2 Cauchy’s integral formula

This is a generalization of the result in Key Point 2:

Example 13

Evaluate ${\oint }_C\frac{z}{z^2+1}dz$   where $C$ is the path shown in Figure 15:

$C_1:\left|z-i\right|=\frac{1}{2}$

Figure 15

Solution

We note that $z^2+1\equiv \left(z+i\right)\left(z-i\right).$

Let $\frac{z}{z^2+1}=\frac{z}{\left(z+i\right)\left(z-i\right)}=\frac{z∕\left(z+i\right)}{z-i}$ .

The numerator $z∕\left(z+i\right)$ is analytic inside and on the path $C_1$ so putting $z_0=i$ in the Cauchy integral formula (Key Point 3)

${\oint }_{C_1}\frac{z}{z^2+1}dz=2\pi i\left[\frac{i}{i+i}\right]=2\pi i.\frac{1}{2}=\pi i$ .

Task!

Evaluate ${\oint }_C\frac{z}{z^2+1}dz$   where $C$ is the path (refer to the diagram)

1. $C_2:|z+i|=\frac{1}{2}$
2. $C_3:\left|z\right|=2$ .

   

1. Use the Cauchy integral formula to find an expression for ${\oint }_{C_2}\frac{z}{z^2+1}dz$ :

   Answer

   $\frac{z}{z^2+1}=\frac{z∕\left(z-i\right)}{z+i}$ . The numerator is analytic inside and on the path $C_2$ so putting $z_0=-i$ in the Cauchy integral formula gives

   ${\oint }_{C_2}\frac{z}{z^2+1}dz=2\pi i\left[\frac{-i}{-2i}\right]=\pi i$ .

2. Now find ${\oint }_{C_3}\frac{z}{z^2+1}dz$ :

   Answer

   By analogy with the previous part,

   ${\oint }_{C_3}\frac{z}{z^2+1}dz={\oint }_{C_1}\frac{z}{z^2+1}dz+{\oint }_{C_2}\frac{z}{z^2+1}dz=\pi i+\pi i=2\pi i.$

2.1 The derivative of an analytic function

If $f\left(z\right)$ is analytic in a simply-connected region then at any interior point of the region, $z_0$ say, the derivatives of $f\left(z\right)$ of any order exist and are themselves analytic (which illustrates what a powerful property analyticity is!). The derivatives at the point $z_0$ are given by Cauchy’s integral formula for derivatives:

$f^{\left(n\right)}\left(z_0\right)=\frac{n!}{2\pi i}{\oint }_C\frac{f\left(z\right)}{{\left(z-z_0\right)}^{n+1}}dz$

where $C$ is any simple closed curve, in the region, which encloses $z_0$ .

Note the case $n=1$ :

$f^{\prime }\left(z_0\right)=\frac{1}{2\pi i}{\oint }_C\frac{f\left(z\right)}{{\left(z-z_0\right)}^2}dz.$

Example 14

Evaluate the contour integral

${\oint }_C\frac{z^3}{{\left(z-1\right)}^2}dz$

where $C$ is a contour which encloses the point $z=1$ .

Solution

Since $f\left(z\right)=\frac{z^3}{{\left(z-1\right)}^2}$ has a pole of order 2 at $z=1$ then ${\oint }_{C}f\left(z\right)dz={\oint }_{C^{\prime }}\frac{z^3}{{\left(z-1\right)}^2}dz$

where $C^{\prime }$ is a circle centered at $z=1$ .

If $g\left(z\right)=z^3$ then ${\oint }_{C}f\left(z\right)dz={\oint }_{C^{\prime }}\frac{g\left(z\right)}{{\left(z-1\right)}^2}dz$

Since $g\left(z\right)$ is analytic within and on the circle $C^{\prime }$ we use Cauchy’s integral formula for derivatives to show that

${\oint }_C\frac{z^3}{{\left(z-1\right)}^2}dz=2\pi i\times \frac{1}{1!}{\left[g^{\prime }\left(z\right)\right]}_{z=1}=2\pi i{\left[3z^2\right]}_{z=1}=6\pi i$ .

Exercise

Evaluate  ${\oint }_C\frac{z}{z^2+9}dz$  where $C$ is the path:

1. $C_1:\left|z-3i\right|=1$
2. $C_2:\left|z+3i\right|=1$
3. $C_3:\left|z\right|=6.$

Answer