it acts a lot like a potential, the difference being that a primitive has complex, not real values.
In calculus, in the context of a function of one real variable, this is often called an antiderivative and
every continuous function has one thanks to the fundamental theorem of calculus. However,
it will be shown below that the situation is not at all the same for functions of a complex
variable.
So what if a function has a primitive? Say F^{′}
(z)
= f
(z)
where f is continuous with values in X a
complex Banach space.
Theorem 15.4.2Suppose γ is continuous and bounded variation, a parametrization of Γ where t ∈
[a,b]
.Suppose f : γ^{∗}→ X is continuous and has a primitive F. Thus F^{′}
(z)
= f
(z)
for some Ω ⊇ γ^{∗}.Then
∫
f (z)dz = F (γ(b))− F (γ (a))
γ
Proof: Define the following function
{
F(γ(t))−F(γ(s))− f (γ (s)) if γ (t) ⁄= γ (s)
h (s,t) ≡ γ(t)−γ(s)
0 if γ (t) = γ (s)
Then h is continuous at points
(s,s)
. To see this note that the top line is
f (γ(s))(γ (t)− γ (s))+ o(γ(t)− γ(s))
---------------------------------− f (γ (s))
γ (t)− γ (s)
= o(γ(t)-− γ-(s))
γ(t)− γ(s)
thus if
(sn,tn)
→
(s,s)
, the continuity of γ requires this to also converge to 0. That is,
∥∥∫ ∥∥
∥∥ f (z)dz − (F (γ(b))− F (γ (a)))∥∥
∥∫γ ∥
∥∥ n∑ ∥∥
= ∥∥ γf (z)dz − F (γ (ti)) − F (γ(ti− 1))∥∥ ≤ ε (1 + V (γ))
k=1
Therefore, since ε is arbitrary,
∫
γ f (z)dz = (F (γ(b)) − F (γ(a))) ■
Probably the most fundamental result in the subject is Cauchy’s theorem which says that the contour
integral of an analytic function over a simple closed curve equals 0. The following is like what was first done
by Cauchy back in the early 1800’s.
Theorem 15.4.3Let γ^{∗}be a simple closed curve with parametrization γ
(t)
. Let f : U_{i}∪ γ^{∗}→ ℂ becontinuous and let f be analytic on U_{i}where U_{i}denotes the inside of the simple closed curve.Then
One can reduce to this case even if the function has values in a Banach space. This is done by
considering ϕ
(f )
for ϕ ∈ X^{′} and then applying the above to ϕ
(f)
which has complex values. However, I do
not wish to emphasize functional analysis techniques any more than necessary so I will give a more direct
way to get this result which is based on the earlier development of Green’s theorem and the Cauchy
Goursat theorem which is the next big result. This is a major theorem which does not depend on the
derivative being continuous. Thus it will also provide the needed generalization which involves not
assuming that z → f^{′}
(z)
is continuous.
If you have two points in ℂ, z_{1} and z_{2}, you can consider γ
(t)
≡ z_{1} + t
(z2 − z1)
for t ∈
[0,1]
to obtain a
continuous bounded variation curve from z_{1} to z_{2}. More generally, if z_{1},
⋅⋅⋅
,z_{m} are points in ℂ you can
obtain a continuous bounded variation curve from z_{1} to z_{m} which consists of first going from z_{1} to z_{2} and
then from z_{2} to z_{3} and so on, till in the end one goes from z_{m−1} to z_{m}. Denote this piecewise linear curve
as γ
(z1,⋅⋅⋅,zm)
. Now let T be a triangle with vertices z_{1},z_{2} and z_{3} encountered in the counter clockwise
direction as shown.
PICT
Denote by ∫_{∂T}f
(z)
dz, the expression, ∫_{γ(z1,z2,z3,z1)
}f
(z)
dz. Consider the following picture.
PICT
Thus
∫ ∑4 ∫
f (z)dz = f (z)dz. (15.21)
∂T k=1 ∂T1k
(15.21)
On the “inside lines” the integrals cancel because there are two integrals going in opposite directions for
each of these inside lines. Recall Theorem 15.3.2 which tells how to evaluate a line integral with a C^{1}
parametrization.
Theorem 15.4.4(Cauchy Goursat) Let f : Ω → X, where Ω is an open subset of ℂ and X is a complexcomplete normed linear space, have the property that f^{′}
(z)
exists for all z ∈ Ω and let T be a trianglecontained in Ω. Then
Since ε is arbitrary, this shows α = 0, a contradiction. Thus ∫_{∂T}f
(w )
dw = 0 as claimed.
■
Note that no assumption of continuity of z → f^{′}
(z)
was needed.
Obviously, there is a version of the above Cauchy Goursat theorem which is valid for a rectangle.
Indeed, apply the Cauchy Goursat theorem for the triangles obtained from a diagonal of the
rectangle. The diagonal will be oriented two different ways depending on which triangle it is a part
of.
PICT
Corollary 15.4.5Let Ω be an open set on which f^{′}
(z)
exists. Here f has values in a complex Banachspace. Then if R is a rectangle contained in Ω along with its inside, then orienting R either way resultsin
∫
f (z)dz = 0.
R
The following is a general version of the Cauchy integral theorem. If f^{′}
(z)
exists on the inside and if f
is continuous on the boundary, then the integral over the bounding curve is 0. Note how the closed curve is
arbitrary, not just a triangle. First recall Lemma 12.4.3.
Lemma 15.4.6Let Γ be a simple closed rectifiable curve. Also let δ > 0 be given such that 2δ issmaller than both the height and width of Γ. Then there exist finitely many non overlappingregions
{Rk}
_{k=1}^{n}consisting of simple closed rectifiable curves along with their interiors whoseunion equals U_{i}∪ Γ. These regions consist of two kinds, those contained in U_{i}and those withnonempty intersection with Γ. These latter regions are called “border” regions. The boundary ofa border region consists of straight line segments parallel to the coordinate axes of the formx = m
(δ)
4
or y = k
(δ)
4
for m,k integers along with arcs from Γ. The non border regions consist ofrectangles. Thus all of these regions have boundaries which are rectifiable simple closed curves. Alsoeach region is contained in a rectangle having sides of length no more than δ. There are atmost
(4V-(Γ ) )
4 δ + 1
border regions. The construction also yields an orientation for Γ and for all these regions, the orientationsfor any line segment shared by two regions are opposite.
Theorem 15.4.7Let U_{i}be the inside of Γ a simple closed rectifiable curve having parameterization γ.Also let f^{′}
(z)
exist in U_{i}and f is continuous on U_{i}∪ Γ. Here we assume only that f has values in acomplex Banach space X. Then
∫
f (z)dz = 0.
γ
Proof:Let ℬ_{δ},ℐ_{δ} be those regions of Lemma 12.4.3 where as earlier ℐ_{δ} are those which have
empty intersection with Γ and ℬ_{δ} are the border regions. Without loss of generality, assume Γ
is positively oriented. Then since the line integrals cancel on adjacent sides of boundaries of
regions,
∫ ∫ ∫
∑ f (z)dz + ∑ f (z)dz = f (z)dz
R ∈ℐδ ∂R R∈ℬδ ∂R γ
In this case the first sum on the left in the above formula equals 0 from Corollary 15.4.5 for any δ > 0.
Recall that there were at most
( )
4 4V-(Γ )+ 1
δ
border regions where each of these border regions is contained in a box having sides of length no
more than δ. Letting ε > 0 be given, suppose δ < 1 is so small that for
|z − w|
< 8δ with
z,w ∈U_{i},
|f (z)− f (w )| < ε
this by uniform continuity of f on this compact set. Let R be a border region. Then picking z_{1} a point in
R,
∫ ∫ ∫
f (z)dz = (f (z)− f (z1))dz + f (z1)dz
∂R ∂R ∂R