Cauchy Theorem and Its Applications

Updated 2 Aug 2026Chapter (PDF)

We have proven that if $f$ is continuous and admits a primitive $F$ on an open set $\Omega$ , then for any path $\gamma \subseteq \bC$ that starts at $w$ and ends at $w_0$ , we have that

\int_\gamma f(z) \, \dd z = F(w_0) - F(w).

From this, we immediately have that if $\gamma$ is a closed path, then the integral evaluates to 0.

This mirrors the fundamental theorem of calculus, with the exception that this doesn't apply to all continuous functions—the condition that $f$ admits a primitive is crucial. In real variables, a continuous function automatically admits a primitive/antiderivative, but this is not necessarily true in complex variables. For example, consider the function $1/z$ that is entire on $\bC^*$ . For the unit circle $\gamma$ , note that

\int_\gamma \frac{1}{z} \, \dd z = \int_0^{2\pi} \frac{ie^{it}}{e^{it}} \, \dd t = 2\pi i \ne 0,

and so by contrapositive, $1/z$ has no primitive on $\bC^*$ in spite of its holomorphicity.

Explicitly finding a primitive for a function is admittedly difficult. We alter our hypotheses to be more tractable, yielding Cauchy's theorem. It states that if $f$ is holomorphic on an open set $\Omega$ , then for any path $\gamma \subseteq \Omega$ whose interior is also contained on $\Omega$ , we have that

\int_\gamma f(z) \, \dd z = 0.

We prove this for toy contours (i.e., for contours where the “interior” is intuitively obvious) and defer the proof of the full statement for simply closed (i.e., Jordan) curves to the future. We start by proving Goursat's theorem, which proves this statement for the case where $\gamma$ is a triangle.

[2.0.1]Theorem(Goursat's Theorem)#

Suppose $f$ is holomorphic on an open set $\Omega$ . Let $T \subseteq \Omega$ be some path that traces out a triangle whose interior also lies in $\Omega$ . Then

\int_T f(z) \, \dd z = 0.

Proof.

Let $T^{(0)}$ be the triangle in question, whose orientation is fixed and considered “positive,” and let $d^{(0)}, p^{(0)}$ represent the diameter and perimeter of these triangles respectively. Consider the midpoints of each side of $T^{(0)}$ and connect each of them by a straight line, partitioning $T^{(0)}$ into four triangles $T_k^{(0)}$ for $k = 1, 2, 3, 4$ . We choose the direction of each triangle in the diagram shown in the next remark block.

Since the contours that aren't on the sides of $T^{(0)}$ cancel out, we know that

\int_{T^{(0)}} f(z) \, \dd z = \sum_{k = 1}^4 \int_{T_k^{(0)}} f(z) \, \dd z.

By the triangle inequality, there exists a $k \in \{1, 2, 3, 4\}$ such that

\abs{\int_{T^{(0)}} f(z) \, \dd z} \le 4 \abs{\int_{T_k^{(0)}} f(z) \, \dd z}.

Let $T^{(1)} = T_k^{(0)}$ , and then let $d^{(1)}, p^{(1)}$ be the diameter and perimeter for $T^{(1)}$ . By geometry, we note that $d^{(1)} = d^{(0)}/2$ and $p^{(1)} = p^{(0)}/2$ . We generate a sequence of triangles

T^{(0)}, T^{(1)}, \dots, T^{(n)}, \dots

with the properties that

\abs{\int_{T^{(0)}} f(z) \, \dd z} \le 4^n \abs{\int_{T^{(n)}} f(z) \, \dd z}, \quad d^{(n)} = d^{(0)}/2^n, \quad p^{(n)} = p^{(0)}/2^n.

Let $\mathcal{T}^{(k)} = \bar{T^{(k)}}$ , a closed and bounded (and thus compact) subspace of $\bC$ in its metric topology. Thus

\mathcal{T}^{(0)} \supseteq \mathcal{T}^{(1)} \supseteq \cdots

forms a chain of compact subsets. Then there is a $z_0 \in \bigcap_{k \in \bZ_+} \mathcal{T}^{(k)}$ , and since $f$ is holomorphic on $\Omega$ , it is holomorphic at $z_0$ . Thus write

f(z) = f(z_0) + f'(z_0)(z - z_0) + \psi(z)(z - z_0),

where $\psi(z) \to 0$ as $z \to z_0$ . Integrating both sides over $T^{(n)}$ , for any $n \in \bZ_+$ , yields

\int_{T^{(n)}} f(z) \, \dd z = f(z_0) \int_{T^{(n)}} \dd z + f'(z_0) \int_{T^{(n)}} (z - z_0) \, \dd z + \int_{T^{(n)}} \psi(z)(z - z_0),

where the first two integrals vanish since the polynomials $1$ and $z - z_0$ are entire (and $T^{(n)}$ is closed). So we are left with

\int_{T^{(n)}} f(z) \, \dd z = \int_{T^{(n)}} \psi(z)(z - z_0).

By the ML inequality, we have that

\int_{T^{(n)}} f(z) \, \dd z \le \sup_{z \in T^{(n)}} |\psi(z)(z - z_0)| \times p^{(n)}.

Note that $|z - z_0| \le d^{(n)}$ , obviously, and so we have

\int_{T^{(n)}} f(z) \, \dd z \le \epsilon_n d^{(n)} p^{(n)} = \epsilon_n d^{(0)} p^{(0)} / 4^n,

where $\epsilon_n = \sup_{z \in T^{(n)}} |\psi(z)|$ , and since $z \to z_0$ as $n \to \infty$ , we have $\epsilon_n \to 0$ as $n \to \infty$ . Finally, we have that

\abs{\int_{T^{(0)}} f(z) \, \dd z} \le 4^n \abs{\int_{T^{(n)}} f(z) \, \dd z} \le \epsilon_n d^{(0)}p^{(0)} \to 0,

and so $\displaystyle \int_T f(z) \, \dd z = 0$ .

[2.0.2]Remark(Figure for Goursat's Theorem)#

[2.0.3]Corollary#

Suppose $f$ is holomorphic on an open set $\Omega$ . Let $R \subseteq \Omega$ be some path that traces out a rectangle whose interior also lies in $\Omega$ . Then

\int_R f(z) \, \dd z = 0.

Proof.

Split the rectangle about the diagonal, yielding two triangles, and apply Goursat's theorem to each.