I have been reading the complex analysis part from the book "Advanced Engineering Mathematics by Erwin Kreyszig" and have been confused regarding this theorem's application.
Independence of Path
If $f(z)$ is analytic in a simply connected domain $D$, then the integral of $f(z)$ is independent of path in $D$.
Later in an example, the author brings up this standard result
$$\oint_{c} (z-z_0)^ndz = \begin{cases}
2\pi i\ for\ n = -1 \\\\
0\ for\ all\ other\ integers
\end{cases}$$
for counterclockwise integration around any simple closed path containing $z_0$ in its interior.
We had proved this using parametrisation of curve earlier for the case of a circle and he extends the argument by saying that we could deform this circle into any required curve.
Later in the chapter, the author also brings up Cauchy's Integral theorem for multiply connected domains
Cauchy’s theorem applies to multiply connected domains. We first explain this for a doubly connected domain D with outer boundary curve $C_1$ and inner $C_2$. If a function $f(z)$ is analytic in any domain $D*$ that contains $D$ and its boundary curves, we claim that
$$\oint_{c_1} f(z)dz = \oint_{c_2} f(z)dz$$
both integrals being taken counterclockwise (or both clockwise, and regardless of whether or not the full interior of $C_2$ belongs to $D*$).
Question
While this makes sense for all $n > -1$ since they are analytic at origin, I don't see how we are applying this for all the lower powers.
- How can we claim so generally that the result would be true for any curve without checking whether the function is holomorphic?
- What exactly is the way to find the integral of $z^-1$ for any general contour?
- Can we actually use Cauchy's integral formula for multiply connected domains and simple consider a domain with inner boundary of a unit circle and any general outer boundary and claim that these two would be equal?
Best Answer
The basics
$$\int_{|z|=r} z^n dz \ = \ r^n \ \int_0^{2\pi} e^{i \ n \phi}\ d(r \ e^{i \phi}) \ = \ i \ r^{n+1} \ \int_0^{2\pi}\ e^{i \ (n+1) \ \phi} \ d\phi \ = \ \cases{ 2\pi \ i \ & n=-1\\ 0},$$
The same result is true for any triangle containing the origin (try it by parametrizing three lines enclosing the origin).
Orientation of the path direction circling the origin is implied, so we orient all boundaries counterclockwise i.e. "mathematically positive" by the rule of thumb, that orientation of mathematicians is always reverse to the common rules.
By a triangulation of the inner of a rectifiable curve making a simple loop around the point $z$ you get the inegral theorem of Cauchy,
$$\frac{1}{2\ \pi \ i}\int_{\mathit C(z)} \frac{f(\xi)}{ \xi-z} \ = \ f(z) \quad \ \Rightarrow \quad \frac{1}{n!} \ \frac{d}{dz}^n \ f(z) \ = \ \frac{1}{2\ \pi \ i}\int_{\mathit C(z)} \frac{f(\xi)}{ (\xi-z)^{n+1}}. $$
Consequences are the residue theorem; the path independence of the complex line integral for functions analytic,i.e. complex differentiable, in the enclosed area between the paths; all inner line integrals cancel by the fact that each inner boundary segment is passed two times in different directions, if a line integral along the boundary is divided into a sum of integrals over the boundary of a partition of the area.