12.2 Contour Integration and Cauchy's Theorems

12.2 Contour Integration and Cauchy's Theorems

Contour integration is a core component of the GRE Mathematics Subject Test. The central theorem of complex analysis is that an analytic function's boundary values completely determine its interior behavior.

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Contours, Parameterization, and Complex Line Integrals

A contour $C$ is a piecewise smooth directed curve $z(t) = x(t) + i y(t)$ for $t \in [a, b]$. The contour integral of $f(z) = u + iv$ is: $\int_C f(z) dz = \int_a^b f(z(t)) z'(t) dt = \int_C (u dx - v dy) + i \int_C (v dx + u dy)$

Canonical Circle Integral

For $C: z(\theta) = z_0 + R e^{i\theta}$ ($0 \le \theta \le 2\pi$), $dz = i R e^{i\theta} d\theta$: $\oint_{|z - z_0| = R} (z - z_0)^m dz = \begin{cases} 2\pi i, & m = -1 \\ 0, & m \in \mathbb{Z}, \, m \neq -1 \end{cases}$ This fundamental identity underlies Cauchy's integral formula and residue calculus.

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The ML-Inequality (Estimation Lemma)

If $f$ is continuous on contour $C$ of length $L$ and $|f(z)| \le M$ on $C$, then: $\left| \int_C f(z) dz \right| \le M L$ For an upper semicircle $C_R$ of radius $R$ ($L = \pi R$), if $|f(z)| \le K / R^p$ with $p > 1$, then $\left| \int_{C_R} f(z) dz \right| \le \frac{\pi K}{R^{p-1}} \to 0$ as $R \to \infty$.

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The Cauchy-Goursat Theorem and Path Independence

Cauchy-Goursat Theorem

If $f(z)$ is analytic on a simply connected domain $D$, then for any simple closed contour $C$ in $D$: $\oint_C f(z) dz = 0$

Key Consequences

1. Path Independence: $\int_{z_1}^{z_2} f(z) dz$ depends solely on endpoints $z_1, z_2 \in D$.
2. Primitives: There exists analytic $F(z)$ with $F'(z) = f(z)$, so $\int_{z_1}^{z_2} f(z) dz = F(z_2) - F(z_1)$.
3. Deformation of Path: If $f$ is analytic between nested contours $C_1$ and $C_2$, then $\oint_{C_1} f(z) dz = \oint_{C_2} f(z) dz$.

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Cauchy's Integral Formulas

Interior Values

If $f(z)$ is analytic on and inside a simple closed counterclockwise contour $C$, and $z_0$ lies inside $C$: $f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} dz \iff \oint_C \frac{f(z)}{z - z_0} dz = 2\pi i f(z_0)$

Formula for Higher Derivatives

Analytic functions are infinitely differentiable, with derivatives: $f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} dz \iff \oint_C \frac{f(z)}{(z - z_0)^{n+1}} dz = \frac{2\pi i}{n!} f^{(n)}(z_0)$

Cauchy's Estimates

If $|f(z)| \le M$ on $|z - z_0| = R$, the $ML$-inequality implies $\left| f^{(n)}(z_0) \right| \le \frac{n! M}{R^n}$.

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Liouville's Theorem and Fundamental Theorem of Algebra

Liouville's Theorem

Theorem: A bounded entire function is constant.

• Proof: If $|f(z)| \le M$ on $\mathbb{C}$, Cauchy's estimate gives $|f'(z_0)| \le M / R \to 0$ as $R \to \infty$, so $f' \equiv 0$.
• Extended Form: If $|f(z)| \le A + B |z|^k$, $f$ is a polynomial of degree at most $k$.

Fundamental Theorem of Algebra

Theorem: Every non-constant polynomial $P(z)$ has at least one root in $\mathbb{C}$.

• Proof: If $P(z) \neq 0$, then $1/P(z)$ is a bounded entire function, hence constant by Liouville, contradiction.

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Maximum Modulus Principle

Let $f(z)$ be analytic on connected domain $D$:

1. Interior Maximum: If $|f(z)|$ attains a local maximum in $D$, $f(z)$ is constant.
2. Boundary Maximum: On a bounded domain with continuous extension to $\overline{D}$, $\max_{z \in \overline{D}} |f(z)| = \max_{z \in \partial D} |f(z)|$.
3. Minimum Modulus: If $f(z) \neq 0$ on $D$, $|f(z)|$ also attains its minimum on $\partial D$.

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Summary of Cauchy Theorems

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Step-by-Step Worked Problems

Problem 1: Higher-Order Derivative Formula

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

Solution:

1. Pole $z_0 = 1$ is inside $|z|=2$. Denominator is $(z - 1)^{n+1}$ with $n = 2$.
2. Let $f(z) = e^{3z}$. Then $f'(z) = 3e^{3z}$ and $f''(z) = 9e^{3z}$.
3. By Cauchy's Derivative Formula: $I = \frac{2\pi i}{2!} f''(1) = \pi i (9e^3) = 9\pi i e^3$

Problem 2: Bounded Entire Function Degree

Let $f$ be entire with $|f(z)| \le 4|z|^2 + 7$. Given $f(0) = 1$, $f(1) = 3$, and $f(-1) = 1$, find $f(2)$.

Solution:

1. By extended Liouville, $f(z) = az^2 + bz + c$.
2. $f(0) = c = 1$.
3. $f(1) = a + b + 1 = 3 \implies a + b = 2$.
4. $f(-1) = a - b + 1 = 1 \implies a - b = 0$.
5. Thus $a = 1, b = 1$, so $f(z) = z^2 + z + 1$.
6. $f(2) = 2^2 + 2 + 1 = 7$.

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GRE Exam Traps & Pitfalls

Trap 1: The $(n!)$ Denominator Error For $\oint \frac{f(z)}{(z - z_0)^k} dz$, the derivative order is $n = k - 1$. Divide by $(k - 1)!$.

Trap 2: Ignoring Singularity Locations Singularities outside contour $C$ contribute zero. Do not include exterior residues or poles.

Trap 3: Zeros and the Minimum Modulus Principle The Minimum Modulus Principle requires $f(z) \neq 0$. If $f$ has an interior zero, the minimum modulus is $0$ in the interior, not on the boundary.