Cauchy Residue Theorem and Cauchy integral formula

Question

Is it true that you can use the Cauchy Residue Theorem and the Cauchy integral formula interchangeably? I believe that the functions that satisfy the conditions of one, will indeed satisfy the conditions of the other?

Answer 1

The connection is as follows:

Consider a closed curve $\Gamma$, and a function $f$ holomorphic in $[\Gamma]$ then according to Cauchy's integral formula:

$$f(z_0) = \frac{1}{2\pi i}\int_{\Gamma} \frac{f(z)}{z-z_0}\text{d}z$$

The following establishes the link with the residue theorem:

$f$ is holomorphic on $[\Gamma]$, consider its Laurent (or even Taylor)-expansion at $z_0$.

$$f(z) = a_0 + a_1(z-z_0)+a_2(z-z_0)^2+\ldots$$

See that $f(z_0) = a_0$.

Now consider the function $\displaystyle \frac{f(z)}{z-z_0}$ which has a simple pole at $z_0$.

Its Laurent expansion will be:

$$\frac{f(z)}{z-z_0} = \frac{a_0}{z-z_0} + a_1+a_2(z-z_0)+\ldots$$

With $\displaystyle \operatorname*{res}_{z=z_0} \frac{f(z)}{z-z_0} = \color{red}{a_0}$.

Which results into according to the residue theorem:

$$\int_\Gamma \frac{f(z)}{z-z_0} \text{d} z = 2\pi i \cdot \color{red}{a_0} \qquad \text{or}\qquad \frac{1}{2\pi i}\int_\Gamma \frac{f(z)}{z-z_0} \text{d} z =a_0 = f(z_0)$$