Cauchy's Theorem for Simply Connected Domains

Tell me and I’ll forget; show me and I may remember; involve me and I’ll understand, Chinese Proverb

Introduction

A complex function $f(z)$ maps $z = x + iy \in \mathbb{C}$ to another complex number. For example: $f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy, f(z) = \frac{1}{z}, f(z) = \sqrt{z^2 + 7}$.

A contour is a continuous, piecewise-smooth curve defined parametrically as: $z(t) = x(t) + iy(t), \quad a \leq t \leq b$.

Definition (Smooth Contour Integral). Let ᵞ be a smooth contour (a continuously differentiable path in the complex plane), $\gamma: [a, b] \to \mathbb{C}$. Let $f: \gamma^* \to \mathbb{C}$ be a continuous complex-valued function defined on the trace $\gamma^*$ of the contour (i.e. along the image of $\gamma$). Then, the contour integral of f along $\gamma$ is defined as $\int_{\gamma} f(z)dz := \int_{a}^{b} f(\gamma(t)) \gamma^{'}(t)dt$.

Properties

• 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 $\oint_{\gamma} f(z)dz = 0$.

Simply and multiply connected domains

Definition. A domain in complex analysis is a subset $D \subseteq \mathbb{C}$ that is simultaneously:

• Nonempty: $D$ contains at least one point.
• Open: For every point $z_0 \in D$, there exists some radius $r > 0$ such that the open disk $D(z_0, r) = \{z \in \mathbb{C} : |z - z_0| < r\}$ is entirely contained in $D$. In other words, every point of $D$ is an interior point.
• Connected: $D$ cannot be partitioned into two disjoint nonempty open subsets. Equivalently, any two points in $D$ can be joined by a continuous path lying entirely within $D$. The set is “all one piece.”

Examples of domains: The entire complex plane $\mathbb{C}$; any open disk $\{z : |z - z_0| < r\}$; any open half-plane such as $\{z : \operatorname{Re}(z) > 0\}$; the interior of any simple closed curve; any open strip $\{z : a < \operatorname{Im}(z) < b\}$.

Non-examples: A closed disk $\{z : |z| \le 1\}$ (not open — boundary points like $z = 1$ have no open neighborhood inside the set); the union of two disjoint open disks (open but not connected); a single point $\{0\}$ (not open); the real line $\mathbb{R} \subset \mathbb{C}$ (not open in $\mathbb{C}$).

Simply Connected Domains

Definition. A domain $D$ is simply connected if it has no holes. Intuitively, $D$ is a “solid” region: there is no missing point, puncture, or gap that a closed loop could wind around. Formally, every simple closed curve (a loop that does not cross itself) lying entirely within $D$ can be continuously shrunk to a single point without ever leaving $D$.

This shrinking process is called a null-homotopy. The closed curve is null-homotopic (or contractible) if such a continuous deformation exists. A domain is simply connected if and only if every closed curve in it is null-homotopic.

Geometric intuition. Imagine the domain is a rubber sheet and the closed curve is a rubber band lying on it. If you can contract the rubber band down to a dot without it ever sliding off the edge of the sheet, the domain is simply connected. If the sheet has a hole (like a ring or a punctured disk), a rubber band looped around that hole gets “stuck” — it cannot shrink past the missing region.

Equivalent characterizations. For a domain $D \subseteq \mathbb{C}$, the following are equivalent:

1. $D$ is simply connected.
2. Every closed curve in $D$ is null-homotopic (contractible to a point within $D$).
3. For every simple closed curve $C$ lying in $D$, the interior $\operatorname{Int}(C)$ is entirely contained in $D$.
4. $\mathbb{C}_\infty \setminus D$ is connected, where $\mathbb{C}_\infty = \mathbb{C} \cup \{\infty\}$ is the Riemann sphere (the complement, viewed on the sphere, is one piece).
5. Every analytic function on $D$ has an antiderivative on $D$.

Examples of simply connected domains:

• The entire complex plane $\mathbb{C}$. There are no missing points at all, so no loop can encircle a hole.
• Any open disk, e.g., $\{z \in \mathbb{C} : |z| < 1\}$. Disks are convex, and every convex set is simply connected.
• Any open half-plane, e.g., $\{z \in \mathbb{C} : \operatorname{Re}(z) > 0\}$. Half-planes are convex.
• The interior of any simple closed curve (by the Jordan Curve Theorem, this is a bounded, connected, simply connected region).
• Any open convex set (a set where the line segment between any two points stays inside the set). Convexity $\implies$ star-shaped $\implies$ simply connected.
• Any open star-shaped set (there exists a center point $z_0$ such that the segment from $z_0$ to any other point of the set lies entirely within the set).
• A slit plane, e.g., $\mathbb{C} \setminus (-\infty, 0]$ (the complex plane with the non-positive real axis removed). Despite the dramatic-looking cut, the complement on the Riemann sphere is connected, so the slit plane is simply connected. This domain is essential for defining the principal branch of $\log z$.

Multiply Connected Domains

Definition. A domain is multiply connected if it is not simply connected, i.e., it has at least one “hole.”. More formally, there exists at least one simple closed curve in the domain that cannot be continuously shrunk to a single point while staying within the domain. The curve will get stuck around a hole.

Geometric intuition. Think of a multiply connected domain as a surface with one or more islands removed. A rubber band looped around an island cannot shrink to a point. Different loops can wind around different islands, and the topology of the domain determines which integrals vanish and which do not.

Examples of multiply connected domains:

• Annulus (ring): $\{z \in \mathbb{C} : 1 < |z| < 3\}$. This is the region between two concentric circles. The “hole” is the closed disk $|z| \le 1$. Any loop that winds around the origin (e.g., the circle $|z| = 2$) cannot be contracted to a point within the annulus, because it would have to cross the inner boundary.
• Punctured disk: $\{z \in \mathbb{C} : 0 < |z| < 1\}$. A disk with just the center point $z = 0$ removed. Despite the hole being a single point, the domain is multiply connected: the loop $|z| = 1/2$ winds around the missing origin and cannot be shrunk.
• Punctured plane: $\mathbb{C} \setminus \{0\}$. The entire complex plane minus the origin. Again doubly connected: loops around the origin cannot contract.
• Plane with several punctures: $\mathbb{C} \setminus \{0, 1\}$. This has two holes (at $0$ and $1$) and is triply connected.

In a simply connected domain where $f$ is analytic:

• $\oint_C f(z)\,dz = 0$ for every closed contour $C$ in $D$ (Cauchy’s Theorem).
• $f$ possesses a global antiderivative $F$ on $D$ (i.e., $F'(z) = f(z)$ for all $z \in D$).
• Contour integrals $\int_\gamma f(z)\,dz$ depend only on the endpoints, not on the path.
• Multi-valued functions like $\log z$ and $z^\alpha$ can be defined as single-valued analytic functions by choosing a branch.

In a multiply connected domain, all of these properties can fail:

• The integral $\oint_C \frac{dz}{z} = 2\pi i \neq 0$ around any loop encircling the origin in $\mathbb{C} \setminus \{0\}$.
• The function $1/z$ has no single-valued antiderivative on $\mathbb{C} \setminus \{0\}$ (any candidate for $\log z$ is multi-valued).
• Contour integrals can depend on the path (specifically, on how many times the path winds around each hole).

Cauchy’s Theorem for simply connected domains

Cauchy’s Theorem for simply connected domains. If a function f is analytic (a function that is complex-differentiable at every point within a domain, i.e., well-behaved and smooth, with no sharp corners, breaks, or singularities like division by zero) throughout a simple connected domain D, then $\oint_C f(z)dz = 0$ for every closed contour C lying in D.

Proof.

Case 1. If C is a simple close contour (a path that starts and ends at the same point but never crosses itself, e.g, a circle, an ellipse, a polygon, or any non-self-intersecting loop) lying entirely within the simply connected domain D.

Since D is simply connected (it has no holes), and the curve C lies completely within D, the interior region $\operatorname{Int}(C)$ must also lie completely within D.

This is precisely the characterization of simple connectedness we have already identified: there is no hole in $D$ that $\operatorname{Int}(C)$ could contain.

Recall, the Cauchy-Goursat theorem. If a function f is analytic at all points interior to and on a simple closed contour, then the integral of f around that contour is zero.

1. $f$ is analytic throughout the entire domain $D$.
2. The interior of $C$ and the boundary $C$ itself are both contained in $D$.
3. Therefore, $f$ is analytic at all points interior to and on $C$.

Since all conditions of the Cauchy-Goursat theorem are met, we can directly apply its conclusion: $\oint_C f(z)dz = 0$

Case 2. If C is closed but self-intersecting contour, meaning a path that starts and ends at the same point but crosses itself a finite number of times, e.g., a figure-eight.

The core idea is to show that the integral over the complex path is just the sum of the integrals over a set of simple closed paths. We can then apply the result from Case 1 to each simple path.

1. Decomposition. Suppose $C$ is a figure-eight contour that intersects itself at a single point $A$. The contour traces out two loops: starting at $A$, it traverses the first loop $C_1$ (say counterclockwise) and returns to $A$, then traverses the second loop $C_2$ (say counterclockwise) and returns to $A$.
   The self-intersection at $A$ naturally divides $C$ into two simple closed contours $C_1$ and $C_2$, each starting and ending at $A$. Neither $C_1$ nor $C_2$ crosses itself (they are simple).

2. Additivity of the contour integral. The integral over $C$ splits by the additivity of path integrals: the contour $C$ is the concatenation of the path along $C_1$ followed by the path along $C_2$. Therefore, $\oint_C f(z)\,dz = \oint_{C_1} f(z)\,dz + \oint_{C_2} f(z)\,dz.$
   Let the path segment that is shared during the intersection be denoted by the line from $A$ to $B$ (in reality, the intersection might just be a single point).
   

   On cancellation of shared segments. If the contour $C$ traverses a segment from $A$ to $B$ as part of $C_1$ and then retraces it from $B$ to $A$ as part of $C_2$, the two integrals over that shared segment cancel: $\int_A^B f(z)\,dz + \int_B^A f(z)\,dz = 0,$ since reversing the direction of integration negates the integral value. After cancellation, only the simple loops remain. This is the mechanism by which self-intersecting contours reduce to sums of simple ones.

3. Apply Case 1. Both $C_1$ and $C_2$ are simple closed contours lying in the simply connected domain $D$. By Case 1: $\oint_{C_1} f(z)\,dz = 0 \qquad \text{and} \qquad \oint_{C_2} f(z)\,dz = 0.$

4. Conclusion: $\oint_C f(z)\,dz = 0 + 0 = 0.$

5. This logic can be extended to a contour $C$ with any finite number of intersections. It can always be decomposed into a finite number of simple closed contours ($C_1, C_2, \ldots, C_n$), and the integral over $C$ will be the sum of the integrals over each $C_k$. Since every individual integral is zero, according to the Cauchy-Goursat theorem, the total sum is zero: $\oint_C f(z)dz = \sum_{k=1}^n \oint_{C_k} f(z)dz = 0$

6. When decomposing a self-intersecting contour, the induced orientations on the sub-loops must be tracked carefully. Some sub-loops may be traversed clockwise (negative orientation) depending on how the original contour crosses itself. However, the Cauchy–Goursat theorem gives $\oint_{C_k} f\,dz = 0$ regardless of orientation (zero is unchanged by a sign flip), so the conclusion $\oint_C f = 0$ holds in all cases.

Examples

• Rational Function with Distant Singularities. Let C be any closed contour (e.g., the circle $|z|=1.5$, a square centered at the origin with side length 2, or a complicated self-intersecting loop, Figure i) that lies entirely within the open disk $D$ defined by |z| < 2. Then, $\oint_C \frac{zsin(z)}{(z²-9)^3}dz = 0$.

1. The domain $D$ is the open disk $|z| < 2$. This is a simply connected domain because it’s a solid disk with no holes in it. Any loop drawn inside it can be shrunk to a single point without leaving the disk.
2. We look for points where $f(z)$ is not analytic (i.e., where it “blows up” or is undefined). This happens when the denominator is zero, $z^2 - 9 = 0$, i.e., $z = \pm 3$.
3. The contour $C$ lies entirely inside this domain. Both singularities $z=3$ and $z=-3$ are outside this domain.
4. Because the function $f(z)$ is analytic everywhere inside and on the contour $C$ (its only “problems”, more formally singularities, are far away), the Cauchy-Goursat theorem applies (all hypotheses of Cauchy’s Theorem for simply connected domains are satisfied), and the integral must be zero.

It does not matter how absurdly complicated, wiggly, or even self-intersecting the contour $C$ is. As long as $C$ stays inside $|z| < 2$, the integral vanishes. The singularities at $z = \pm 3$ are .

• Contour in a Half-Plane Avoiding Boundary Singularities. Let C be a semi-circle of radius $R=0.5$ in the upper half-plane, centered at $z=2i$, and closed by the straight line segment from $2i - 0.5$ to $2i + 0.5$ that lies entirely within the simply connected open domain Im(z) > 1. Then, $\oint_C \frac{e^z}{z^2+1} dz = 0$.

1. The domain $D = \{z : \operatorname{Im}(z) > 1\}$ is the open half-plane above the horizontal line $y = 1$. It is a simply connected open set.
2. We look for points where f(z) is not analytic (i.e., where it “blows up” or is undefined), $z^2 + 1 = 0$, i.e., at $z = +i$ and $z = -i$.
3. The contour C is centered at $2i$ (or $y=2$) and stays well within the region $y > 1$. The singularity $z = +i$ (or $y=1$) sits on the boundary of D, but not inside it. The singularity $z = -i$ (or $y=-1$) is far outside the domain.
4. Because the function $f(z)$ is analytic everywhere inside and on the contour $C$ (its only “problems”, more formally singularities, are far away), the Cauchy-Goursat theorem applies, and the integral must be zero, $\oint_C \frac{e^z}{z^2+1}\,dz = 0.$.

The singularity $z = i$ lies on the boundary of $D$, not inside $D$. This is sufficient! Cauchy's Theorem requires $f$ to be analytic on the (open) domain $D$, and boundary points are not in an open set. The careful choice of domain $\operatorname{Im}(z) > 1$ (strict inequality) excludes the singularity at $z = i$.

• Entire Function — Any Contour in the Entire Plane. Let C be any simple closed contour $C$ in the entire complex plane (e.g., a circle, a square, a triangle, a figure-eight, or any complicated, wiggly, self-intersecting loop) that lies entirely in the complex plane. Then, $\oint_C (z^4 + 8z^3 -7z^2 + 3z + 4cos(z) -8sin(z))dz = 0.$

1. The domain $D$ is the entire complex plane $\mathbb{C}$. The entire plane is a simply connected (it has no missing points, no “edges”, and no “holes”) set.
2. We look for points where f(z) is not analytic (i.e., where it “blows up” or is undefined). However, this function is a combination of polynomials, the cosine and sine functions. They are all analytic everywhere (entire). Finite sums and scalar multiples of entire functions are entire.
3. Any closed contour $C$ trivially lies in $\mathbb{C}$.
4. Since $f(z)$ is analytic everywhere, it is automatically analytic inside and on any closed contour $C$ you can possibly draw or imagine. Therefore, the integral is always zero, $\oint_C f(z)\,dz = 0.$

For entire functions, the theorem is very powerful: the integral over any closed contour vanishes, with no restrictions on the contour's shape, size, location, or self-intersections. This is because $\mathbb{C}$ is simply connected and the function is analytic everywhere — there is simply nothing that can obstruct the contraction of a loop.

Alternatively, using antiderivatives, since $\mathbb{C}$ is simply connected and $f$ is entire, $f$ has a global antiderivative: $F(z) = \frac{z^5}{5} + 2z^4 - \frac{7z^3}{3} + \frac{3z^2}{2} + 4\sin z + 8\cos z.$ By the Fundamental Theorem of Calculus for contour integrals, $\oint_C f(z)\,dz = F(z_{\text{end}}) - F(z_{\text{start}}) = 0$ for any closed contour (since the start and end points coincide).