Significance of Cauchy's theorem.

Question

I am studying complex analysis. In the beginning of the chapter on integration theory there is a theorem known as Cauchy's theorem. It states that:

If $\Omega\subset \mathbb C$ be open and $f:\Omega\to \mathbb C$ be a holomorphic function and $C$ be a closed curve in $\Omega$ whose interior is also in the set, then $\int\limits_C f(z)dz=0$.

I want to know the significance of this theorem. I have tried to understand on my own but I want to make sure if it is all I need to understand. Cauchy's theorem basically says that if I integrate a holomorphic function along a closed curve, then the integral will be $0$, this basically means that given any two points in the set $\Omega$, I can choose whichever path I want between them for integrating between the two points without affecting the result. That means integral of a holomorphic function is path independent. Am I correct? Is there anything I am yet to realize about this theorem?

Answer 1

The significance of Cauchy's theorem is in all of it's extreme consequences.

Take for example, the fact that complex differentiability immediately implies infinite differentiability. In fact, if a function, $f(z)$, is differentiable on any $\overline{B(z_0,r)}$ not only will $f(z)$ be infinitely differentiable at $z_0$, it is guaranteed that the Taylor series expansion at $z_0$ will agree with $f(z)$ on all of $B(z_0,r)$ and consequently will have a radius of convergence of at least $r$.

How is this possible you might ask? We normally think of Taylor series as expressions entirely defined by all the derivatives at the center. But Cauchy's consequences demonstrates that a Taylor series expansion can be calculated by the boundary values of the ball alone (no need to know anything about $z_0$ or any of it's derivatives) (this is just a non-obvious fact):

$$f(z) = \sum_{k=0}^\infty \left(\frac{1}{2\pi i}\int_C \frac{f(w)}{(w-z_0)^{k+1}}dz \right) (z-z_0)^k $$

In fact, $f$ does not need any values of it's derivatives anywhere in the ball, to calculate it's Taylor series. This is just simply remarkable. This is of course an extreme consequence of Cauchy's theorem which is precluded by Cauchy's integral formula, which states (as we might now expect) that only the boundary values alone are sufficient to calculate all interior points of a holomorphic function; that is for any $z$ in the interior:

$$f(z) = \frac{1}{2\pi i}\int_C \frac{f(w)}{w-z}dw$$

That's right, the boundary values alone are all that is needed to find ALL the values of $f$ within it's interior.

This is a direct consequence of Cauchy's theorem. To provide a fairly simply proof, I do need to point out that Cauchy's theorem does not require analytic everywhere in the domain; it only requires analytic everywhere but a finite number of points on which it is continuous. This is simply because continuity doesn't effect the value an integral will have; i.e., if two contours are geometrically "close" to each other then their integrals will be close in value as well (this follows straight from the definition of contour integration); hence given any contour that happens to go through/around a few points of continuity we could simply perturb the contour a little so that we do not go through/around any of the continuity points, concluding that integrals of closed contours are still zero even with the presences of a few points of continuity.

Cauchy's integral formula now follows immediately. Given any fixed $z$ we have that the function $g(w)=\frac{f(w)-f(z)}{w-z}$ (obviously we take $g(z)=f'(z)$ to maintain continuity) is holomorphic everywhere and continuous at $z$. Hence,

$$\int_C \frac{f(w)-f(z)}{w-z} dw = 0$$

$$f(z)\int_C \frac{1}{w-z} dw = \int_C \frac{f(w)}{w-z} dw$$

$$f(z) = \frac{1}{2\pi i} \int_C \frac{f(w)}{w-z}dw$$

Answer 2

What you did is mostly correct. There is however one important thing missing: yes, if you have two points $z_0$ and $w_0$ and if you take two paths $\gamma$ and $\mu$ from $z_0$ to $w_0$, then the integrals $\int_\gamma f(z)\,\mathrm dz$ and $\int_\mu f(z)\,\mathrm dz$ will be equal provided that you can find an homotopy $H$ between them such that the range of $H$ is contained in $\Omega$. To be more precise: here, an homotopy is a continuous map from $[a,b]\times[0,1]$ (where $[a,b]$ is the domain of $\gamma$ and $\eta$) such that

• $(\forall t\in[a,b]):H(t,0)=\gamma(t)$;
• $(\forall t\in[a,b]):H(t,1)=\eta(t)$
• $(\forall u\in[0,1]):H(a,u)=z_0$;
• $(\forall u\in[0,1]):H(b,u)=w_0$.

That is, you can deform (in a continuous way) the path $\gamma$ into the path $\eta$ without leaving $\Omega$ and in such a way that all the intermediate paths (those of the form $t\mapsto H(t,u)$, for a fixed $u\in[0,1]$) go from $z_0$ to $w_0$. Then, yes, the integral will be the same for both paths.

Answer 3

Cauchy's theorem applies to things other than closed curves. It applies to boundaries of open sets. If $\Omega$ is a bounded open set with nice enough boundary, $f$ is $C^1$ on a neighborhood of $\overline{\Omega}$ and holomorphic on $\Omega$, then $\int_{\partial \Omega}f\,dz = 0$. One use of the theorem is to conclude that if you have a open set $D$ with $\overline{D} \subset \Omega$, then $$0 = \int_{\partial(\Omega \setminus \overline{D})}f\,dz = \int_{\partial \Omega}f\,dz - \int_{\partial D}f\,dz = 0.$$

Answer 4

Suppose you have a function that is analytic everywhere except at one point, and you want to evaluate the integral along a contour that includes this point in the interior of the contour.

Cauchy's theorem allows you to adjust the contour to one that is easier to integrate. In fact, for every contour that includes this point in the interior, the integral will be the same.

Effectively this means that meromorphic functions (functions that are holomorphic everywhere except for a set of isolated points) can be evaluated solely based on the points where they fail to be holomorphic.