Cauchy's Theorem

Why Closed-Curve Integrals Matter

The Evaluation by a Primitive theorem gives a simple reason for some closed-curve integrals to vanish: if $f$ has a primitive $F$, the integral around a closed curve is $F$ at the endpoint minus $F$ at the same point. But a holomorphic function need not come with a known primitive. Cauchy’s theorem provides a way to establish zero integrals directly from holomorphicity.

We begin with triangle boundaries. The key proof technique is subdivision: repeatedly divide a triangle into smaller triangles and use complex differentiability at a point where the selected triangles accumulate. This argument is often called the Goursat proof of Cauchy’s theorem. It requires only that the function be holomorphic, not that its derivative be continuous.

Definition (Positively Oriented Triangle Boundary): Let $T$ be a nondegenerate triangle in $\mathbb{C}$. Its positively oriented boundary, denoted $\partial T$, is the piecewise linear curve that traverses the three sides in counterclockwise order. The notation $\int_{\partial T} f(z)\,dz$ means the line integral along this curve.

Cauchy’s Theorem for Triangles

The theorem below is the central local result. The triangle and its boundary must lie in a region where $f$ is holomorphic. In particular, differentiability at every point of the triangle, including its boundary, is part of the hypothesis.

Theorem (Cauchy’s Theorem for Triangles, or Cauchy–Goursat Theorem): Let $U\subseteq\mathbb{C}$ be open, and let $f:U\to\mathbb{C}$ be holomorphic. If $T$ is a nondegenerate closed triangle contained in $U$, then $$ \int_{\partial T} f(z)\,dz=0. $$

Proof. Divide $T$ by joining the midpoints of its sides. This produces four smaller triangles, each with half the diameter and half the perimeter of $T$. Give each of their boundaries the positive orientation. The integral over $\partial T$ is the sum of the four integrals over these boundaries: every internal side occurs twice, once in each direction, and its two integrals cancel by the Additivity and Reversal proposition.

At least one of the four smaller triangles has a boundary integral whose modulus is at least one quarter of the modulus of the integral over $\partial T$. Otherwise, the triangle inequality applied to the sum of the four integrals would make the original integral’s modulus strictly smaller than itself. Select such a triangle and repeat the subdivision. This gives nested closed triangles $T_0,T_1,T_2,\ldots$, with $T_0=T$, such that

\left|\int_{\partial T_n} f(z)\,dz\right| \geq \frac{1}{4^n}\left|\int_{\partial T}f(z)\,dz\right|.

Write $D$ and $P$ for the diameter and perimeter of $T$. The diameter and perimeter of $T_n$ are $D/2^n$ and $P/2^n$. Since the triangles are nested, closed, and their diameters tend to zero, they have exactly one common point; call it $z_0$. This point belongs to $T$, so $f$ is complex differentiable there.

By complex differentiability at $z_0$, for $z$ near $z_0$ we can write

$$ f(z)=f(z_0)+f'(z_0)(z-z_0)+r(z)(z-z_0), $$

where $r(z)\to 0$ as $z\to z_0$. Set $r(z_0)=0$. Because every point of $T_n$ is at distance at most $D/2^n$ from $z_0$, the numbers

\varepsilon_n=\sup_{z\in T_n}|r(z)|

tend to zero. On the boundary of $T_n$, the constant term and the linear term in the expansion have zero integrals. Indeed, they have the primitives $f(z_0)z$ and $f'(z_0)(z-z_0)^2/2$, respectively, so this follows from the Evaluation by a Primitive theorem. The remaining term can be bounded using the length estimate for a line integral. Every boundary point $z$ satisfies $|z-z_0|\leq D/2^n$, so

\left|\int_{\partial T_n} f(z)\,dz\right| \leq \varepsilon_n\frac{D}{2^n}\frac{P}{2^n} = \frac{\varepsilon_n DP}{4^n}.

Combining the lower and upper bounds gives

\left|\int_{\partial T}f(z)\,dz\right| \leq \varepsilon_n DP

for every $n$. Since $\varepsilon_n\to0$, the original integral must be zero. This proves the theorem. $\square$

The argument’s crucial step is the local expansion at $z_0$. It controls the error after subtracting a constant and a linear function, and the remaining error is small enough, relative to the shrinking perimeter, to force the original integral to vanish. No continuity of $f'$ was used.

From Triangles to Convex Regions

A convex region contains the straight line segment joining any two of its points. That geometric property lets the triangle theorem produce a primitive throughout the region. We use “domain” to mean a nonempty, open, connected subset of $\mathbb{C}$; a convex open set is a domain.

Theorem (Cauchy’s Theorem on a Convex Domain): Let $U\subseteq\mathbb{C}$ be an open convex set, and let $f:U\to\mathbb{C}$ be holomorphic. Then $f$ has a primitive on $U$. Consequently, for every piecewise $C^1$ curve $\gamma:[a,b]\to U$, $$ \int_\gamma f(z)\,dz=F(\gamma(b))-F(\gamma(a)) $$ for that primitive $F$. In particular, the integral around every closed piecewise $C^1$ curve in $U$ is zero.

Proof. Fix a point $a\in U$. For $z\in U$, define $F(z)$ to be the integral of $f$ along the straight segment from $a$ to $z$. The segment lies in $U$ by convexity. We show that $F'(z)=f(z)$.

Take $z\in U$ and a nonzero complex number $h$ small enough that $z+h\in U$. Since $U$ is convex, the closed triangle with vertices $a$, $z$, and $z+h$ lies in $U$. If the three points are not collinear, they form a triangle; the triangle theorem, together with additivity and reversal, gives

F(z+h)-F(z)=\int_{[z,z+h]} f(w)\,dw,

where $[z,z+h]$ denotes the segment from $z$ to $z+h$. If the three points are collinear, the same identity follows by cancellation of the overlapping, oppositely traversed segments. Parametrize the segment by $w=z+th$, $0\leq t\leq1$. The definition of the line integral then gives

\frac{F(z+h)-F(z)}{h} = \int_0^1 f(z+th)\,dt.

Continuity of $f$ at $z$ implies that the right-hand side tends to $f(z)$ as $h\to0$: all points $z+th$ are within distance $|h|$ of $z$, uniformly for $0\leq t\leq1$. Hence $F'(z)=f(z)$, and $F$ is a primitive of $f$ on $U$. The Evaluation by a Primitive theorem now gives the asserted formula for every piecewise $C^1$ curve. If $\gamma$ is closed, then $\gamma(a)=\gamma(b)$, so the endpoint difference is zero. $\square$

This argument shows how the local triangle theorem yields a global conclusion when the region’s geometry permits segments between points. Convexity is a convenient sufficient condition. More general versions of Cauchy’s theorem apply to other regions under appropriate topological hypotheses, but those require additional arguments about the region.

Worked Examples

Worked Example: The Boundary of a Triangle

Let $T$ have vertices $0$, $1$, and $i$, in counterclockwise order, and let $f(z)=e^z$. The exponential function is holomorphic on all of $\mathbb{C}$, so the closed triangle $T$ lies in a region where $f$ is holomorphic. Cauchy’s theorem for triangles applies directly:

\int_{\partial T}e^z\,dz=0.

The conclusion is an exact integral value without separately parametrizing the three sides. It does not depend on the triangle’s shape; the same theorem applies to any nondegenerate triangle in $\mathbb{C}$.

Worked Example: A Closed Curve in a Convex Region

Consider the square boundary with vertices $0$, $1$, $1+i$, and $i$, traversed counterclockwise, and let $f(z)=1/(z+2)$. The function is holomorphic on the convex open half-plane $U=\{z:\operatorname{Re}z>-1\}$, which contains the square. The convex-domain theorem therefore gives

\int_{\partial T}\frac{1}{z+2}\,dz=0.

The pole at $-2$ is outside $U$. The example illustrates why the region matters: the theorem is applied where the function is holomorphic, not merely where it is defined at individual points of the curve.

Worked Example: A Singularity Inside the Circle

Let $\gamma(t)=e^{it}$, $0\leq t\leq2\pi$, be the unit circle traversed counterclockwise, and take $f(z)=1/z$. Along the curve, $f$ is defined. Since $\gamma'(t)=ie^{it}$, the line integral is

\int_\gamma \frac{1}{z}\,dz = \int_0^{2\pi}\frac{1}{e^{it}}\,ie^{it}\,dt = \int_0^{2\pi}i\,dt = 2\pi i.

There is no contradiction with Cauchy’s theorem. The function $1/z$ is not holomorphic on any open region containing the whole disk bounded by $\gamma$, because it is undefined at $0$. Holomorphicity along the curve alone is not enough to apply the theorem.

What the Theorem Does—and Does Not—Say

Cauchy’s theorem is a statement about a function on a region containing the curves or triangles under consideration. A common error is to check only that the integrand is defined on the boundary. The $1/z$ example shows why that is insufficient: the enclosed singularity prevents applying the theorem to the disk, and the integral is nonzero.

A second point is that a zero integral around a closed curve and path independence are closely related when a primitive exists. On a convex domain, the triangle theorem constructs that primitive, so any two piecewise $C^1$ paths in the domain with the same endpoints have equal integrals. Indeed, each integral equals the same endpoint difference. The result is powerful both as a way to evaluate integrals and as a foundation for subsequent results in complex analysis.

Check Your Understanding

Use the statements and arguments in this tutorial to answer the following questions.

What hypotheses are needed to apply Cauchy’s theorem to a triangle boundary?
Why does subdivision produce a nested sequence of triangles whose boundary integrals retain at least a quarter of the preceding modulus at each step?
In the triangle proof, why do the constant and linear terms in the expansion of $f$ at the common point have zero boundary integrals?
How does convexity help construct a primitive from the triangle theorem?
Why does the value $2\pi i$ for the integral of $1/z$ around the unit circle not contradict Cauchy’s theorem?

What You'll Learn