4.7: Extensions of Cauchy's theorem
( \newcommand{\kernel}{\mathrm{null}\,}\)
Cauchy’s theorem requires that the function f(z) be analytic on a simply connected region. In cases where it is not, we can extend it in a useful way.
Suppose R is the region between the two simple closed curves C1 and C2. Note, both C1 and C2 are oriented in a counterclockwise direction.
If f(z) is analytic on R then
∫C1−C2f(z) dz=0.
- Proof
-
The proof is based on the following figure. We ‘cut’ both C1 and C2 and connect them by two copies of C3, one in each direction. (In the figure we have drawn the two copies of C3 as separate curves, in reality they are the same curve traversed in opposite directions.)
With C3 acting as a cut, the region enclosed by C1+C3−C2−C3 is simply connected, so Cauchy's Theorem 4.6.1 applies. We get
∫C1+C3−C2−C3f(z) dz=0
The contributions of C3 and −C3 cancel, which leaves ∫C1−C2f(z) dz=0. QED
This clearly implies ∫C1f(z) dz=∫C2f(z) dz.
Let f(z)=1/z. f(z) is defined and analytic on the punctured plane. What values can ∫Cf(z) dz take for C a simple closed curve (positively oriented) in the plane?
Solution
We have two cases (i) C1 not around 0, and (ii) C2 around 0
Case (i): Cauchy’s theorem applies directly because the interior does not contain the problem point at the origin. Thus,
∫C1f(z) dz=0.
Case (ii): we will show that
∫C2f(z) dz=2πi.
Let C3 be a small circle of radius a centered at 0 and entirely inside C2.
By the extended Cauchy theorem we have
∫C2f(z) dz=∫C3f(z) dz=∫2π0i dt=2πi.
Here, the lline integral for C3 was computed directly using the usual parametrization of a circle.
Answer to the question
The only possible values are 0 and 2πi.
We can extend this answer in the following way:
If C is not simple, then the possible values of
∫Cf(z) dz
are 2πni, where n is the number of times C goes (counterclockwise) around the origin 0.
n is called the winding number of C around 0. n also equals the number of times C crosses the positive x-axis, counting ±1 for crossing from below and -1 for crossing from above.
A further extension: using the same trick of cutting the region by curves to make it simply connected we can show that if f is analytic in the region R shown below then
∫C1−C2−C3−C4f(z) dz=0.
That is, C1−C2−C3−C4 is the boundary of the region R.
Orientation
It is important to get the orientation of the curves correct. One way to do this is to make sure that the region R is always to the left as you traverse the curve. In the above example. The region is to the right as you traverse C2,C3 or C4 in the direction indicated. This is why we put a minus sign on each when describing the boundary.