kuniga.me > NP-Incompleteness > Cauchy Integral Theorem

26 Apr 2024

This is our third post in the series with my notes on complex integration, corresponding to *Chapter 4* in Ahlfors’ Complex Analysis.

The Cauchy integral theorem provides conditions under which the integral over a closed curve is zero.

The previous posts from the series:

In the previous post [3] we ended with the following *Corollary 2* stating that:

The complex line integral $\int_\gamma f(z)dz$, defined in $\Omega$, depends only on the endpoints of $\gamma$ if and only if $f$ is the derivative of some holomorphic function in $\Omega$.

Another corollary is the following:

**Corollary 1.** Let $f(z)$ be a function defined in $\Omega$. Then

If and only if $f$ is the derivative of some holomorphic function $F$ in $\Omega$.

The general idea of Cauchy’s theorem that we’ll cover in this post is that we only need $f$ itself to be holomorphic in $\Omega$, for special types of the region $\Omega$.

A result we haven’t proved yet says that the derivative of a holomorphic function is itself holomorphic, but not all holomorphic functions are derivatives of of a holomorphic function. So Cauchy’s theorem is a stronger result.

We first consider the case where $\Omega$ is a rectangle $R$ defined by

\[a \le x \le b, c \le y \le d\]The curve we’ll use is the border of $R$, denoted by $\partial R$, with a clock-wise orientation as depicted in *Figure 1*. In other words, it’s the segments $(a, c) \rightarrow (b, c)$, $(b, c) \rightarrow (b, d)$, $(b, d) \rightarrow (a, d)$ and $(a, d) \rightarrow (a, c)$.

**Theorem 1.** If the function $f(z)$ is holomorphic in $R$, then

So far we have $\abs{z - z^*} \lt \delta$, but we can find a tigher upper bound. The maximum distance between two points in a rectangle is its diagonal. Let $\Delta$ be the diagonal of $R$. Every time we pick a subrectangle the diagonal is halved, so the diagonal of $R_n$ is $2^{-n} \Delta$. Thus if $z$ and $z^*$ are in $R_n$, $\abs{a - z^*} \le 2^{-n} \Delta$. We have then: $$(1.5) \quad \abs{f(z) - f(z^*) + f'(z*) (z - z*)} \lt \epsilon \abs{z - z*} \le \epsilon 2^{-n} \Delta$$ In [2] we saw in the last example that for any curve $\gamma$, $$\int_\gamma (z - a)^{n}dz = 0$$ For $n \gt 0$. This let's us conclude the following: $$(1.6) \quad \int_{\partial R_n} (z - z^*) dz = 0$$ By replacing $a$ with $z^*$, $\gamma$ with $\partial R$, and setting $n = 1$. Similarly, if we do it for $n = 0$: $$(1.7) \quad \int_{\partial R_n} dz = 0$$ Since $-f(z*)$ exists, we can multiply it by $(1.7)$ and still get a 0: $$(1.8) \quad -f(z*) \int_{\partial R_n} dz = 0$$ Similarly, $f'(z*)$ exists and we can multiple it by $(1.6)$ and still get a 0: $$(1.9) \quad f'(z*) \int_{\partial R_n} (z - z^*) dz = 0$$ Since $(1.8)$ and $(1.9)$ are zero we can add them to $\eta(R_n)$ and obtain [4]: $$\eta(R_n) = \int_{\partial R_n} f(z)dz = \int_{\partial R_n} f(z)dz - f(z*) \int_{\partial R_n} dz + \int_f'(z*) {\partial R_n} (z - z^*) dz$$ Moving them under one integral (we can also move $f(z*)$ and $f'(z*)$ inside since they're constants with respect to $z \in R_n$): $$\eta(R_n) = \int_{\partial R_n} f(z) - f(z*) + f'(z*) (z - z^*) dz$$ All this trickery so that we get to the form of the inequality $(1.5)$. However, that inequality is with respect to the modulus, so we can use

The proof is very clever but I don’t have a good intuition on why it works. Anyway, we can generalize the theorem a bit by allowing points in $R$ for which $f(z)$ isn’t holomorphic:

**Theorem 2.** Let $f(z)$ be holomorphic in $R’$, obtained from the rectangle $R$ by removing a finite set of points $\xi_j$. Then if

for all $j$, then

\[(1) \quad \int_{\partial R} f(z) dz = 0\]This lets us reduce the theorem to the case where we remove exactly one point $\xi$ from $R$ (the case with zero points is

Using the same argument, we can subdivide a rectangle $R'$ containing $\xi$ such that the subrectangle containing $\xi$, denoted by $R_0$, is:

- A square of size $L$
- $\xi$ lies on its center
- L is infinitesimally small
- Satisfying: $$(2.1) \quad \int_{\partial R'} f(z)dz = \int_{\partial R_0} f(z)dz$$

So we’re saying that if $f(z)$ is not holomorphic at specific points in $R$ but it tends to 0 there, integrating over its boundary is still yields 0.

We now consider the case where $\Omega$ is the open circle $\abs{z - z_0} \lt \rho$, which we’ll denote by $\Delta$. We have the following result:

**Theorem 3.** If $f(z)$ is holomorphic in $\Delta$, then

for any closed curve $\gamma$ in $\Delta$.

We'll keep $(x_0, y_0)$ fixed and assume $(x', y')$ is variable. We can then define a function of $x'$ and $y'$: $$F(x', y') = \int_{\sigma} f(z) dz$$ Note that $\sigma$ is implicitly a function of $x'$ and $y'$ since its endpoint is $(x', y')$. And as discussed in [3] (section

One question that came to mind when trying to deal with these corner cases: can't we simply choose a different starting point for these cases? I believe the answer to be no. Then we wouldn't be able to treat $x_0$ and $y_0$ as constants and they would be a function of $x$ and $y$.

Note that it's fine for the curve $\sigma$ to be a function of $x$ and $y$ (as it is in fact) since we don't make any assumption about its constancy in $(3.1)$.

In the same way we generalized *Theorem 1.* to allow for points in the rectangle where we allow $f$ to be non-holomorphic, we can generalize *Theorem 3.*

**Theorem 4.** Let $f(z)$ be holomorphic in the region $\Delta’$ obtained by omitting a finite number of points $\xi_j$ from the open disk $\Delta$. If $f(z)$ is such that

then

\[\int_{\gamma} f(z) dz = 0\]for any closed curve $\gamma$ in $\Delta’$.

The first difference is that equation $(3.1)$ will now look like: $$F(x', y') = \int_{x_0}^{x'} f(x, y_1) dx + i \paren{\int_{y_0}^{y_1} f(x_0, y) dy + \int_{y_1}^{y'} f(x', y) dy}$$ To determine $\frac{\partial F}{\partial y}(x', y')$, we'll need to compute $F(x', y' + h)$, with $h \rightarrow 0$. We can use the exact same curve as we did for $F(x', y')$ except that the last segment will now go to $y' + h$, that is: $$F(x', y') = \int_{x_0}^{x'} f(x, y_1) dx + i \paren{\int_{y_0}^{y_1} f(x_0, y) dy + \int_{y_1}^{y' + h} f(x', y) dy}$$ By considering the limit $(3.2)$: $$\frac{\partial F}{\partial y}(x', y') = \lim_{h \rightarrow 0} \frac{F(x', y' + h) - F(x', y')}{h}$$ We'll arrive at the same conclusion that $$\frac{\partial F}{\partial y}(x', y') = i f(x', y')$$ If we call the blue segments $\sigma$ and the red ones $\overline{\sigma}$, we'll still conclude that $$(4.1) \quad F(x', y') = \int_{\sigma} f(z) dz = \int_{-\overline{\sigma}} f(z) dz$$ Because the 2 rectangles cancel out. Computing $\partial F/\partial x (x', y')$ is trickier however. That's because when we consider the point $(x' \pm h, y')$, the curve utilized is the 2-segment one because the path to $(x' \pm h, y')$ doesn't contain a $\xi_j$. This is illustrated in

So when we consider the difference $F(x' + h, y') - F(x', y')$ we need to be careful. The trick is to use rectangles to reduce the differences. Taking the example for

It took me a long time to figure out the proof of *Theorem 4*. In [1], Ahlfors provides almost no details besides *Figure 4.1* included in the proof of *Theorem 4*.

On my first read of the book I thought I had understood the proof but once I tried to plug it into the definition of derivative as a limit, I realized I didn’t understand it properly.