It is often convenient to have a way of bounding the size of an integral, and one such method is the
ML
inequality.
Suppose
M
is a positive real number such that the function
f
(
z
)
satisfies

f
(
z
)
 ≤
M
for all
z
on the
image of a curve
C
in the plane.
If we let
L
be the length of the image of the curve
C
, we have the
ML
inequality:
Z
C
f
≤
M
·
L.
Example 4.7.
Suppose
f
(
z
) =
3
z
z
2
+ 2
, and
C
is the arc of the circle

z

= 2
that lies in the first quadrant.
We must find an appropriate bound
M
for the function
f
(
z
)
for
z
on the arc of the circle.
Note that when we estimate the size of a quotient, we look for the maximum of the numerator and the
minimum of the denominator.
So

f
(
z
)

=
3
z
z
2
+ 2
≤
3

z


z

2

2
=
6
2
= 3 =
M,
where we have used one form of the triangle inequality: Since

z

= 2
, so

z
2

=

z

2
= 4
>
2
, so

z
2
+ 2
 ≥ 
z

2

2 =

z

2

2 = 4

2 = 2
on the circle

z

= 2
.
Then,
Z
C
f
≤
M
·
L
= 3
·
2
π
2
4
= 3
π,
since the length of one quarter of a circle of radius 2 is equal to
π
.
Mathematics 366: Complex Variables I
Study Guide
89
Example 4.8.
Suppose
C
is the straight line segment from 0 to
1 +
i
and
f
(
z
) =
e
z
.
Since

e
z

=
e
x
and the real exponential function is a strictly increasing function, we have

f
(
z
)
 ≤
e
1
=
e
=
M
on the image of
C
(the maximum value of
e
x
occurs when
x
= 1
, its largest value on the image of
C
).
Thus,
Z
C
f
≤
M
·
L
=
e
·
√
2
,
where we have found the length of
C
by using Pythagoras’ theorem.
Incidentally, this integral can be found exactly, since the function
e
z
has a complex antiderivative, namely
itself!
Remark
We would like to emphasize the fact that the estimate that we obtain from the
ML
inequality only gives an
upper bound for the possible size of the integral, and does not have to be close to its actual value.
To illustrate, suppose in
Example 4.7
, above, we took
C
to be the whole circle

z

= 2
.
Then, its length would be
4
π
, and the same
M
= 3
would still work.
So the
ML
inequality tells us that the modulus of the integral is less than or equal to
12
π
. However, the
actual value of this integral is
6
πi
, so the estimate is not very close! We discuss how to compute this value
later.
We could create more extreme examples, but you should get the point.
4.3
Contour Integration and Green’s Theorem
In this section, we undertake our first foray into the theory behind Cauchy’s theorem.
Indications
1.
Read Section 4.3, “Contour Integration and Green’s Theorem,” pages 172–179.
2.
Read the comments below.
3.
Do Exercises 2–7 and 17–20 on pages 180–181.
Comments
For a closed contour, by default, we always traverse counterclockwise exactly once. For example,

z

= 1
,
we can parameterize as
e
iθ
and let
θ
goes from 0 to
2
π
, then the corresponding point
z
=
e
iθ
will start from
Mathematics 366: Complex Variables I
Study Guide
90
z
= 0
, moving along the unit circle counterclockwise for exactly one loop. When a curve is given a
counterclockwise direction or can only be traversed counterclockwise, we say it is
positively oriented
.