Contour Integration

Why It Matters

Contour integration is the computational heart of complex analysis. It transforms the problem of evaluating difficult or impossible real integrals into a routine procedure: find singularities, compute residues, sum them up. Cauchy's integral theorem reveals that for analytic functions, the integral depends only on the endpoints (not the path), while Cauchy's integral formula reconstructs a function's value at any interior point purely from its boundary values — a remarkable result with no real-variable analog. These tools evaluate integrals like $\int_{-\infty}^{\infty} \frac{\cos x}{x^2+1}dx$ in minutes, solve boundary value problems in physics, and form the basis of the residue theorem that underpins all of advanced complex analysis. Without contour integration, Fourier analysis, quantum field theory, and much of mathematical physics would lack their most elegant computational methods.

Core Definitions

DfContour (Path)

A contour $\gamma$ is a piecewise smooth curve in $\mathbb{C}$ , parameterized as $z(t) = x(t) + iy(t)$ for $a \leq t \leq b$ . The curve is smooth if $z'(t)$ is continuous and $z'(t) \neq 0$ . A contour is closed if $z(a) = z(b)$ , and simple (non-self-intersecting) if it crosses itself at most at the endpoints.

DfContour Integral

The integral of $f$ along a contour $\gamma: z(t)$ , $a \leq t \leq b$ , is:

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

This reduces complex line integration to a real definite integral.

DfSimply Connected Domain

A domain $D$ is simply connected if every closed curve in $D$ can be continuously deformed to a point without leaving $D$ . Equivalently, $D$ has no "holes." The complement of a disk $\mathbb{C} \setminus \overline{D(a,r)}$ is not simply connected; the interior of a circle is.

DfWinding Number (Index)

The winding number of a closed contour $\gamma$ around a point $z_0$ not on $\gamma$ is:

\operatorname{Ind}_\gamma(z_0) = \frac{1}{2\pi i}\oint_\gamma \frac{dz}{z - z_0}

It counts how many times $\gamma$ winds around $z_0$ (positive = counterclockwise).

DfRemovable Singularity

A singularity $z_0$ of $f$ is removable if $\lim_{z \to z_0} (z - z_0)f(z) = 0$ . In this case, $f$ can be redefined at $z_0$ to be analytic.

DfSimple Pole

A singularity $z_0$ of $f$ is a simple pole if $\lim_{z \to z_0} (z - z_0)f(z) = c \neq 0$ . The residue is this limit: $\operatorname{Res}(f, z_0) = c$ .

Key Formulas

Contour Integral Definition

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

Here,

$z(t)$ =Parameterization of the contour
$z'(t)$ =Derivative of parameterization (velocity)

Cauchy's Integral Theorem

\oint_C f(z)\,dz = 0

Here,

$f$ =Analytic inside and on closed contour C
$C$ =Closed contour in a simply connected domain

Cauchy's Integral Formula

f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}\,dz

Here,

$z_0$ =Interior point of contour C
$f$ =Analytic on and inside C

Generalized Cauchy Integral Formula

f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{n+1}}\,dz

Here,

$n$ =Order of derivative
$z_0$ =Interior point of C

Residue Definition (Simple Pole)

\operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)

Here,

$z_0$ =Simple pole of f

Residue Theorem

\oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}(f, z_k)

Here,

$z_k$ =Singularities inside contour C
$\operatorname{Res}(f, z_k)$ =Residue of f at z_k

Residue for P/Q at Simple Pole

\operatorname{Res}\left(\frac{P}{Q}, z_0\right) = \frac{P(z_0)}{Q'(z_0)}

Here,

$P(z)$ =Analytic (or polynomial) numerator
$Q(z)$ =Analytic denominator with simple zero at z_0

Residue for Pole of Order n

\operatorname{Res}(f, z_0) = \frac{1}{(n-1)!}\lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}}\left[(z-z_0)^n f(z)\right]

Here,

$n$ =Order of the pole
$z_0$ =Location of the pole

Parameterization of Standard Contours

C_R: z = Re^{i\theta}, \quad 0 \leq \theta \leq 2\pi \quad \text{(circle of radius R)}

Here,

$R$ =Radius of the circle
$\theta$ =Angle parameter from 0 to 2π

Important Theorems

ThCauchy's Integral Theorem (Simple Form)

If $f$ is analytic inside and on a simple closed contour $C$ , and the domain interior to $C$ is simply connected, then:

\oint_C f(z)\,dz = 0

Proof Sketch: By Green's theorem, $\oint_C u\,dx - v\,dy + i\oint_C v\,dx + u\,dy$ . The Cauchy-Riemann equations make both integrands have zero curl, so both contour integrals vanish. $\square$

General version (Goursat): The theorem holds even without assuming $f'$ is continuous — analyticity alone suffices.

ThCauchy's Integral Formula

If $f$ is analytic inside and on a simple closed contour $C$ , and $z_0$ is interior to $C$ , then:

f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}\,dz

Proof Sketch: Define $g(z) = \frac{f(z) - f(z_0)}{z - z_0}$ for $z \neq z_0$ and $g(z_0) = f'(z_0)$ . Then $g$ is analytic inside $C$ . By Cauchy's theorem, $\oint_C g(z)\,dz = 0$ . Expanding and rearranging yields the formula. $\square$

Geometric meaning: The value of an analytic function at an interior point is completely determined by its values on the boundary.

ThResidue Theorem

If $f$ is analytic inside and on a simple closed contour $C$ except at isolated singularities $z_1, z_2, \ldots, z_n$ inside $C$ , then:

\oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}(f, z_k)

Proof Sketch: For each singularity $z_k$ , draw a small circle $\gamma_k$ around it. By Cauchy's theorem applied to the multiply connected region (deforming the contour), the integral over $C$ equals the sum of integrals over the small circles. Each integral over $\gamma_k$ equals $2\pi i \operatorname{Res}(f, z_k)$ by the Laurent series expansion. $\square$

ThDeformation of Contour

If $f$ is analytic in a region between two simple closed contours $C_1$ and $C_2$ (with $C_2$ inside $C_1$ ), and $f$ has no singularities between them, then:

\oint_{C_1} f(z)\,dz = \oint_{C_2} f(z)\,dz

provided both contours are oriented the same way (counterclockwise).

Significance: You can deform contours to pass around singularities conveniently, as long as you don't cross any singularities.

ThEstimation Lemma (ML Inequality)

If $|f(z)| \leq M$ on a contour $\gamma$ of length $L$ , then:

\left|\int_\gamma f(z)\,dz\right| \leq ML

Application: This is used to show that integrals over large arcs vanish as the arc radius $R \to \infty$ , which is essential for improper real integrals.

ThJordan's Lemma

If $g(z)$ is analytic in the upper half-plane and $|g(z)| \to 0$ uniformly as $|z| \to \infty$ in the upper half-plane, then for $a > 0$ :

\lim_{R \to \infty} \int_{C_R} g(z)e^{iaz}\,dz = 0

where $C_R$ is the upper semicircle of radius $R$ .

Application: Justifies discarding the semicircular arc when evaluating Fourier-type integrals $\int_{-\infty}^{\infty} f(x)e^{iax}\,dx$ .

Worked Examples

Example 1: Direct Parameterization

Evaluate $\oint_C \frac{dz}{z}$ where $C$ is the unit circle $|z| = 1$ , oriented counterclockwise.

Step 1: Parameterize: $z = e^{i\theta}$ , $dz = ie^{i\theta}\,d\theta$ , $0 \leq \theta \leq 2\pi$ .

Step 2: Substitute:

\oint_C \frac{dz}{z} = \int_0^{2\pi} \frac{ie^{i\theta}}{e^{i\theta}}\,d\theta = i\int_0^{2\pi} d\theta = 2\pi i

Alternatively: By the residue theorem, $z = 0$ is inside $C$ with $\operatorname{Res}(1/z, 0) = 1$ . So $\oint = 2\pi i \cdot 1 = 2\pi i$ . ✓

Example 2: Cauchy's Integral Formula

Evaluate $\oint_{|z|=2} \frac{e^z}{z - 1}\,dz$ .

Step 1: Identify: $f(z) = e^z$ is entire (analytic everywhere), and $z_0 = 1$ is inside $|z| = 2$ .

Step 2: Apply Cauchy's integral formula: $\oint_{|z|=2} \frac{e^z}{z-1}\,dz = 2\pi i \cdot f(1) = 2\pi i \cdot e^1 = 2\pi e i$ .

Verification: The function $e^z/(z-1)$ has a simple pole at $z = 1$ with residue $e^1 = e$ . The residue theorem gives $2\pi i \cdot e$ . ✓

Example 3: Generalized Cauchy Formula

Evaluate $\oint_{|z|=3} \frac{\sin z}{(z - \pi)^3}\,dz$ .

Step 1: $f(z) = \sin z$ is entire, $z_0 = \pi$ is inside $|z| = 3$ , $n = 2$ (since denominator is $(z-\pi)^{2+1}$ ).

Step 2: By the generalized Cauchy formula:

\oint \frac{\sin z}{(z-\pi)^3}\,dz = \frac{2\pi i}{2!} f''(\pi) = \pi i \cdot (-\sin\pi) = \pi i \cdot 0 = 0

Alternatively: $\operatorname{Res}\left(\frac{\sin z}{(z-\pi)^3}, \pi\right) = \frac{1}{2!}\lim_{z\to\pi}\frac{d^2}{dz^2}[\sin z] = \frac{1}{2}(-\sin\pi) = 0$ .

Example 4: Residue at a Simple Pole — Rational Function

Evaluate $\oint_{|z|=2} \frac{z}{(z-1)(z+1)}\,dz$ .

Step 1: Singularities at $z = 1$ and $z = -1$ , both inside $|z| = 2$ . Both are simple poles.

Step 2: Residue at $z = 1$ :

\operatorname{Res}(f, 1) = \lim_{z\to 1}(z-1)\frac{z}{(z-1)(z+1)} = \lim_{z\to 1}\frac{z}{z+1} = \frac{1}{2}

Step 3: Residue at $z = -1$ :

\operatorname{Res}(f, -1) = \lim_{z\to-1}(z+1)\frac{z}{(z-1)(z+1)} = \lim_{z\to-1}\frac{z}{z-1} = \frac{-1}{-2} = \frac{1}{2}

Step 4: Apply residue theorem:

\oint = 2\pi i\left(\frac{1}{2} + \frac{1}{2}\right) = 2\pi i \cdot 1 = 2\pi i

Example 5: Residue at a Pole of Order 2

Evaluate $\oint_{|z|=2} \frac{e^z}{z^2}\,dz$ .

Step 1: $z = 0$ is a pole of order 2 (double pole) inside $|z| = 2$ .

Step 2: Use the formula for a pole of order $n = 2$ :

\operatorname{Res}(f, 0) = \frac{1}{(2-1)!}\lim_{z\to 0}\frac{d}{dz}\left[z^2 \cdot \frac{e^z}{z^2}\right] = \lim_{z\to 0}\frac{d}{dz}[e^z] = \lim_{z\to 0} e^z = 1

Step 3: $\oint = 2\pi i \cdot 1 = 2\pi i$ .

Alternatively: Use the generalized Cauchy formula with $f(z) = e^z$ , $z_0 = 0$ , $n = 1$ : $\frac{1!}{2\pi i}\oint \frac{e^z}{z^2}\,dz = f'(0) = 1$ , so $\oint = 2\pi i$ . ✓

Example 6: Contour Deformation

Evaluate $\oint_{|z|=3} \frac{dz}{z^2 + 1}$ by deforming to small circles around the singularities.

Step 1: $z^2 + 1 = (z-i)(z+i)$ , so poles at $z = i$ and $z = -i$ , both inside $|z| = 3$ .

Step 2: Deform: draw small circles $\gamma_1$ around $i$ and $\gamma_2$ around $-i$ . By the deformation principle:

\oint_{|z|=3} = \oint_{\gamma_1} + \oint_{\gamma_2}

Step 3: $\operatorname{Res}(f, i) = \lim_{z\to i}\frac{z-i}{(z-i)(z+i)} = \frac{1}{2i}$

\operatorname{Res}(f, -i) = \lim_{z\to -i}\frac{z+i}{(z-i)(z+i)} = \frac{1}{-2i} = -\frac{1}{2i}

Step 4: $\oint = 2\pi i\left(\frac{1}{2i} - \frac{1}{2i}\right) = 2\pi i \cdot 0 = 0$

Verification: Since $1/(z^2+1)$ is analytic at $z = \infty$ (behaves like $1/z^2$ ), and the sum of all residues including the residue at infinity is zero, the residues at $\pm i$ cancel. ✓

Practice Problems

Problem 1: Evaluate a Contour Integral

Evaluate $\oint_{|z|=2} \frac{z+1}{z(z-1)}\,dz$ .

Solution

Step 1: Singularities at $z = 0$ (inside $|z|=2$ ) and $z = 1$ (inside $|z|=2$ ). Both are simple poles.

Step 2: Residue at $z = 0$ :

\operatorname{Res}(f, 0) = \lim_{z\to 0}\frac{z+1}{z-1} = \frac{1}{-1} = -1

Step 3: Residue at $z = 1$ :

\operatorname{Res}(f, 1) = \lim_{z\to 1}\frac{z+1}{z} = \frac{2}{1} = 2

Step 4: $\oint = 2\pi i(-1 + 2) = 2\pi i$

Problem 2: Cauchy Formula for Derivatives

Evaluate $\oint_{|z|=3} \frac{\cos z}{(z - \pi/2)^2}\,dz$ .

Solution

Step 1: $f(z) = \cos z$ is entire, $z_0 = \pi/2$ is inside $|z| = 3$ , $n = 1$ .

Step 2: Generalized Cauchy formula:

\frac{1!}{2\pi i}\oint \frac{\cos z}{(z-\pi/2)^2}\,dz = f'(\pi/2) = -\sin(\pi/2) = -1

Step 3: $\oint = 2\pi i \cdot (-1) = -2\pi i$

Problem 3: Residue at a Pole of Order 3

Find the residue of $f(z) = \frac{e^z}{(z-1)^3}$ at $z = 1$ .

Solution

Step 1: Pole of order $n = 3$ .

Step 2: Apply formula:

\operatorname{Res}(f, 1) = \frac{1}{2!}\lim_{z\to 1}\frac{d^2}{dz^2}\left[(z-1)^3 \cdot \frac{e^z}{(z-1)^3}\right] = \frac{1}{2}\lim_{z\to 1}\frac{d^2}{dz^2}[e^z] = \frac{1}{2}e^1 = \frac{e}{2}

Problem 4: Integral Over a Semicircular Contour

Evaluate $\int_{-\infty}^{\infty} \frac{dx}{x^2 + 4}$ using contour integration.

Solution

Step 1: Consider $f(z) = \frac{1}{z^2 + 4} = \frac{1}{(z+2i)(z-2i)}$ . Close the contour in the upper half-plane with a large semicircle $C_R$ .

Step 2: For the arc: $|f(z)| \leq \frac{1}{R^2 - 4}$ for $|z| = R$ , so $\left|\int_{C_R}\right| \leq \frac{\pi R}{R^2 - 4} \to 0$ as $R \to \infty$ .

Step 3: Only $z = 2i$ is in the upper half-plane. $\operatorname{Res}(f, 2i) = \frac{1}{4i}$ .

Step 4: By residue theorem:

\int_{-\infty}^{\infty}\frac{dx}{x^2+4} = 2\pi i \cdot \frac{1}{4i} = \frac{\pi}{2}

Problem 5: Sum of Residues

Evaluate $\oint_{|z|=4} \frac{z^2}{(z-1)(z-2)(z-3)}\,dz$ .

Solution

Step 1: All three poles $z = 1, 2, 3$ are inside $|z| = 4$ . All are simple poles.

Step 2: Residues (using $P(z)/Q'(z)$ with $Q(z) = (z-1)(z-2)(z-3)$ ):

Q'(z) = (z-2)(z-3) + (z-1)(z-3) + (z-1)(z-2)

\operatorname{Res}(f, 1) = \frac{1^2}{(1-2)(1-3)} = \frac{1}{(-1)(-2)} = \frac{1}{2}

\operatorname{Res}(f, 2) = \frac{4}{(2-1)(2-3)} = \frac{4}{(1)(-1)} = -4

\operatorname{Res}(f, 3) = \frac{9}{(3-1)(3-2)} = \frac{9}{(2)(1)} = \frac{9}{2}

Step 3: Sum of residues: $\frac{1}{2} - 4 + \frac{9}{2} = \frac{1 - 8 + 9}{2} = \frac{2}{2} = 1$

Step 4: $\oint = 2\pi i \cdot 1 = 2\pi i$

Common Mistakes

Mistake	Correction	Example
Forgetting the $2\pi i$ factor in the residue theorem	The residue theorem states $\oint = 2\pi i \sum \operatorname{Res}$ , not just $\sum \operatorname{Res}$	$\oint \frac{dz}{z} = 2\pi i$ , not $1$
Using the wrong orientation	Counterclockwise is positive; clockwise gives a minus sign	$\oint_{clockwise} = -2\pi i \sum \operatorname{Res}$
Including singularities outside the contour	Only sum residues of singularities inside $C$	If $z_0 \notin \operatorname{int}(C)$ , don't include it
Wrong residue formula for higher-order poles	For order $n$ : $\frac{1}{(n-1)!}\lim \frac{d^{n-1}}{dz^{n-1}}[(z-z_0)^n f]$ , not just the limit	Need the derivative for $n \geq 2$
Forgetting the ML inequality	Large arcs vanish only if $\|f\| \cdot L \to 0$ ; must verify the arc contribution vanishes	Check $\|f(z)\| = O(1/R^p)$ with $p > 0$
Incorrect parameterization	$dz = z'(t)dt$ , not just $dz = dt$	For $z = Re^{i\theta}$ : $dz = iRe^{i\theta}d\theta$
Misidentifying pole order	Factor the denominator carefully; check the limit	$\frac{1}{(z-1)^2(z+1)}$ has a pole of order 2 at $z=1$

Interview / Exam Questions

Q1: State Cauchy's integral theorem. What are its hypotheses, and why is the simply connected condition important?

A1: Cauchy's theorem states: if $f$ is analytic inside and on a simple closed contour $C$ , and the interior of $C$ is simply connected, then $\oint_C f(z)\,dz = 0$ . The simply connected condition ensures there are no singularities "hidden" inside the contour. For example, $\oint_{|z|=1} \frac{dz}{z} = 2\pi i \neq 0$ because $1/z$ has a pole at $z = 0$ inside the contour — the domain $\mathbb{C} \setminus \{0\}$ is not simply connected. The theorem holds because the Cauchy-Riemann equations make the integrand's "curl" vanish, and Green's theorem converts the contour integral to a double integral of this zero curl.

Q2: Why does Cauchy's integral formula imply that analytic functions are infinitely differentiable?

A2: Cauchy's formula $f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}dz$ expresses $f$ as an integral of an analytic integrand. Differentiating under the integral sign (justified by analyticity) gives $f'(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^2}dz$ , and repeated differentiation yields the generalized formula for $f^{(n)}$ . Since this works for all $n$ , $f$ is infinitely differentiable. In real analysis, a function can be differentiable once but not twice (e.g., $f(x) = x^{3/2}$ ). Complex differentiability is far more restrictive.

Q3: How do you compute the residue of $f(z) = \frac{P(z)}{Q(z)}$ at a simple pole $z_0$ ?

A3: If $z_0$ is a simple zero of $Q(z)$ (i.e., $Q(z_0) = 0$ but $Q'(z_0) \neq 0$ ), and $P(z_0) \neq 0$ , then:

\operatorname{Res}(f, z_0) = \frac{P(z_0)}{Q'(z_0)}

This comes from $f(z) = \frac{P(z)}{Q'(z_0)(z-z_0)} + \text{analytic terms}$ near $z_0$ , so the residue (coefficient of $1/(z-z_0)$ ) is $P(z_0)/Q'(z_0)$ .

Example: $\operatorname{Res}\left(\frac{z^2}{z^4+1}, e^{i\pi/4}\right) = \frac{e^{i\pi/2}}{4e^{3i\pi/4}} = \frac{e^{i\pi/2}}{4e^{3i\pi/4}} = \frac{e^{-i\pi/4}}{4}$ .

Q4: Explain the deformation of contour principle and give an example.

A4: If $f$ is analytic in a region between two simple closed contours $C_1$ (outer) and $C_2$ (inner), and $f$ has no singularities between them, then $\oint_{C_1} f\,dz = \oint_{C_2} f\,dz$ (both counterclockwise). This is because the region between $C_1$ and $C_2$ is multiply connected, but Cauchy's theorem applied to this annular region shows the two contour integrals are equal.

Example: To compute $\oint_{|z|=3} \frac{dz}{z(z-1)}$ , we can deform to two small circles: one around $z = 0$ and one around $z = 1$ , each giving $2\pi i$ times the local residue.

Q5: When is the residue theorem more useful than direct parameterization?

A5: The residue theorem is superior when: (1) The contour encloses singularities — you only need the residues, not the full parameterization. (2) The integral has multiple poles — the theorem converts integration to algebra (summing residues). (3) You need to evaluate integrals over large contours (as $R \to \infty$ ) — residues capture the essential behavior while the contour contribution vanishes. Direct parameterization is simpler for elementary functions with no singularities inside the contour (where the integral is zero by Cauchy's theorem) or for very simple cases like $\oint \frac{dz}{z}$ .

Q6: What is the ML inequality, and when do you use it?

A6: The ML inequality states: if $|f(z)| \leq M$ on a contour $\gamma$ of length $L$ , then $\left|\int_\gamma f(z)\,dz\right| \leq ML$ . It's used to bound contour integrals and show that contributions from arcs vanish. For example, to show $\int_{C_R} \frac{e^{iz}}{z^2+1}dz \to 0$ as $R \to \infty$ (where $C_R$ is a semicircle): on $C_R$ , $|e^{iz}| = e^{-R\sin\theta} \leq 1$ and $|z^2+1| \geq R^2 - 1$ , so $|f| \leq \frac{1}{R^2-1}$ and $L = \pi R$ . Thus $\left|\int_{C_R}\right| \leq \frac{\pi R}{R^2-1} \to 0$ .

Quick Reference

Formula Summary

Formula	Expression	Notes
Contour integral	$\int_\gamma f\,dz = \int_a^b f(z(t))z'(t)dt$	Parameterize and integrate
Cauchy's theorem	$\oint_C f\,dz = 0$	$f$ analytic on/inside simply connected $C$
Cauchy's formula	$f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}dz$	Reconstruct interior value from boundary
Generalized Cauchy	$f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}dz$	Derivatives from boundary values
Residue theorem	$\oint_C f\,dz = 2\pi i\sum \operatorname{Res}(f, z_k)$	Sum of residues × $2\pi i$
Simple pole residue	$\operatorname{Res}(f, z_0) = \lim_{z\to z_0}(z-z_0)f(z)$	Or $P(z_0)/Q'(z_0)$ for $P/Q$
Pole of order $n$	$\operatorname{Res} = \frac{1}{(n-1)!}\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}[(z-z_0)^n f]$	Need $(n-1)$ -th derivative
ML inequality	$\\|\int_\gamma f\,dz\\| \leq ML$	$M = \max\\|f\\|$ , $L =$ length of $\gamma$
Winding number	$\operatorname{Ind}_\gamma(z_0) = \frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-z_0}$	Counts loops around $z_0$
Deformation	$\oint_{C_1} f\,dz = \oint_{C_2} f\,dz$	If no singularities between $C_1$ and $C_2$

Cross-References

091 - Complex Numbers — Polar form and De Moivre's theorem are used to parameterize circular contours.
092 - Complex Functions — Analyticity and singularity classification (from 092) determine which residues to compute.
094 - Residue Theory Applications — The residue theorem applied to specific integral types (trigonometric, improper, keyhole).
095 - Applications — Fourier transform inversion uses contour integration; Green's functions in physics are built from Cauchy integrals.
Vector Calculus (Topic 19): Green's theorem connects real line integrals to complex contour integrals; Stokes' theorem generalizes these ideas.
Differential Equations — Cauchy integral formula provides solutions to differential equations via integral representations.

Why It Matters

Core Definitions

DfContour (Path)

DfContour Integral

DfSimply Connected Domain

DfWinding Number (Index)

DfRemovable Singularity

DfSimple Pole

Key Formulas

Contour Integral Definition

Cauchy's Integral Theorem

Cauchy's Integral Formula

Generalized Cauchy Integral Formula

Residue Definition (Simple Pole)

Residue Theorem

Residue for P/Q at Simple Pole

Residue for Pole of Order n

Parameterization of Standard Contours

Important Theorems

ThCauchy's Integral Theorem (Simple Form)

ThCauchy's Integral Formula

ThResidue Theorem

ThDeformation of Contour

ThEstimation Lemma (ML Inequality)

ThJordan's Lemma

Worked Examples

Example 1: Direct Parameterization

Example 2: Cauchy's Integral Formula

Example 3: Generalized Cauchy Formula

Example 4: Residue at a Simple Pole — Rational Function

Example 5: Residue at a Pole of Order 2

Example 6: Contour Deformation

Practice Problems

Problem 1: Evaluate a Contour Integral

Solution

Problem 2: Cauchy Formula for Derivatives

Solution

Problem 3: Residue at a Pole of Order 3

Solution

Problem 4: Integral Over a Semicircular Contour

Solution

Problem 5: Sum of Residues

Solution

Common Mistakes

Interview / Exam Questions

Quick Reference

Formula Summary

Cross-References

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