Cauchy's Theorem for a Rectangle

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Introduction

• Cauchy Integral Formula. If a function f is analytic in a simply connected domain D and γ is a simply closed contour (positive orientated) in D. Then, for any point $z_0$ inside γ we have $f(z_0) = \frac{1}{2\pi i}\cdot \int_{\gamma} \frac{f(z)}{z-z_0}dz$ (Figure D).

• The Residue Theorem states that the integral of a function f(z) around a simple (doesn't cross itself), and positively oriented (counter-clockwise) contour γ is equal to 2πi times the sum of the residues of all singularities inside the contour, $\int_{\gamma} f(z)dz = 2\pi i \sum Res(f, z_k)$ where $z_k$ is a reside inside γ. f must be analytic inside the contour γ, except for a finite number of isolated singularities.

1. For a simple pole at a point $z_k$ (i.e., the denominator has a simple root) is calculated using the formula $Res(f, z_k) = \lim_{z \to z_k}(z-z_k)f(z)$.
2. For a pole of order m, $Res(f, z_k) = \frac{1}{(m-1)!}\lim_{z \to z_k} \frac{d^{m-1}}{dz^{m-1}}[(z-z_k)^mf(z)]$

• Proposition. Linearity of the contour integral, $\int_{\gamma} (af(z) + bg(z))dz = a\int_{\gamma} f(z)dz + b\int_{\gamma} g(z)dz$
• Proposition. Properties of contour integrals under path manipulation. Let γ be a contour, defined by a function $\gamma: [a, b] \to \mathbb{C}$ and let f be a continuous functions $f: \gamma^* \to \mathbb{C}$ defined along the image of ᵞ. Then,

1. Reversal of orientation. $\int_{-\gamma} f(z)dz = -\int_{\gamma} f(z)dz$ where $-\gamma:[a,b] \to \mathbb{C}$ is the reverse of the contour (traversing the same path in the opposite direction) defined by $(- \gamma)(t)=\gamma(a+b-t)$. This property states that reversing the direction of a contour changes or flips the sign of the integral. Intuitively, this is analogous to how reversing the limits of integration in real analysis changes the sign: $\int_a^b f(x)dx = -\int_b^a f(x)dx$. In complex analysis, the orientation of the contour matters because the integral depends on the direction in which we traverse the path.
2. Additivity under subdivision. Suppose a < c < b, let split $\gamma$ into two sub-contours $\gamma_1 = \gamma|_{[a, c]} \text{ and } \gamma_2 = \gamma|_{[c, b]}$, then $\int_{\gamma} f(z)dz = \int_{\gamma_1} f(z)dz + \int_{\gamma_2} f(z)dz$. This property states that integrating over the whole contour is the same as integrating over the pieces successively. This is the complex analogue of the additive property of definite integrals: $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx $ for a < c < b.
3. Invariance of Contour Integrals Under Reparameterization. Let $\tilde{\gamma}$ be a contour, defined by a function $\tilde{\gamma}: [c, d] \to \mathbb{C}$. If $\tilde{\gamma}$ is another parameterization of the same oriented path $\gamma$ meaning there exists a one-to-one, continuously differentiable map $\psi: [c,d]\to[a,b]$ with a positive derivative $\psi'(t)>0$ such that $\tilde{\gamma}(t) = \gamma(\psi(t))$, then $\int_{\gamma} f(z)dz = \int_{\tilde{\gamma}} f(z)dz$

• Deformation of Contours. If two contours $\gamma_1$ and $\gamma_2 $ are homotopic (i.e., one can be continuously deformed into the other without crossing any singularities of f) in a domain where f(z) is analytic, then: $\int_{\gamma_1} f(z)dz = \int_{\gamma_2} f(z)dz.$

• Fundamental Theorem of Calculus for Contours. Suppose $\gamma$ is a contour (piecewise smooth path) from a to b, f is defined on a domain D containing $\gamma^*$ (the image of $\gamma$) and admits a primitive (antiderivative) F on D (i.e., $F'(z) = f(z)$), then $\int_{\gamma} f(z)dz = F(\gamma(b)) - F(\gamma(a))$. In particular, if $\gamma$ is a closed contour (i.e., $\gamma(a)=\gamma(b)$), this integral evaluates to zero, $\int_{\gamma} f(z)dz = 0.$

• Estimation Theorem or the Triangle Inequality for Integrals. The triangle inequality for integrals in complex analysis states thatfor any continuous complex function $f:[a,b] \to \mathbb{C}$ on a closed real interval [a,b] (f(t) = u(t) + iv(t), t a real parameter), the following holds: $∣\int_a^b f(t)dt| \leq \int_a^b |f(t)|dt$.

• Estimation Lemma (ML Inequality) for contour integrals. For any continuous complex function $f:[a,b] \to \mathbb{C}$ on a closed real interval [a,b] (f(z) = u(x, y) + iv(x, y)) with f bounded by some constant M along the entire contour, |f(z)| ≤ M for all $z \in \gamma^*$ (the image/trace of the contour in the complex plane), the following holds: $∣\int_\gamma f(z)dz| \leq M \cdot l(\gamma)$ where l(γ) is the arc length of the contour γ given by $\int_a^b |\gamma^{'}(t)|dt = \int_a^b \sqrt{x'(t)^2 + y'(t)^2} \text{ where } \gamma(t) = x(t) + iy(t)$.

• Jordan’s curve theorem. Any simple closed curve (a continuous loop in the plane that does not intersect itself) separates the plane into two disjoint connected regions: one interior (bounded) and one exterior (unbounded). The curve itself is the boundary of both regions. In other words, it partitions the plane into exactly three disjoint sets:

1. The Curve itself ($\gamma$).
2. The Interior (Int($\gamma$)) is the finite area enclosed by the curve (e.g., if you draw a circle on a piece of paper, the interior region would be everything inside the circle). It is a bounded, simply connected region.
3. The Exterior (Ext($\gamma$)) is the infinite area outside the curve (e.g., using the same circle example, the exterior region would include all points on the paper that are not inside the circle). It is an unbounded region.

• Definition. Let $\gamma: [α, β] → \mathbb{C}$ be a piecewise smooth, closed curve (so γ(α) = γ(β)), the winding number of γ about $z_0$, a complex point not on the curve $z_0$ ∉ γ([α, β]), also known as the index of $z_0$ with respect to γ, is defined as $n(\gamma, z_0) = Ind_{\gamma}(z_0) = \frac{1}{2\pi i} \oint_{\gamma} \frac{dz}{z - z_0} = \frac{1}{2\pi i} \int_{\alpha}^{\beta} \frac{\gamma'(t)}{\gamma(t) - z_0}dt$ (the total number of counterclockwise turns that the object makes around $z_0$).

Cauchy’s Theorem in Complex Analysis

• Cauchy’s theorem (Classical “Green’s theorem” version). Let $\Omega \subset \mathbb{C}$ be an open domain. Suppose f = u + iv is analytic in $\Omega$ and its partial derivatives ( $u_x,u_y,v_x,v_y$) are continuous in $\Omega$. If $\gamma$ is a positively oriented, piecewise-smooth $C^1$, simple closed contour with $\gamma^* \cup \operatorname{Int}(\gamma) \subset \Omega$ (its path and interior both lie inside Ω), then $\oint_{\gamma} f(z)dz = 0.$

• Cauchy’s Theorem (Cauchy–Goursat). This is the more powerful version, as it removes the need for continuous partial derivatives. If f is analytic in an open set containing a simple closed contour γ and its interior $\gamma^*\cup\operatorname{Int}(\gamma)$, then $\oint_{\gamma} f(z)dz = 0$.

Cauchy’s Theorem for a Rectangle

Cauchy’s Theorem for a Rectangle. Let R be a rectangle region $R = \{x + iy : a \leq x \leq b,\; c \leq y \leq d\}$. Let f be holomorphic on an open setcontaining R (so f is analytic on and in a neighborhood of R). Then, the positively oriented boundary ($\partial R$) contour integral of f vanishes,$\boxed{\oint_{\partial R} f(z)\,dz = 0.}$

Theorem (Cauchy’s Theorem for a Triangle). Let T be a triangular region. Let f be holomorphic on an open set containing T (so f is analytic on and in a neighborhood of T). Then, the positively oriented boundary ($\partial T$) contour integral of f vanishes, $\oint_{\partial T} f(z)\,dz = 0.$

Proof.

The proof relies on the method of bisection. While the logic applies equally well to a triangle, for the sake of simplicity, we will formalize the proof for the rectangular region $R$.

1. The Bisection Process.

Let’s name the initial rectangle $R_0 = R$. We denote its perimeter as $l$ and its diagonal as $d$.

We subdivide $R_0$ into four equally sized subrectangles (Figure 1) by drawing lines parallel to its sides through its center (Figure 1). We observe that the contour integrals along the inner boundaries cancel each other out due to opposite orientations (every newly created interior edge is traversed twice in opposite directions).

Consequently, the integral over the entire boundary of ∂R can be expressed as the sum of integrals over these four smaller rectangles, $\int_{\partial R}f(z)dz = \int_{\partial R_1}f(z)dz + \int_{\partial R_2}f(z)dz + \int_{\partial R_3}f(z)dz + \int_{\partial R_4}f(z)dz$. Or using a different notation, $\int_{\partial R_0}f(z)dz = \int_{\partial R_1^{(1)}}f(z)dz + \int_{\partial R_1^{(2)}}f(z)dz + \int_{\partial R_1^{(3)}}f(z)dz + \int_{\partial R_1^{(4)}}f(z)dz$

Similarly, we can divide the triangular contour shown in Figure 4, then: $\int_{\partial T}f(z)dz = \int_{\partial T_1}f(z)dz + \int_{\partial T_2}f(z)dz + \int_{\partial T_3}f(z)dz + \int_{\partial T_4}f(z)dz$

Applying the triangle inequality, we get: $\left|\int_{\partial R_0}f(z)dz\right| \le \sum_{k=1}^{4} \left|\int_{\partial R_1^{(k)}}f(z)dz\right|$

By a simple averaging argument (or the Pigeonhole Principle), at least one of these four subrectangles must contribute at least one-fourth of the total absolute value. Let’s call this specific subrectangle $R_1$. Therefore: $\left|\int_{\partial R_1}f(z)dz\right| \ge \frac{1}{4}\left|\int_{\partial R_0}f(z)dz\right|$

Set $a_k:=\left| \int _{\partial R_1^{(k)}}f(z)\, dz\right| \geq 0,\quad k=1,\dots ,4,$ and let $S:=\sum _{k=1}^4a_k,$ so $\left| \int _{\partial R_0}f(z)\, dz\right| \leq S.$ Now, by the Pigeonhole Principle (or just basic averaging), if all four $a_k$ were strictly less than S/4, then $\sum _{k=1}^4a_k<4\cdot \frac{S}{4}=S,$ which is impossible.

2. Constructing the Nested Sequence.

We can apply this reasoning recursively. We divide $R_1$ into four even smaller rectangles, select the one that satisfies the inequality, and call it $R_2$.

Continuing by induction, we obtain an infinite sequence of rectangles $\{R_n\}_{n=0}^\infty$ with the following properties:

1. Nested Structure: $R_{n+1} \subseteq R_n$ for all $n$.
2. Integral Bound: $\left|\int_{\partial R_n}f(z)dz\right| \ge \frac{1}{4^n}\left|\int_{\partial R_0}f(z)dz\right|$.
3. Shrinking Geometry: The perimeter of $R_n$ is $l_n = \frac{l}{2^n}$, and the diagonal of $R_n$ is $d_n = \frac{d}{2^n}$

3. Cantor’s Intersection Theorem (Metric Space Version).

Let (X, d) be a complete metric space (like $\mathbb{R}^n$). Suppose $\{S(n)\}_{n=1}^\infty$ is a sequence of non-empty, closed, and bounded subsets of X such that:

• The sequence is nested: $S(n+1) \subseteq S(n)$ for all n,
• The diameter of the sets tends (shrinks) to zero: $\lim_{n \to \infty} \text{diameters}(S(n)) = 0.$

Then, their intersection $\bigcap_{n=1}^\infty S(n)$ contains exactly one point, $z^*$.

Because $\mathbb{C}$ (equivalent to $\mathbb{R}^2$) is a complete metric space, we can apply Cantor’s Intersection Theorem. Since our sequence $\{R_n\}_{n=0}^\infty$ consists of non-empty, closed, and bounded subsets that are nested, and their diameters (the diagonals $d_n$) shrink to zero as $n \to \infty$, their infinite intersection must contain exactly one point. Let’s call this unique point $z^*$: $\bigcap_{n=0}^\infty R_n = \{z^*\}$.

4. Applying Analyticity

Since $z^*$ is inside $R_0$, and $f$ is analytic on an open set containing $R_0$, we know $f$ is complex differentiable at $z^*$ (analyticity implies complex differentiability at $z^*$). By the definition of the derivative: $\lim_{z \to z^*} \frac{f(z) - f(z^*)}{z-z^*} = f'(z^*) \iff \lim_{z \to z^*} \frac{f(z) - f(z^*) - f'(z^*)(z-z^*)}{z-z^*} = 0$

This implies that for any arbitrarily small $\epsilon > 0$, there exists a $\delta > 0$ such that for all $z$ satisfying $|z - z^*| \le \delta$, the following holds: $|f(z) - f(z^*) - f'(z^*)(z-z^*)| \le \epsilon|z-z^*|, (\star).$

Because the diameters of our rectangles shrink to zero, we can find a sufficiently large index $N$ such that for all $n \ge N$, the entire rectangle $R_n$ is contained within the $\delta$-neighborhood of $z^*$. Thus, for any $z$ on the boundary $\partial R_n$, the above $\epsilon$-inequality holds.

5. The Final Estimation

Let’s integrate the function over the boundary of this tiny rectangle $R_n$. By the Fundamental Theorem of Calculus (or noting that polynomials have global antiderivatives), the integrals of constant and linear terms over any closed loop vanish:

By the Fundamental Theorem of Calculus, we know that $\int_{\partial R_n} 1dz = 0, \int_{\partial R_n} zdz = 0$

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Constant and linear terms vanish. $\int_{\partial S} 1dz = 0, \int_{\partial S} zdz = 0$ for any closed contour S because their respective antiderivatives are z and $\frac{z²}{2}$.

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Therefore, subtracting zero from our integral doesn’t change its value: $\int_{\partial R_n}f(z)dz = \int_{\partial R_n} \left[ f(z) - f(z^*) - f'(z^*)(z-z^*) \right] dz$

Now, we take the absolute value and apply the standard Estimation Lemma ($ML$-inequality). The maximum value of the integrand ($M$) is bounded by $\epsilon|z-z^*|, (\star)$, and the length of the contour ($L$) is the perimeter $l_n$:

$$\left|\int_{\partial R_n}f(z)dz\right| \le \int_{\partial R_n} \epsilon|z-z^*| |dz|$$

Since both $z$ and $z^*$ lie within $R_n$, the distance $|z-z^*|$ can never exceed the diagonal of the rectangle, $d_n$. $\left|\int_{\partial R_n}f(z)dz\right| \le \epsilon \cdot d_n \cdot l_n$

Substitute the known values for $d_n$ and $l_n$: $\left|\int_{\partial R_n}f(z)dz\right| \le \epsilon \left(\frac{d}{2^n}\right) \left(\frac{l}{2^n}\right) = \frac{\epsilon \cdot d \cdot l}{4^n}$

Finally, we chain this result back to our original geometric inequality from Step 2: $\frac{1}{4^n}\left|\int_{\partial R_0}f(z)dz\right| \le \left|\int_{\partial R_n}f(z)dz\right| \le \frac{\epsilon \cdot d \cdot l}{4^n}$

Multiplying the entire inequality by $4^n$ yields: $\left|\int_{\partial R_0}f(z)dz\right| \le \epsilon \cdot d \cdot l$

Because $d$ and $l$ are fixed constants of our initial rectangle, and $\epsilon$ can be chosen to be arbitrarily close to $0$, the only way this inequality holds is if the absolute value is exactly zero. $\left|\int_{\partial R}f(z)dz\right| = 0 \implies \int_{\partial R}f(z)dz = 0\, \blacksquare$

Remarks and extensions

• Once we believe the rectangle case, we get every simple polygon for free by a tiling or triangulation argument. In other words, $\int_{\partial C}f(z)dz = 0$ for any positively oriented, simple closed polygon C when f is analytic on an open set containing $C \cup \operatorname{Int}(C)$ (so f is analytic on and in a neighborhood of C) (Figure 5).

There are several forms of Cauchy’s theorem. If C is a closed polygonal contour lying entirely within a simply connected domain D, f is a complex function analytic in D, then $\int_{\partial C}f(z)dz = 0$.

Proof by triangulation:

1. Tile or triangulate the interior of the polygon C, meaning that the closed polygon C can be decomposed into a finite number of rectangles (tiles) or triangles, $T_1, T_2, \ldots, T_m$ whose union covers $C \cup \operatorname{Int}(C)$.
2. Orient Boundaries Positively. Assign counterclockwise orientation to each $\partial T_k$, consistent with C’s positive orientation.
3. Sum the integrals. Every interior edge is shared by two adjacent triangles and is traversed in opposite directions — so these cancel each other. Only the outer boundary survives: $\oint_{\partial C} f(z)\,dz = \sum_{k=1}^{m}\oint_{\partial T_k} f(z)\,dz.$
4. By the rectangle (triangle) theorem each individual integral is zero (f holomorphic inside $C_k$); hence their sum is zero; hence $\oint_{\partial C} f(z)\,dz = \sum_{k=1}^{m} 0 = 0. \quad\square$

Whether $C$ is a tidy square, an irregular hexagon, or a jagged polygon with thousands of vertices, analyticity inside forces the integral to zero.

• Cauchy’s Theorem for Rectifiable Jordan Curves. Let $\Gamma$ be a rectifiable Jordan curve and let f be analytic on an open set $\Omega$ containing $\Gamma \cup \operatorname{Int}(\Gamma)$. Then, $\oint_{\Gamma} f(z)dz = 0.$

The idea is to approximate $\Gamma$ by inscribed polygons and use the polygon result. The uniform continuity of $f$ and the finite length of $\Gamma$ ensure the approximation error vanishes in the limit.

Examples

• For entire functions (analytic everywhere in ℂ) such as f(z) = $e^z, z^2+7, sin(z)$, the integral over any rectangle, triangle, or any closed polygon P vanishes, $\int_{\partial P}f(z)dz = 0.$ This follows because entire functions satisfy the analyticity condition everywhere and any such contour lies in a simply connected domain (the entire complex plane).
• Functions with isolated singularities. For f(z) = 1/z, this function is analytic everywhere except at z = 0 and the theorem does not apply if 0 ∈ Int(R).
  If the rectangle R does not contain 0 in its interior (i.e., $0 \notin \operatorname{Int}(R)$), then f is analytic on an open set containing R, and thus, $\int_{\partial R}\frac{1}{z}dz = 0$.
  If 0 is in the interior of R ($0 \in \operatorname{Int}(R)$), then the theorem does not apply. By the residue theorem, $\int_{\partial R}\frac{1}{z}dz = 2\pi i$ when the rectangle winds once counterclockwise around 0. This illustrates how singularities inside contours usually lead to non-zero integrals.
• Non-simply connected domains: Let $D = \mathbb{C}\setminus\{0\}$ (punctured plane). This domain is not simply connected. Consider f(z) = 1/z, which is analytic in D. For the unit circle C (a closed contour in D), $\int_{\partial R}\frac{1}{z}dz = 2\pi i \ne 0$. This demonstrates that the simply connected condition is essential for Cauchy’s theorem to hold.
• Exponential with singularity: For $f(z) = e^z/(z-a)$, if a is inside a triangular contour T, then $\int_{\partial T}e^z/(z-a)dz = 2\pi i e^a$ by Cauchy’s integral formula. If a is outside T, then $\int_{\partial T}e^z/(z-a)dz = 0$ since f is analytic on an open set containing T.