1 Cauchy’s theorem

1.1 Simply-connected regions

A region is said to be simply-connected if any closed curve in that region can be shrunk to a point without any part of it leaving a region. The interior of a square or a circle are examples of simply connected regions. In Figure 11 (a) and (b) the shaded grey area is the region and a typical closed curve is shown inside the region. In Figure 11 (c) the region contains a hole (the white area inside). The shaded region between the two circles is not simply-connected; curve $C_1$ can shrink to a point but curve $C_2$ cannot shrink to a point without leaving the region, due to the hole inside it.

Figure 11

As a straightforward example note that ${\oint }_C{z}^{2}dz=0$ , where $C$ is the unit circle, since $z^2$ is analytic everywhere (see Section 261). Indeed ${\oint }_C{z}^{2}dz=0$ for any simple contour: it need not be circular.

Consider the contour shown in Figure 12 and assume $f\left(z\right)$ is analytic everywhere on and inside the contour $C$ .

Figure 12

Then by analogy with real line integrals

${\int \; <!--nolimits\; -->}_{AEB}f\left(z\right)dz+{\int \; <!--nolimits\; -->}_{BDA}f\left(z\right)dz={\oint }_{C}f\left(z\right)dz=0$ by Cauchy’s theorem.

Therefore

${\int \; <!--nolimits\; -->}_{AEB}f\left(z\right)dz=-{\int \; <!--nolimits\; -->}_{BDA}f\left(z\right)dz={\int \; <!--nolimits\; -->}_{ADB}f\left(z\right)dz$

(since reversing the direction of integration reverses the sign of the integral).

This implies that we may choose any path between $A$ and $B$ and the integral will have the same value providing $f\left(z\right)$ is analytic in the region concerned.

Integrals of analytic functions only depend on the positions of the points $A$ and $B$ , not on the path connecting them. This explains the ‘coincidences’ referred to previously in Section 26.4.

Task!

Using ‘simple’ integration evaluate ${\int \; <!--nolimits\; -->}_i^{1+2i}coszdz$ , and explain why this is valid.

Answer

We now investigate what occurs when the closed path of integration does not necessarily lie within a simply-connected region. Consider the situation described in Figure 13.

Figure 13

Let $f\left(z\right)$ be analytic in the region bounded by the closed curves $C_1$ and $C_2$ . The region is cut by the line segment joining $A$ and $B$ .

Consider now the closed curve $AEABFBA$ travelling in the direction indicated by the arrows. No line can cross the cut $AB$ and be regarded as remaining in the region. Because of the cut the shaded region is simply connected . Cauchy’s theorem therefore applies (see Key Point 2).

Therefore

${\oint }_{AEABFBA}f\left(z\right)dz=0$ since $f\left(z\right)$ is analytic within and on the curve $AEABFBA$ .

Note that

${\int \; <!--nolimits\; -->}_{AB}f\left(z\right)dz=-{\int \; <!--nolimits\; -->}_{BA}f\left(z\right)dz,$ being a simple change of direction.

Also, we can divide the closed curve into smaller sections:

i.e.

${\oint }_{C_1}f\left(z\right)dz-{\oint }_{C_2}f\left(z\right)dz=0$

(since we assume that closed paths are travelled anticlockwise).

Therefore ${\oint }_{C_1}f\left(z\right)dz={\oint }_{C_2}f\left(z\right)dz$ .

This allows us to evaluate ${\oint }_{C_1}f\left(z\right)dz$ by replacing $C_1$ by any curve $C_2$ such that the region between them contains no singularities (see Section 261) of $f\left(z\right)$ . Often we choose a circle for $C_2$ .

Example 12

Determine   ${\oint }_C\frac{6}{z\left(z-3\right)}dz$   where $C$ is the curve  $\left|z-3\right|=5$  shown in Figure 14.

Figure 14

Solution

We observe that $f\left(z\right)=\frac{6}{z\left(z-3\right)}$ is analytic everywhere except at $z=0$ and $z=3$ .

Let $C_1$ be the circle of unit radius centred at $z=3$ and $C_2$ be the unit circle centered at the origin. By analogy with the previous example we state that

${\oint }_C\frac{6}{z\left(z-3\right)}dz={\oint }_{C_1}\frac{6}{z\left(z-3\right)}dz+{\oint }_{C_2}\frac{6}{z\left(z-3\right)}dz.$

(To show this you would need two cuts: from $C$ to $C_1$ and from $C$ to $C_2$ .)

The remaining parts of this problem are presented as two Tasks.

Task!

Expand $\frac{6}{z\left(z-3\right)}$ into partial functions.

Answer

Thus:

${\oint }_C\frac{6}{z\left(z-3\right)}dz={\oint }_{C_1}\frac{2}{z-3}dz-{\oint }_{C_1}\frac{2}{z}dz+{\oint }_{C_2}\frac{2}{z-3}dz-{\oint }_{C_2}\frac{2}{z}dz=I_1-I_2+I_3-I_4.$

Task!

Find the values of $I_1,I_2,I_3,I_4$ , using Key Point 1 (page 35):

1. Find the value of $I_1$ :

   Answer

   Using Key Point 1 we find that $I_1=2\times 2\pi i=4\pi i$ .

2. Find the value of $I_2$ :

   Answer

   The function $\frac{1}{z}$ is analytic inside and on $C_1$ so that $I_2=0$ .

3. Find the value of $I_3$ :

   Answer

   The function $\frac{1}{z-3}$ is analytic inside and on $C_2$ so $I_3=0$ .

4.  Find the value of $I_4$ :

   Answer

   $I_4=4\pi i$ again using Key Point 1.

5. Finally, calculate $I=I_1-I_2+I_3-I_4$ :

   Answer

   ${\oint }_C\frac{6dz}{z\left(z-3\right)}=4\pi i-0+0-4\pi i=0.$

Exercises

1. Evaluate  ${\int \; <!--nolimits\; -->}_{1+i}^{2+3i}sinzdz.$
2. Determine  ${\oint }_C\frac{4}{z\left(z-2\right)}dz$  where $C$ is the contour  $\left|z-2\right|=4.$

Answer