Properties of Contour Integrals: Linearity, Path Manipulation, and Deformation

All those who seem stupid, they are, and also so are half of those who do not, Quevedo

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: $z(t) = x(t) + iy(t), \quad a \leq t \leq b$, e.g., a straight line between two points $z_0$ and $z_1$, $\gamma(t) = z_0 + t(z_1 - z_0), t \in [0, 1]$; the Unit Circle, $\gamma(t) = e^{it} = \cos t + i \sin t, t \in [0, 2\pi]$.

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 $\int_{\gamma} f(z)dz := \int_{a}^{b} f(\gamma(t)) \gamma^{'}(t)dt$.

This section develops the fundamental properties of contour integrals: linearity, behavior under path manipulation, key theorems (Cauchy’s theorem, the Cauchy integral formula, the residue theorem), and the principle of contour deformation.

Properties

Proposition. Linearity of the contour integral. Let γ be a contour, defined by a function $\gamma: [a, b] \to \mathbb{C}$ and let f and g be continuous complex functions $f, g: \gamma^* \to \mathbb{C}$ both defined along the image of ᵞ. Then, the contour integral satisfies the following linearity properties:

1. Additivity of integrands. $\int_{\gamma} (f(z) \plusmn g(z))dz = \int_{\gamma} f(z)dz \plusmn \int_{\gamma} g(z)dz$.
2. Homogeneity (scaling). $\int_{\gamma} af(z)dz = a\int_{\gamma} f(z)dz$ for any a ∈ ℂ.
3. In other words, linearity, $\int_{\gamma} (af(z) + bg(z))dz = a\int_{\gamma} f(z)dz + b\int_{\gamma} g(z)dz$

The Cauchy integral formula is one of the most powerful results in complex analysis. It shows that the values of a holomorphic function inside a contour are completely determined by its values on the contour.

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

Interpretation. The value of $f$ at any interior point $z_0$ is determined by a weighted average of $f$ over the boundary contour.

The formula can be used to calculate integrals: $\int_{\gamma} \frac{f(z)}{z-z_0}dz = 2\pi i \cdot f(z_0)$.

• Generalized Cauchy Integral Formula. The previous formula extends to higher derivatives: $\boxed{f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\,dz,}$ or equivalently: $\oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\,dz = \frac{2\pi i}{n!}\,f^{(n)}(z_0).$

This shows that all derivatives of a holomorphic function at $z_0$ are determined by the values of $f$ on any enclosing contour — and, as a corollary, that holomorphic functions are infinitely differentiable.

• Theorem. Residue Theorem. Let $f$ be holomorphic inside and on a simple closed contour $\gamma$ (doesn’t cross itself, oriented counterclockwise), except for finitely many isolated singularities $z_1, z_2, \ldots, z_N$ inside $\gamma$. Then, the integral of f around a simple $\gamma$ is equal to $2\pi i$ times the sum of the residues of all singularities inside the contour, $\boxed{\int_{\gamma} f(z)dz = 2\pi i \sum_{k=1}^{N} Res(f, z_k)}$.

Think of integrating an analytic function around a closed loop as a “free trip” —you always end up back where you started with a net result of zero (the integral is zero by Cauchy’s theorem).

However, if your function has singularities (points where it “blows up,” like poles), your path is no longer free. The Residue Theorem says that for every singularity you circle, you must “pay a toll.” The residue is the exact value of that toll. The total value of the integral is simply 2πi times the sum of all the “tolls” for the singularities you enclosed on your trip.

1. For a simple pole (order 1) at a point $z_0$ (i.e., the denominator has a simple root) is calculated using the formula $\operatorname{Res}(f, z_0) = \lim_{z \to z_0}(z - z_0)f(z).$.
2. For a pole of order m, $\operatorname{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z \to z_0}\frac{d^{m-1}}{dz^{m-1}}\left[(z - z_0)^m f(z)\right].$

From the Laurent series: If $f(z) = \sum_{n=-\infty}^{\infty} a_n(z-z_0)^n$ near $z_0$, then $\operatorname{Res}(f, z_0) = a_{-1}$ — the coefficient of $(z-z_0)^{-1}$.

Examples

• Two Simple Poles. Compute $\oint_\gamma \frac{z-1}{z(z-2)}dz$ where $\gamma$ is the circle |z| = 3. The function has singularities where the denominator is zero, namely $z_1 = 0, z_2 = 0$. Both are simple poles (order 1) inside the contour (∣0∣ < 3, ∣2∣ < 3 respectively).

Calculate the Residues: $Res(f,0) = \lim_{z \to 0}(z - 0)\frac{z-1}{z(z-2)} ​= \lim_{z \to 0} \frac{z-1}{z-2} = \frac{-1}{-2}= \frac{1}{2}.$

$​Res(f,2) = \lim_{z \to 2}(z - 2)\frac{z-1}{z(z-2)} ​= \lim_{z \to 2} \frac{z-1}{z} = \frac{1}{2}.$

$\oint_\gamma \frac{z-1}{z(z-2)}dz =[\text{Residue Theorem}] 2\pi i \sum Res(f, z_k) = 2\pi i[\frac{1}{2}+\frac{1}{2}] = 2\pi i$.

• Figure-Eight Contour (Cauchy Integral Formula). Compute $\oint_\gamma \frac{2z + 1}{z(z-1)^2}dz$ where $\gamma$ is the curve shown in Figure A (it is traversed in the 8 shape). The contour is not simple, so we break $\gamma$ into $\gamma_1$ encloses only z = 1, oriented counterclockwise (positive), and $\gamma_2$ encloses only z = 0, oriented clockwise (negative) as shown in Figure B. Both of these two new contours are simple.

$$ \begin{aligned} \oint_\gamma \frac{2z + 1}{z(z-1)^2}dz &=\oint_{\gamma_1} \frac{2z + 1}{z(z-1)^2}dz + \oint_{\gamma_2} \frac{2z + 1}{z(z-1)^2}dz \\[2pt] &=[\text{Reversing the contour, ᵞ₂ is -orientated}] \oint_{\gamma_1} \frac{2z + 1}{z(z-1)^2}dz - \oint_{-\gamma_2} \frac{2z + 1}{z(z-1)^2}dz\\[2pt] &\oint_{\gamma_1} \frac{\frac{2z + 1}{z}}{(z-1)^2}dz - \oint_{-\gamma_2} \frac{\frac{2z + 1}{(z-1)^2}}{z}dz\\[2pt] &=\text{Cauchy's Integral formula}\\[2pt] &f(z) = \frac{2z + 1}{z} = 2 + \frac{1}{z}, n = 1, z_0 = 1, f'(z) = -\frac{1}{z^2}; g(z) = \frac{2z + 1}{(z-1)^2}, z_0 = 0\\[2pt] &=\frac{2\pi i}{1!} \cdot f'(1) - 2\pi i \cdot g(0)\\[2pt] &=2\pi i \cdot -\frac{1}{1^2} - 2\pi i \cdot \frac{2\cdot 0 + 1}{(0-1)^2}\\[2pt] &=2\pi i \cdot -\frac{1}{1^2} - 2\pi i \cdot \frac{2\cdot 0 + 1}{(0-1)^2}\\[2pt] &=-2\pi i -2\pi i = -4\pi i. \end{aligned} $$

• Compute $\oint_\gamma \frac{z}{z^2+4}dz$ where $\gamma$ is a circle centered at $0$ enclosing both singularities at $z = \pm 2i$ (Figure C).

The contour encloses both singularities. Split $\gamma$ into two simple pieces by introducing a “bridge” (a line segment $\gamma_3$ traversed in both directions, whose contributions cancel) as shown in Figure D. Piece 1 encloses only $z = 2i$; Piece 2 encloses only $z = -2i$. Thus, $\gamma = \text{piece1 + piece2} = \gamma_1 + \gamma_3 -\gamma_3 + \gamma_2$ . Both of these two new pieces are simple and each surround just one singularity.

$$ \begin{aligned} \oint_\gamma \frac{z}{z^2+4}dz &=\oint_\gamma \frac{z}{(z-2i)(z+2i)}dz \\[2pt] &=\oint_{piece1} \oint_{piece2} \frac{\frac{z}{z+2i}}{z-2i}dz + \frac{\frac{z}{z-2i}}{z+2i}dz\\[2pt] &=\oint_{\gamma_1 + \gamma_3} \frac{\frac{z}{z+2i}}{z-2i}dz + \oint_{-\gamma_3 + \gamma_2} \frac{\frac{z}{z-2i}}{z+2i}dz\\[2pt] &=\text{Cauchy's Integral formula}\\[2pt] &=2\pi i \frac{2i}{2i+2i} + 2\pi i \frac{-2i}{-2i-2i}\\[2pt] &=2\pi i (\frac{1}{2}+\frac{1}{2})= 2\pi i. \end{aligned} $$

• Cauchy’s Theorem via Direct Computation. Compute $\int_{\gamma} z^2dz$ where $\gamma$ is a contour formed by joining $\gamma_1$ the line segment on the real axis from −R to R, $\gamma_1(t) = Rt, -1 \le t \le 1$ and the half upper circle centered at 0 and radius R, $\gamma_2(t) = Re^{it}, 0 \le t \le \pi$.

The quickest way to solve this is by applying Cauchy’s Integral Theorem. The function being integrated is f(z) = z². As a polynomial, it is analytic (differentiable everywhere) in the entire complex plane. The path $\gamma$ (a line segment joined to a semicircle) is a simple closed contour.

Cauchy’s Integral Theorem states that the integral of an analytic function over any simple closed contour is zero. Therefore, we can immediately conclude: $\int_{\gamma} z^2dz = 0$.

We can confirm this result by calculating the integral along the two separate paths, γ₁ and γ₂, and adding them together.

$\int_{\gamma} z^2dz = \int_{\gamma_1} z^2dz + \int_{\gamma_2} z^2dz$

$\int_{\gamma_1} z^2dz = \int_{-1}^1 (Rt)^2 \cdot R dt = R^3 \int_{-1}^1 t^2dt = R^3 \frac{t^3}{3}\Big|_{-1}^{1} = \frac{2R^3}{3}$

$\int_{\gamma_2} z^2dz = \int_{0}^{\pi} (Re^{it})^2 \cdot iRe^{it} dt = iR^3\int_{0}^{\pi} e^{3it} = iR^3 \frac{e^{3it}}{3i}\Big|_{0}^{\pi} = \frac{R^3}{3}(e^{3i\pi} - e^{0}) = \frac{R^3}{3}(-1-1) = -\frac{2R^3}{3}$

$\int_{\gamma} z^2dz = \int_{\gamma_1} z^2dz + \int_{\gamma_2} z^2dz = \frac{2R^3}{3} -\frac{2R^3}{3} = 0.$ The direct calculation confirms the result provided by Cauchy’s Theorem.

One of the most useful principles in complex analysis is that contour integrals are invariant under continuous deformation of the path, provided the function remains analytic throughout the deformation.

• Deformation of Contours. If two contours $\gamma_1$ and $\gamma_2 $ are two simple closed homotopic contours (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: $\boxed{\int_{\gamma_1} f(z)dz = \int_{\gamma_2} f(z)dz}.$

Key Idea: If a function f(z) is analytic in a region D, the integral over any closed contour in D is zero (by Cauchy’s theorem). Deforming a contour is valid only if the entire region between the original and the deformed contour is free of singularities.

Visual proof via a “bridge”

The Cauchy-Goursat theorem states that if a function f is analytic at all points inside and on a simple closed contour γ, then: $\oint_{\gamma} f(z)dz = 0$.

However, when dealing with functions that are not analytic everywhere, such as ($f(z) = \frac{1}{z}$), the principle of deformation of contours becomes essential. f(z) is analytic everywhere except at z = 0, where it has a simple pole. If a contour encloses this singularity, the integral may be nonzero. For instance, over the unit circle centered at the origin: $\oint_{\gamma}\frac{1}{z} = 2\pi i$.

To illustrate this principle, consider two contours: ($\gamma_1$), a square centered at the origin, and ($\gamma_2$), a unit circle also centered at the origin (Figure A). While the integral of ($f(z) = \frac{1}{z}$) over the unit circle has been previously calculated as $2\pi i$, the integral over the square requires a different approach.

Using the principle of deformation of contours, one can replace ($\gamma_1$) and ($\gamma_2$) with new contours that are entirely contained within regions where the function remains analytic. For instance, if an inner circle with radius ($\frac{1}{2}$) is considered, the integral along this path will also yield ($2\pi i$) because the function remains analytic between the two curves (Figure B).

The theoretical foundation of the deformation principle can be visualized by constructing a new contour that connects points on $\gamma_1$ and $\gamma_2$. By creating a “bridge” between these contours, one can define a closed path that encircles a simply connected region:

1. Construct a New Path: We start with an outer contour $\gamma_1$ (red, counter-clockwise) and an inner contour $\gamma_2$ (green, also counter-clockwise). We create a new, single, non-intersecting closed path γ by adding a “bridge” (two lines connecting the contours).
2. Trace the Full Contour γ: The new path travels: (i) along $\gamma_1$ (red, counter-clockwise); (ii) down the bridge; (iii) along $\gamma_2$ in the opposite (clockwise) direction; (iv) up the bridge, back to the start.
3. Apply Cauchy-Goursat: The function f(z) is analytic everywhere in the region enclosed by this new path γ (the region between the two original contours contains no singularities). Therefore, by the Cauchy-Goursat Theorem: $\oint_{\gamma}f(z) = 0$.
4. Deconstruct the Integral: The integral over the full path γ is the sum of its parts: $\oint_{\gamma}f(z) = \oint_{\gamma_1}f(z) + \oint_{bridge}f(z) - \oint_{\gamma_2}f(z) - \oint_{bridge}f(z) = 0$. If you traverse γ₂ clockwise, pick up a minus sign.
5. Simplify and Conclude. The two bridge integrals are traversed in opposite directions, so they cancel. What remains is $\oint_{\gamma_1}f(z) - \oint_{\gamma_2}f(z) = 0 \leadsto \oint_{\gamma_1}f(z) = \oint_{\gamma_2}f(z)$

Consequently, the integral along ($\gamma_1$) equals the integral along ($\gamma_2$) (Figure C). An integral is unchanged under continuous contour deformation provided the integrand stays analytic throughout the region swept out.

Examples

The integral $ \int_{\gamma} \frac{1}{z} dz$ is the same for any closed contour $\gamma$ enclosing the origin. The function f(z) = 1/z has a singularity at z = 0. For any closed contour γ that encloses the origin (with positive orientation), the integral is: $\oint_\gamma \frac{dz}{z} = 2\pi i.$

This is the same whether $\gamma$ is:

• The unit circle $|z| = 1$. This is a standard contour positively oriented (counterclockwise), enclosing the origin and avoiding the singularity. $ \int_{\gamma_1} \frac{1}{z} dz =[\text{The Residue Theorem}] 2\pi i \sum Res(f, z_k) = 2\pi i \sum Res(f, 0) = 2\pi i$
• $\gamma_2, \frac{x^2}{4}+\frac{y^2}{9}=1$ is an ellipse centered at the origin. The vertices are located at (0, ±3) and the co-vertices are at (±2, 0). Since it can be continuously deformed into the circle without crossing z = 0, the deformation principle applies: $ \int_{\gamma_2} \frac{1}{z} dz = \int_{\gamma_1} \frac{1}{z} dz = 2\pi i$.
• A square with vertices at $\pm 2 \pm 2i$,
• Or any other contour enclosing $z = 0$.

  For a simple pole at a point $z_k$ (i.e., the denominator has a simple root) is calculated using the formula $Res(f, z_k) = \lim_{z \to z_k}(z-z_k)f(z)$. In our particular case, $Res(f, z_0) = \lim_{z \to 0}(z-0)\frac{1}{z} = 1.$

Conversely, if $\gamma$ does not enclose the origin, then $1/z$ is analytic inside $\gamma$, and Cauchy’s theorem gives $\oint_\gamma dz/z = 0$.