Cauchy's integral formula

Cauchy’s integral formula gives us new approaches to compute contour integrals. For a function $f(z)$ that is analytic on and inside a closed simple curve $C$:

$$f(z_0)=\frac{1}{2\pi i}{\oint }_C\frac{f(z)}{z-z_0}dz$$

Similarly, for derivatives of complex-valued functions:

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

Computations

One of the key implications of this is that if our contour integral can be manipulated to where the denominator is $(z-z_0{)}^n$, even where $z_0=0$ or $n=1$, then we can use Cauchy’s integral formulas. For example:

$$2\pi if(z_0)={\oint }_C\frac{f(z)}{z-z_0}dz$$

and the equivalent same thing for higher orders.

For complicated looking integrands, we are able to factor and exploit a division trick to get something we can use Cauchy’s integral formula with. Take an example:

$${\oint }_C\frac{z^2+3z+2i}{z^2+3z-4};C:|z|=2$$ $${\oint }_C\frac{z^2+3z+2i}{(z+4)(z-1)}dz={\oint }_C\frac{\frac{z^2+3z+2i}{z+4}}{z-1}dz$$

since the denominator would be analytic within the contour $C$, we can now use the formula.

Intuitive understanding

An integral is an infinite sum, so we can intuitively think of it as a kind of partial fraction decomposition.

$$-\frac{1}{2\pi i}(\frac{F(z_1)}{z_1-s}+\cdots +\frac{F(z_n)}{z_n-s})$$

The integral formula says that $F$ is the sum of functions that look like:

$$\frac{1}{z-z_0}$$