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

Silence is a source of great strength, Lao Tzu

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, $\int_{\gamma} (af(z) + bg(z))dz = a\int_{\gamma} f(z)dz + b\int_{\gamma} g(z)dz$

Proof:

Proof of Additivity. Consider the integral of the sum or difference of f and g:

$\int_\gamma \big(f(z)\pm g(z)\big)dz=\int_a^b \big(f(\gamma(t))\pm g(\gamma(t))\big) \gamma'(t) dt$

By the linearity of the ordinary complex integral (which itself follows from the linearity of real Riemann integrals applied to real and imaginary parts), we can distribute the integral:

$=\int_a^b f(\gamma(t))\,\gamma'(t)dt \pm \int_a^b g(\gamma(t)) \gamma'(t) dt$

Recognizing these as contour integrals of f and g respectively:

$=\int_\gamma f(z)\,dz \;\pm\; \int_\gamma g(z)\,dz.$

Proof of Homogeneity. Now consider the integral of a scalar multiple of f:

$\int_\gamma a\,f(z) dz = \int_a^b \big(a\,f(\gamma(t))\big) \gamma'(t) dt$

Since a is a constant, we can factor it out of the integral (again, by the linearity of the ordinary complex integral):

$=a\int_a^b f(\gamma(t))\gamma'(t)dt$

This is exactly a times the contour integral of f:

$=a\int_\gamma f(z) dz.$

Key insights:

1. Inheritance of Properties: The linearity of contour integrals directly inherits from the linearity of ordinary complex integrals, which in turn inherits from the linearity of Riemann integration for real-valued functions.
2. Practical Significance. These properties are fundamental for breaking down complex integrals into simpler components, factoring out constants to simplify computations, and establishing further theoretical results in complex analysis.
3. Geometric Interpretation. Linearity reflects the intuitive notion that integrating a sum of functions should equal the sum of their integrals, and scaling a function scales its integral proportionally.

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 $\gamma_1 = \gamma|_{[a, \gamma_1]} \text{ and } \gamma_2 = \gamma|_{[c, b]}$, 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 its 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 $\int_{\gamma} f(z)dz = \int_{\tilde{\gamma}} f(z)dz$

This proposition establishes that contour integrals are invariant under orientation-preserving reparameterizations of the contour. The key conditions are: (i) $\tilde{\gamma}$ is a reparameterization of γ via ψ; (ii) ψ has a positive continuous derivative (ensuring ψ is strictly increasing and differentiable). The positive derivative condition is crucial as it guarantees: (1) ψ is a bijection between [c, d] and [a, b]; (2) The substitution in the proof is valid; (3) The orientation of the contour is preserved.

The integral depends only on the geometric path taken and its orientation, not on how it’s parameterized. Reversing the orientation will multiply the result by -1, but any other smooth reparameterization that preserves the orientation leaves the integral unchanged. This property allows flexibility in choosing parameterizations for contour integrals and mirrors the substitution rule for definite integrals in real analysis, where the integral remains invariant under appropriate substitutions. However, while reversing orientation changes the sign of the integral, reparameterization with a positive derivative preserves both the value and the orientation.

Proof.

1. Reversal of orientation

The reversed contour $-\gamma$ is defined by $(-\gamma)(t) = \gamma(a + b - t)$ which traverses rhw path $\gamma$ in the opposite direction.

The contour integral along $-\gamma$ is: $\int_{-\gamma} f(z) dz = \int_a^b f((-\gamma)(t)) \cdot (-\gamma)'(t) dt$

First, compute the derivative of $-\gamma$: $(-\gamma)'(t) = \frac{d}{dt} [\gamma(a + b - t)] =[\text{By the chain rule}] \gamma'(a + b - t) \cdot (-1) = -\gamma'(a + b - t)$

Substitute into the integral: $\int_{-\gamma} f(z) dz = \int_a^b f(\gamma(a + b - t)) \cdot \left(-\gamma'(a + b - t)\right) dt = -\int_a^b f(\gamma(a + b - t)) \gamma'(a + b - t) dt$

Next, let’s make a change of variables. Let $u = a + b - t$. Then: $du = -dt$. When $t = a, u = b$. When $t = b$, $u = a$

The integral transforms as: $-\int_a^b f(\gamma(a + b - t)) \gamma'(a + b - t) dt = -\int_b^a f(\gamma(u)) \gamma'(u) (-du) = \int_b^a f(\gamma(u)) \gamma'(u) du$
Reversing the limits: $= -\int_a^b f(\gamma(u)) \gamma'(u) du$

$f(\gamma(u))\gamma'(u) = \phi(u) + i\psi(u)$ where $\phi, \psi: [a,b] \to \mathbb{R}$ are real-valued.Then, $\int_b^a f(\gamma(u))\gamma'(u)\,du = \int_b^a \phi(u)\,du + i\int_b^a \psi(u)\,du$. For real integrals: $\int_b^a \phi(u)\,du = -\int_a^b \phi(u)\,du$. Same for $\psi$. Therefore, $\int_b^a f(\gamma(u))\gamma'(u)\,du = -\int_a^b \phi(u)\,du - i\int_a^b \psi(u)\,du = -\int_a^b f(\gamma(u))\gamma'(u)\,du$

Recognize that this is the negative of the contour integral along γ:$= -\int_{\gamma} f(z) dz$

Thus, we have shown: $\boxed{\int_{-\gamma} f(z) dz = -\int_{\gamma} f(z) dz}$ as required.

Examples:

• If $\gamma$ is the line between 0 and 1 along the real axis (it can be parametrized by $\gamma(t) = 0 + t\cdot (1-0) = t, 0 \le t \le 1, \gamma'(t) = 1$), then $-\gamma$ is the line between 1 and 0 (it can be parametrized by $-\gamma(t) = 1 + t\cdot (0-1) = 1 - t, 0 \le t \le 1, -\gamma'(t) = -1$).
  For the analytic function f(z) = z we compute the line integral directly, $\int_{-\gamma} zdz =[\text{Reversing the Contour/orientation}] -\int_{\gamma} zdz = -\int_0^1 tdt = -\frac{t^2}{2}\Big|_{0}^{1} = -\frac{1}{2}$

• Let $\gamma$ be the quarter-circle arc from 1 to i along the first quadrant of the unit circle (it can be parametrized by $\gamma(t)=e^{it}$ for $0 \le t \le \pi/2$), then $-\gamma$ is the same quarter-circle arc from i back to 1 (it can be parametrized by $-\gamma(t) = e^{i(\pi/2 - t)}$ for $0 \le t \le \pi/2$).

For the analytic function f(z) = z and we compute the line integral directly: $\int_{-\gamma} zdz =[\text{Reversing the Contour/orientation}] -\int_{\gamma} zdz = -\int_0^{\frac{\pi}{2}} e^{it}ie^{it}dt = -i\int_0^{\frac{\pi}{2}} e^{2it}dt = -i\left[ \tfrac{e^{2it}}{2i} \right]_0^{\pi/2} = -\tfrac{1}{2}\left(e^{i\pi} - e^{0}\right) = -\tfrac{1}{2}\left(-1 - 1\right) = 1.$
Alternatively, the antiderivative is $F(z) = z^2/2$, so: $\int_\gamma z\,dz = F(i) - F(1) = \frac{i^2}{2} - \frac{1}{2} = \frac{-1}{2} - \frac{1}{2} = -1.$ Then, $\int_{-\gamma} zdz =[\text{Reversing the Contour/orientation}] -\int_{\gamma} zdz = 1$.

Direct Computation on $-\gamma$.

$\int_{-\gamma} zdz = \int_0^{\frac{\pi}{2}} e^{i(\pi/2 - t)}\cdot (-ie^{i(\pi/2 - t)})dt = -i\int_0^{\frac{\pi}{2}} e^{2i(\pi/2 - t)}dt = -i\int_0^{\frac{\pi}{2}} e^{i\pi}e^{-2it}d =[\text{Since } e^{\pi i} = -1]$

$= i\int_0^{\frac{\pi}{2}}e^{-2it}dt = \left[ \tfrac{e^{-2it}}{-2} \right]_0^{\frac{\pi}{2}} = \frac{1}{2}-\frac{-1}{2} = 1.\;\checkmark$

2. Additivity under Subdivision

Let c be a point such that a < c < b. Define: $\gamma_1 = \gamma|_{[a, \gamma_1]}$ (the restriction of γ to [a, c]) and $\gamma_2 = \gamma|_{[c, b]}$ (the restriction of γ to [c, b]).

The contour integral along γ is: $\int_{\gamma} f(z) dz = \int_a^b f((\gamma)(t)) \cdot (\gamma)'(t) dt$

By the additive property of definite integrals (for real integrals), a basic real-calculus fact, $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$, we can split the integral at t = c:

$ \int_a^b f((\gamma)(t)) \cdot (\gamma)'(t) dt = \int_a^c f((\gamma)(t)) \cdot (\gamma)'(t) dt + \int_c^b f((\gamma)(t)) \cdot (\gamma)'(t) dt$

The first and second integrals are exactly the contour integral on the subpaths γ₁ and γ₂ respectively. Therefore, $\int_{\gamma} f(z) dz = \int_{\gamma_1} f(z) dz + \int_{\gamma_2} f(z) dz$ which is exactly the stated property.

• Example. Compute $\int_{\gamma} f(z)$ where f(z) = $\bar{z}$ and $\gamma$ is a path that can be divided into two separate paths, $\gamma_1$ a line between 0 to 1 along the real axis (it can be parametrized by $\gamma_1(t) = 0 + t·(1-0) = t, 0 \le t \le 1, \gamma_1^{'}(t)=1$) and $\gamma_2$, a second line from 1 to 2 + i (it can be parametrized by $\gamma_2(t) = 1 + t·(1+i) = (1 + t) + it, 0 \le t \le 1, \gamma_2^{'}(t)=1+i$).

$$ \begin{aligned} \int_{\gamma} f(z)dz &=\int_{\gamma_1} \bar{z}dz + \int_{\gamma_2} \bar{z}dz \\[2pt] &=\int_{0}^{1} t·1dt + \int_{0}^{1} ((1+t)-it)·(1+i)dt\\[2pt] &\text{Simplifying the integrand: } ((1+t)−it)(1+i)=(1+t)(1)+(1+t)(i)−it(1)−it(i)\\[2pt] &=1+t+i+it−it−i^2t = 1+t+i+t = 1+2t+i\\[2pt] &=\frac{t^2}{2}\Big|_{0}^{1}+[t+t^2+it]\big|_{0}^{1}\\[2pt] &=\frac{1}{2}-0 + (1 + 1^2 + i·1) - (0 + 0 + 0) = \frac{1}{2} + 2 + i = \frac{5}{2}+i. \end{aligned} $$

3. Invariance of Contour Integrals Under Reparameterization. Proof.

1. Let $\gamma: [a, b] \to \mathbb{C}$ be a smooth contour, let $\psi: [c, d] \to [a, b]$ be a smooth, strictly increasing bijective function (meaning $\psi'(t) > 0$ for all $t$), such that $\psi(c) = a$ and $\psi(d) = b$.
2. We define our new reparameterized contour as $\tilde{\gamma}(t) = \gamma(\psi(t))$ for $t \in [c, d]$.
3. By the definition of a contour integral, we evaluate $f(z)$ along the new path $\tilde{\gamma}$: $\int_{\tilde{\gamma}} f(z) dz = \int_{c}^{d} f(\tilde{\gamma}(t)) \tilde{\gamma}'(t) dt$
4. Using the Chain Rule, we know that the derivative of our composite function is $\tilde{\gamma}'(t) = \gamma'(\psi(t))\psi'(t)$. Substituting this into our integral gives: $\int_{\tilde{\gamma}} f(z) dz = \int_{c}^{d} f(\gamma(\psi(t))) \gamma'(\psi(t)) \psi'(t) dt$.
5. Now, we apply the real-variable Change of Variables Theorem. We make the substitution $\tau = \psi(t)$. Taking the derivative yields $d\tau = \psi'(t) dt$.Next, we update the limits of integration:
   When $t = c$, $\tau = \psi(c) = a$.
   When $t = d$, $\tau = \psi(d) = b$
6. Because $\psi'(t) > 0$, the function $\psi$ is strictly increasing, meaning the reparameterization strictly preserves the forward direction (orientation) of the curve. Substituting our new variables and limits into the integral: $\int_{\tilde{\gamma}} f(z) dz = \int_{a}^{b} f(\gamma(\tau)) \gamma'(\tau) d\tau$.
7. Notice that the right side is exactly the formal definition of the contour integral along the original path $\gamma$. Therefore, $\int_{\tilde{\gamma}} f(z) dz = \int_{\gamma} f(z) dz$

Key Insights

1. Reversal of Orientation: The proof relies on a change of variables that effectively reverses the parameterization. The negative sign arises from both the derivative of the reversed contour and the reversal of integration limits.
2. Additivity under Subdivision: The proof directly uses the additive property of real integrals. It allows us to break complex contours into simpler pieces for computation and it is indeed essential for defining integrals over piecewise smooth contours.
3. Geometric Interpretation:

• Reversal: Integrating “backward” along a path gives the negative result, reflecting the directional nature of complex integration.
• Subdivision: The integral over a path is the sum of integrals over its segments, analogous to how we break down real integrals.

Examples.

• Let the original curve be $\gamma(t) = t + it$, $t \in [1, 2]$, which traces the line segment from 1 + i to 2 + 2i and $f(z) = z$.

Define the reparameterization function $\psi : [1, \sqrt{2}] \to [1, 2], \psi(t) = t^2$ so that $\psi(1) = 1, \psi(\sqrt{2}) = 2, \psi'(t) = 2t \gt 0, \forall t \in [1, \sqrt{2}]$ (since t ≥ 1), and $\psi$ is continuous on $[1, \sqrt{2}]$

The reparameterized curve is $\tilde{\gamma} = \gamma \circ \psi$, so $\tilde{\gamma}(t) = \gamma(\psi(t)) = \gamma(t^2) = t^2 + it^2$ for $t \in [1, \sqrt{2}]$

This traces the same line segment from 1+i to 2+2i, but with a different parameterization. By the proposition, for any continuous function f defined on this line segment: $\int_{\gamma} f(z)dz = \int_{\tilde{\gamma}} f(z)dz$

Let f(z) = z, which is continuous on the line segment. Compute the Integral Along γ: $\int_{\gamma} f(z)dz = \int_1^2 (t + it)(1 + i) dt = \int_1^2 (1 + i)^2\cdot t dt = [\text{Since } (1 + i)^2 = 1 + 2i + i^2 = 2i] \int_1^2 t \cdot 2i dt = $

$2i \int_1^2 t dt = 2i \left[ \frac{t^2}{2} \right]_1^2 = 2i \cdot \frac{4 - 1}{2} = 2i \cdot \frac{3}{2} = 3i.$

Compute the Integral Along $\tilde{\gamma}$. $\int_{\tilde{\gamma}} f(z)dz = \int_1^{\sqrt{2}} (t^2 + i t^2) \cdot 2t(1 + i) dt = \int_1^{\sqrt{2}} t^2(1 + i) \cdot 2t(1 + i) dt = \int_1^{\sqrt{2}} 2t^3 (1 + i)^2 dt = [\text{Since } (1 + i)^2 = 1 + 2i + i^2 = 2i] \int_1^{\sqrt{2}} 2t^3 \cdot 2i dt.$

$= 4i \int_1^{\sqrt{2}} t^3 dt = 4i \left[ \frac{t^4}{4} \right]_1^{\sqrt{2}} = i \left[ t^4 \right]_1^{\sqrt{2}} = i \left( (\sqrt{2})^4 - 1^4 \right) = i(4 - 1) = 3i.$ Both integrals yield the same value. This confirms the invariance of contour integrals under reparameterization.

• Let the original curve be $\gamma(t) = e^{it}$ for t ∈ [0, π] which traces the upper half of the unit circle from 1 to −1.

Define the reparameterization function $\psi : [0, 2π] \to [0, π], \psi(t) = \frac{t}{2}$ so that $\psi(0) = 0, \psi(2π) = π, \psi'(t) = \frac{1}{2} \gt 0 ∀t \in [0, 2π]$ and $\psi$ is continuous on $[0, 2π]$.

The reparameterized curve is $\tilde{\gamma} = \gamma \circ \psi$, so $\tilde{\gamma}(t) = \gamma(\psi(t)) = \gamma(\frac{t}{2}) = e^{i(t/2)}$ for $t \in [0, 2\pi]$ which traces the same upper half of the unit circle.

By the proposition, for any continuous function f defined on this line segment: $\int_{\gamma} f(z)dz = \int_{\tilde{\gamma}} f(z)dz ↭ \int_{\gamma} f(z)dz = \int_0^{2\pi} f(e^{i(t/2)}) \cdot \left(i \cdot \frac{1}{2} e^{i(t/2)}\right) dt$

Now, let f(z) = $\frac{1}{z}$, which is continuous on the curve (since the curve does not pass through z = 0). Compute the integral along $\gamma$.

$\int_\gamma f(z) dz = \int_0^\pi \frac{1}{e^{it}} \cdot i e^{it} dt = \int_0^\pi i dt = i \left[ t \right]_0^\pi = i\pi.$

Compute the integral along $\tilde{\gamma}$.

$\int_{\tilde{\gamma}} f(z) dz = \int_0^{2\pi} \frac{1}{e^{i t/2}} \cdot \frac{i}{2} e^{i t/2} dt = \int_0^{2\pi} \frac{i}{2} dt = \frac{i}{2}\left[ t \right]_0^2\pi = \frac{i}{2} \cdot 2\pi = i\pi$.

Both integrals yield the same value. This confirms the invariance of contour integrals under reparameterization.