1 Complex integrals
If is a single-valued, continuous function in some region in the complex plane then we define the integral of along a path in (see Figure 7) as
.
Figure 7
Here we have written and in real and imaginary parts:
Then we can separate the integral into real and imaginary parts as
We often interpret real integrals in terms of area; now we define complex integrals in terms of line integrals over paths in the complex plane. The line integrals are evaluated as described in HELM booklet 29.
Example 10
Obtain the complex integral:
where is the straight line path from to . See Figure 8.
Figure 8
Solution
Here, since is constant ( ) along the given path then , implying that and . Also, as is constant, .
Therefore,
Task!
Evaluate where is the straight line path from to .
First obtain expressions for and by finding an appropriate expression for along the path:
Along the path , implying that and . Also . Now find limits on :
The limits on are: to Now evaluate the integral:
Task!
Evaluate where is the straight line path from to .
We first need to find the equation of the line
in the Argand plane.
We note that both points lie on the line
so the complex equation of the straight line is
giving
and
. Also
.
.
Next, we see that the limits on are to . We are now in a position to evaluate the integral:
Note that this result is the sum of the integrals along and . You might have expected this.
A more intricate example now follows.
Example 11
Evaluate where is that part of the unit circle going anticlockwise from the point to the point . See Figure 9.
Figure 9
Solution
First, note that and giving
This is obtained by simply expressing the integral in real and imaginary parts. These integrals cannot be evaluated in this form since and are related. Instead we re-write them in terms of the single variable .
Note that on the unit circle: so that and .
The expressions and can be expressed in terms of since
.
Now as the point moves from to along the path the parameter changes from to . Hence,
We can simplify these daunting-looking integrals by using the trigonometric identities:
.
We obtain (choosing and in both expressions):
.
Also
Now we can complete the evaluation of our integral:
In the last Task we integrated over a given path. We had to perform some intricate mathematics to get the value. It would be convenient if there was a simpler way to obtain the value of such complex integrals. This is explored in the following Tasks.
Task!
Evaluate
We obtain again, which is the same result as from the previous Task.
It would seem that, by carrying out an analogue of real integration (simply integrating the function and substituting in the limits) we can obtain the answer much more easily. Is this coincidence?
If you return to the first Task of this Section you will note:
the result we obtained earlier.
We shall investigate these ‘coincidences’ in Section 26.5.
As a variation on this example, suppose that the path is the entire circumference of the unit circle travelled in an anti-clockwise direction. The limits are and . Hence
Is there an underlying reason for this result? (We shall see in Section 26.5.)
Another technique for evaluating integrals taken around the unit circle is shown in the next example, in which we need to evaluate
where is the unit circle.
Note the use of since we have a closed path; we could have used this notation earlier.
Task!
Evaluate where is the unit circle.
First show that a point on the unit circle can be written and hence find in terms of :
On the unit circle a point is such that and hence which, using De Moivre’s theorem, can be seen to be
. Now evaluate the integral .
We now quote one of the most important results in complex integration which incorporates the last result.
Key Point 1
If is an integer and is the circle centre and radius , that is, it has equation then
Note that the result is independent of the value of .