# Cauchy's Integral Theorem

Cauchy's Integral Theorem is one of two fundamental results in complex analysis due to Augustin Louis Cauchy. It states that if $f : \mathbb{C} \to \mathbb{C}$ is a complex-differentiable function in some simply connected region $R \subset \mathbb{C}$, and $C$ is a path in $R$ of finite length whose endpoints are identical, then $$\int\limits_C f(z) dz = 0 .$$ The other result, which is arbitrarily distinguished from this one as Cauchy's Integral Formula, says that under the same premises if $C$ is a circle of center $z_0$ with counteclockwise direction, then $$f(z_0) = \frac{1}{2\pi i } \int\limits_C \frac{f(z)}{z- z_0} dz .$$

## Proofs

### Proof 1

We will prove the theorem for the case when $C$ is a triangle. We leave it as an exercise to verify that all other paths can be sufficiently approximated with triangles.

Lemma. Cauchy's Integral Theorem holds when $f(z)$ is a constant function.

Proof. Let $C$ be a triangle; let $X$, $Y$, and $Z$ be its vertices, Let $h$ a mapping of $[0,3]$ onto $C$ such that $h([0,1]) = XY$, $h([1,2]) = YZ$, $h([2,3]) = ZX$, and such that $h$ is linear on each of these subintervals. Let $c$ be the constant value of $f$. Then \begin{align*} \int\limits_C f(z)dz &= \int\limits_0^1 f(h(t))h'(t) dt + \int\limits_1^2 f(h(t))h'(t) dt +\int\limits_2^3 f(h(t)) h'(t) dt \\ &= c (B-A) + c(C-B) + c(B-C) = 0 . \qquad \blacksquare \end{align*}

Now we prove the main result.

We construct a sequence parths $C_0 = C, C_1, \dotsc$, recursively, as follows. Let $X, Y, Z$ be the vertices of $C_n$, and let $X'$, $Y'$, $Z'$ be the respective midpoints of $YZ$, $ZX$, $XY$. Then \begin{align*} \int\limits_{C_n} f(z)dz &= \int\limits_{XY} f(z)dz +\int\limits _{YZ} f(z)dz + \int\limits_{ZX} f(z)dz \\ &= \int\limits_{XZ'} f(z)dz + \int\limits_{Z'Y}f(z)dz + \int\limits_{ YX'}f(z)dz + \int\limits_{X'Z}f(z)dz + \int\limits_{ZY'}f(z)dz + \int\limits_{Y'X} f(z)dz \\ &= \int\limits_{XZ'Y'}f(z)dz + \int\limits_{Z'YX'}f(z)dz + \int\limits_{ X'ZY'}f(z)dz + \int\limits_{X'Y'Z'}f(z)dz . \end{align*} We choose $C_{n+1}$ from the paths $XZ'Y'$, $Z'YX'$, $X'ZY'$, $X'Y'Z'$ so that the quantity $\biggl\lvert \int\limits_{C_{n+1}} f(z) dz \biggr\rvert$ is maximal. Then $$\biggl\lvert \int\limits_{C_n} f(z)dz \biggr\rvert \le 4 \cdot \biggl\lvert \int\limits_{C_{n+1}} f(z)dz \biggr\rvert ,$$ so that $$\biggl\lvert \int\limits_{C} f(z)dz \biggr\rvert \le 4^n \cdot \biggl\lvert \int\limits_{C_{n}} f(z)dz \biggr\rvert .$$

Let $L(n)$ denote the longest side length in the triangle $C_n$; let $P(n)$ denote the perimeter of $C_n$. Then $L(n) = 2^{-n} L(0)$ and $P(n) = 2^{-n} P(0)$.

Let $R_n$ denote the closed region bounded by $C_n$. Since $R_0, R_1, \dotsc$ is a descending chain of nonempty closed sets, the set $\bigcap_{k=0}^\infty R_k$ is not empty, so let us choose some $z_0$ that is an element of $R_k$, for all $k \ge 0$.

The function $f$ is differentiable at the point $z_0$, so for every $\epsilon > 0$, there exists an $\delta$ such that for all $z$ for which $\lvert z- z_0 \rvert < \delta$, $$\left\lvert f(z) - f(z_0) \right\rvert < \epsilon \left\lvert z- z_0 \right\rvert .$$ If we pick $n$ such that $L(n) \le \delta$, then this inequality holds for all $z \in C_n$. By the lemma, $$\int\limits_{C_n} f(z)dz = \int\limits_{C_n} \bigl[ f(z) - f(z_0) \bigr] dz .$$ Now, let $h : [0, P(n)] \to C_n$ be a parameterization of $C_n$ such that $\lvert h'(t) \rvert = 1$ for all $t \in [0, P(n)]$. Then \begin{align*} \biggl\lvert \int\limits_{C_n} \bigl[ f(z) - f(z_0) \bigr] dz \biggr\rvert &= \biggl\lvert \int\limits_0^{P(n)} \bigl[ f(h(t)) - f(z_0) \bigr] h'(t) dt \biggr\rvert \\ &\le \int\limits_0^{P(n)} \bigl\lvert f[h(t)] - f(z_0) \bigr\rvert dt \\ &< \epsilon \int\limits_0^{P(n)} \bigl\lvert h(t) - z_0 \bigr\rvert dt \\ &\le \epsilon \int\limits_0^{P(n)} L(n) dt \\ &= \epsilon P(n) L(n) = \epsilon 4^{-n} P(0) L(0) . \end{align*} Therefore $$\biggl\lvert \int\limits_C f(z)dz \biggr\rvert \le 4^n \biggl\lvert \int\limits_{C_n} f(z)dz \biggr\rvert < 4^n \epsilon 4^{-n}P(0) L(0) = \epsilon P(0)L(0) .$$ Since $\epsilon$ can be arbitrarily small, it follows that $$\biggl\lvert \int\limits_{C} f(z) dz \biggr\rvert = 0,$$ whence $$\int\limits_C f(z) dz = 0 ,$$ as desired. $\blacksquare$

### Proof 2

We use Green's Theorem.

Let $x,y$ denote the real numbers such that $z = x+iy$. Let Let $A$ and $B$ be the functions mapping $\Re R \times \Im R$ into $\mathbb{R}$ such that $f(z) = A(x,y) + iB(x,y)$. Then \begin{align*} \int\limits_C f(z) dz &= \int\limits_C f(z) (dx + i dy) = \int\limits_C \bigl[ A(x,y) + i B(x,y) \bigr] (dx + i dy) \\ &= \int\limits_C \bigl[ A(x,y) dx - B(x,y) dy \bigr] + i \int\limits_C \bigl[ A(x,y)dy + B(x,y)dx \bigr] . \end{align*} Now, since $f(z)$ is complex-differentiable, \begin{align*} \frac{\partial A}{\partial y} &= - \frac{\partial B}{\partial x} ,\\ \frac{\partial B}{\partial y} &= \frac{\partial A}{\partial x} . \end{align*} Let $D$ be the region bounded by $C$. Then by Green's theorem, $$\int\limits_C \bigl[ A(x,y) dx - B (x,y)dy \bigr] = \iint\limits_D \left( -\frac{\partial B}{\partial x} - \frac{\partial A}{\partial y} \right) = 0 ,$$ and similarly, $$\int\limits_C \bigl[ B(x,y)dx + A(x,y) dy \bigr] = \iint\limits_D \left( \frac{\partial A}{\partial x} - \frac{\partial B}{\partial y} \right) = 0 .$$ Thus Cauchy's theorem holds. $\blacksquare$

## Meaning

The Cauchy Integral Theorem guarantees that the integral of a function over a path depends only on the endpoints of a path, provided the function in question is complex-differentiable in all the areas bounded by the paths. Indeed, if $P_1$ and $P_2$ are two paths from $A$ to $B$, then $$\int\limits_{P_1} f(z)dz - \int\limits_{P_2} f(z)dz = \int\limits_{P_1 - P_2} f(z)dz = 0 .$$