1 Cauchy’s theorem
1.1 Simplyconnected regions
A region is said to be simplyconnected 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 simplyconnected; 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
Key Point 2
Cauchy’s Theorem
The theorem states that if $f\left(z\right)$ is analytic everywhere within a simplyconnected region then:
${\oint}_{C}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz=0$
for every simple closed path $C$ lying in the region.
This is perhaps the most important theorem in the area of complex analysis.
As a straightforward example note that ${\oint}_{C}{z}^{2}\phantom{\rule{0.3em}{0ex}}dz=0$ , where $C$ is the unit circle, since ${z}^{2}$ is analytic everywhere (see Section 261). Indeed ${\oint}_{C}{z}^{2}\phantom{\rule{0.3em}{0ex}}dz=0$ 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
$\phantom{\rule{2em}{0ex}}{\int}_{AEB}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz+{\int}_{BDA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\oint}_{C}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz=0$ by Cauchy’s theorem.
Therefore
$\phantom{\rule{2em}{0ex}}{\int}_{AEB}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\int}_{BDA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\int}_{ADB}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz$
(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 ${\int}_{i}^{1+2i}cosz\phantom{\rule{1em}{0ex}}dz$ , and explain why this is valid.
$\phantom{\rule{2em}{0ex}}{\int}_{i}^{1+2i}cosz\phantom{\rule{1em}{0ex}}dz={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}sinz\right]}_{i}^{1+2i}=sin\left(1+2i\right)sin\phantom{\rule{1em}{0ex}}i$ .
This way of determining the integral is legitimate because $cosz$ is analytic (everywhere).
We now investigate what occurs when the closed path of integration does not necessarily lie within a simplyconnected 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
$\phantom{\rule{2em}{0ex}}{\oint}_{AEABFBA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz=0$ since $f\left(z\right)$ is analytic within and on the curve $AEABFBA$ .
Note that
$\phantom{\rule{2em}{0ex}}{\int}_{AB}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\int}_{BA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz,$ being a simple change of direction.
Also, we can divide the closed curve into smaller sections:
$$\begin{array}{rcll}{\oint}_{AEABFBA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz& =& {\int}_{AEA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz+{\int}_{AB}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz+{\int}_{BFB}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz+{\int}_{BA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz& \text{}\\ & & & \text{}\\ & =& {\int}_{AEA}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz+{\int}_{BFB}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz=0.& \text{}\end{array}$$i.e.
$\phantom{\rule{2em}{0ex}}{\oint}_{{C}_{1}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz{\oint}_{{C}_{2}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz=0$
(since we assume that closed paths are travelled anticlockwise).
Therefore ${\oint}_{{C}_{1}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\oint}_{{C}_{2}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz$ .
This allows us to evaluate ${\oint}_{{C}_{1}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz$ 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 ${\oint}_{C}\frac{6}{z\left(z3\right)}\phantom{\rule{0.3em}{0ex}}dz$ where $C$ is the curve $\leftz3\right=5$ shown in Figure 14.
Figure 14
Solution
We observe that $f\left(z\right)=\frac{6}{z\left(z3\right)}$ 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
$\phantom{\rule{2em}{0ex}}{\oint}_{C}\frac{6}{z\left(z3\right)}\phantom{\rule{0.3em}{0ex}}dz={\oint}_{{C}_{1}}\frac{6}{z\left(z3\right)}\phantom{\rule{0.3em}{0ex}}dz+{\oint}_{{C}_{2}}\frac{6}{z\left(z3\right)}\phantom{\rule{0.3em}{0ex}}dz.$
(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 $\frac{6}{z\left(z3\right)}$ into partial functions.
Let $\phantom{\rule{2em}{0ex}}\frac{6}{z\left(z3\right)}\equiv \frac{A}{z}+\frac{B}{z3}\equiv \frac{A\left(z3\right)+Bz}{z\left(z3\right)}$ . Then $A\left(z3\right)+Bz\equiv 6.$
$\text{If}\phantom{\rule{1em}{0ex}}z=0\phantom{\rule{2em}{0ex}}A\left(3\right)=6\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{1em}{0ex}}A=2.$ If $z=3\phantom{\rule{2em}{0ex}}B\times 3=6\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{1em}{0ex}}B=2.$
$\therefore $ $\phantom{\rule{2em}{0ex}}\frac{6}{z\left(z3\right)}\equiv \frac{2}{z}+\frac{2}{z3}.$
Thus:
$\phantom{\rule{2em}{0ex}}{\oint}_{C}\frac{6}{z\left(z3\right)}\phantom{\rule{0.3em}{0ex}}dz={\oint}_{{C}_{1}}\frac{2}{z3}\phantom{\rule{0.3em}{0ex}}dz{\oint}_{{C}_{1}}\frac{2}{z}\phantom{\rule{0.3em}{0ex}}dz+{\oint}_{{C}_{2}}\frac{2}{z3}\phantom{\rule{0.3em}{0ex}}dz{\oint}_{{C}_{2}}\frac{2}{z}\phantom{\rule{0.3em}{0ex}}dz={I}_{1}{I}_{2}+{I}_{3}{I}_{4}.$
Task!
Find the values of ${I}_{1},{I}_{2},{I}_{3},{I}_{4}$ , using Key Point 1 (page 35):

Find the value of
${I}_{1}$
:
Using Key Point 1 we find that ${I}_{1}=2\times 2\pi i=4\pi i$ .

Find the value of
${I}_{2}$
:
The function $\frac{1}{z}$ is analytic inside and on ${C}_{1}$ so that ${I}_{2}=0$ .

Find the value of
${I}_{3}$
:
The function $\frac{1}{z3}$ is analytic inside and on ${C}_{2}$ so ${I}_{3}=0$ .

Find the value of
${I}_{4}$
:
${I}_{4}=4\pi i$ again using Key Point 1.

Finally, calculate
$I={I}_{1}{I}_{2}+{I}_{3}{I}_{4}$
:
${\oint}_{C}\frac{6\phantom{\rule{0.3em}{0ex}}dz}{z\left(z3\right)}=4\pi i0+04\pi i=0.$
Exercises
 Evaluate ${\int}_{1+i}^{2+3i}sinz\phantom{\rule{0.3em}{0ex}}dz.$
 Determine ${\oint}_{C}\frac{4}{z\left(z2\right)}\phantom{\rule{0.3em}{0ex}}dz$ where $C$ is the contour $\leftz2\right=4.$
 ${\int}_{1+i}^{2+3i}sinz\phantom{\rule{0.3em}{0ex}}dz={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}cosz\right]}_{1+i}^{2+3i}=cos\left(1+i\right)cos\left(2+3i\right)$ since $sinz$ is analytic everywhere.

$f\left(z\right)=\frac{4}{z\left(z2\right)}$ is analytic everywhere except at $z=0$ and $z=2$ .
Call $I={\oint}_{C}\frac{4}{z\left(z2\right)}\phantom{\rule{0.3em}{0ex}}dz={\oint}_{{C}_{1}}\frac{4}{z\left(z2\right)}\phantom{\rule{0.3em}{0ex}}dz+{\oint}_{{C}_{2}}\frac{4}{z\left(z2\right)}\phantom{\rule{0.3em}{0ex}}dz$ .
Now $\phantom{\rule{1em}{0ex}}\frac{4}{z\left(z2\right)}\equiv \frac{2}{z}+\frac{2}{z2}$ so that
$$\begin{array}{rcll}I& =& {\oint}_{{C}_{1}}\frac{2}{z2}\phantom{\rule{0.3em}{0ex}}dz{\oint}_{{C}_{1}}\frac{2}{z}\phantom{\rule{0.3em}{0ex}}dz+{\oint}_{{C}_{2}}\frac{2}{z2}\phantom{\rule{0.3em}{0ex}}dz{\oint}_{{C}_{2}}\frac{2}{z}\phantom{\rule{0.3em}{0ex}}dz& \text{}\\ & =& {I}_{1}+{I}_{2}+{I}_{3}+{I}_{4}& \text{}\end{array}$$${I}_{2}$ and ${I}_{3}$ are zero because of analyticity.
${I}_{1}=2\times 2\pi i=4\pi i$ , by Key Point 1 and ${I}_{4}=4\pi i$ likewise.
Hence $I=4\pi i+0+04\pi i=0.$