12.3 Laurent Series, Singularities, and the Residue Theorem

12.3 Laurent Series, Singularities, and the Residue Theorem

Residue calculus is the most computational topic in complex analysis on the GRE Mathematics Subject Test. Expanding functions into Laurent series on annular domains classifies isolated singularities and provides rapid evaluation of complex and real integrals.

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Taylor and Laurent Series Expansions

Taylor Series in $\mathbb{C}$

If $f(z)$ is analytic in $|z - z_0| < R$, its power series representation is: $f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n \quad \text{with} \quad a_n = \frac{f^{(n)}(z_0)}{n!}$ The radius of convergence $R$ is the distance to the nearest singularity of $f(z)$.

Laurent Series in Annuli

If $f(z)$ is analytic in an annulus $R_1 < |z - z_0| < R_2$, it expands uniquely into: $f(z) = \sum_{n=-\infty}^\infty a_n (z - z_0)^n = \underbrace{\sum_{n=0}^\infty a_n (z - z_0)^n}_{\text{Analytic Part}} + \underbrace{\sum_{k=1}^\infty \frac{a_{-k}}{(z - z_0)^k}}_{\text{Principal Part}}$ Coefficients are $a_n = \frac{1}{2\pi i} \oint_C \frac{f(w)}{(w - z_0)^{n+1}} dw$ for any circle $C$ in the annulus.

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Classification of Isolated Singularities

An isolated singularity $z_0$ is classified by the principal part of its Laurent series in $0 < |z - z_0| < R$:

1. Removable Singularities

• Condition: Principal part is zero ($a_{-k} = 0$ for all $k \ge 1$).
• Limit: $\lim_{z \to z_0} f(z) = L \in \mathbb{C}$ exists and is finite.
• Riemann's Theorem: If $f$ is bounded near $z_0$, the singularity is removable. Defining $f(z_0) = L$ restores analyticity.
• Example: $\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \dots$ at $z = 0$.

2. Poles of Order $m \ge 1$

• Condition: Principal part has $m$ terms: $\frac{a_{-m}}{(z - z_0)^m} + \dots + \frac{a_{-1}}{z - z_0}$ with $a_{-m} \neq 0$.
• When $m = 1$, $z_0$ is a simple pole.
• Limit: $\lim_{z \to z_0} |f(z)| = \infty$.
• Criterion: $\lim_{z \to z_0} (z - z_0)^m f(z) = L \neq 0, \infty$.

3. Essential Singularities

• Condition: Principal part has infinitely many non-zero negative terms.
• Limit: $\lim_{z \to z_0} f(z)$ does not exist (neither finite nor infinite).
• Casorati-Weierstrass Theorem: The image of any punctured neighborhood of $z_0$ is dense in $\mathbb{C}$.
• Picard's Great Theorem: $f(z)$ assumes every complex value infinitely often, with at most one exception (e.g., $e^{1/z}$ omits $0$).

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Residue Theory and Computation Formulas

The residue of $f(z)$ at $z_0$ is the coefficient $a_{-1}$ of $(z - z_0)^{-1}$: $\operatorname{Res}(f, z_0) = a_{-1} = \frac{1}{2\pi i} \oint_C f(z) dz$

Residue Computation Rules

1. Simple Pole: $\operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z)$
2. Quotient Rule for Simple Poles: If $f(z) = \frac{p(z)}{q(z)}$ with $p(z_0) \neq 0$, $q(z_0) = 0$, $q'(z_0) \neq 0$: $\operatorname{Res}(f, z_0) = \frac{p(z_0)}{q'(z_0)}$
3. Pole of Order $m \ge 2$: $\operatorname{Res}(f, z_0) = \frac{1}{(m - 1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right]$
4. Series Expansion: Directly expand Maclaurin series to identify $a_{-1}$ for essential singularities.

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The Cauchy Residue Theorem

Theorem: If $f(z)$ is analytic on and inside simple closed counterclockwise contour $C$, except at isolated singularities $z_1, \dots, z_N$ inside $C$: $\oint_C f(z) dz = 2\pi i \sum_{k=1}^N \operatorname{Res}(f, z_k)$

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Applications to Real Integrals

1. Trigonometric Integrals over $[0, 2\pi]$

Set $z = e^{i\theta}$, $d\theta = \frac{dz}{iz}$, $\cos\theta = \frac{z + z^{-1}}{2}$, $\sin\theta = \frac{z - z^{-1}}{2i}$ to convert $\int_0^{2\pi} R(\cos\theta, \sin\theta) d\theta$ to a unit circle contour integral $\oint_{|z|=1} \tilde{R}(z) \frac{dz}{iz}$.

2. Rational Improper Integrals

If $\deg(Q) \ge \deg(P) + 2$ and $Q(x) \neq 0$ on $\mathbb{R}$: $\int_{-\infty}^\infty \frac{P(x)}{Q(x)} dx = 2\pi i \sum_{\operatorname{Im}(z_k) > 0} \operatorname{Res}\left( \frac{P}{Q}, z_k \right)$

3. Fourier Integrals and Jordan's Lemma

For $\int_{-\infty}^\infty f(x) e^{iax} dx$ ($a > 0$):

• Jordan's Lemma: If $|f(z)| \to 0$ on upper semicircle $C_R$, then $\lim_{R \to \infty} \int_{C_R} f(z) e^{iaz} dz = 0$.

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Singularity Classification Table

| Type | Principal Part | Limit $\lim_{z \to z_0} |f(z)|$ | Punctured Disk Image | Example | | :--- | :--- | :--- | :--- | :--- | | Removable | No terms | Finite $L \ge 0$ | Bounded set | $\frac{\sin z}{z}$ at $z = 0$ | | Simple Pole | One term: $\frac{a_{-1}}{z - z_0}$ | Approaches $\infty$ | Neighborhood of $\infty$ | $\frac{1}{z - 2}$ at $z = 2$ | | Order $m$ Pole | $m$ terms: $\frac{a_{-m}}{(z - z_0)^m} + \dots$ | Approaches $\infty$ | Neighborhood of $\infty$ | $\frac{1}{(z - i)^m}$ at $z = i$ | | Essential | Infinitely many negative terms | Does not exist | Dense in $\mathbb{C}$ | $e^{1/z}$ at $z = 0$ |

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Step-by-Step Worked Problems

Problem 1: Essential Singularity Residue

Find the residue of $f(z) = z^3 \cos(1/z)$ at $z = 0$.

Solution:

1. Expand $\cos(1/z) = 1 - \frac{1}{2 z^2} + \frac{1}{24 z^4} - \dots$.
2. Multiply by $z^3$: $f(z) = z^3 - \frac{z}{2} + \frac{1}{24 z} - \dots$.
3. The residue is the coefficient $a_{-1}$ of $1/z$, which is $\frac{1}{24}$.

Problem 2: Real Trigonometric Integral

Evaluate $I = \int_0^{2\pi} \frac{d\theta}{5 + 3\cos\theta}$.

Solution:

1. With $z = e^{i\theta}$, $5 + 3\cos\theta = \frac{3z^2 + 10z + 3}{2z}$.
2. Integral transforms to $\oint_{|z|=1} \frac{2}{i(3z + 1)(z + 3)} dz$.
3. Enclosed pole is $z = -1/3$. Using $p/q'$: $\operatorname{Res} = \frac{2/i}{6(-1/3) + 10} = \frac{2/i}{8} = -\frac{i}{4}$
4. By Residue Theorem: $I = 2\pi i \left(-\frac{i}{4}\right) = \frac{\pi}{2}$.

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GRE Exam Traps & Pitfalls

Trap 1: Summing Poles Outside Contour In improper integrals, sum residues ONLY for poles in the upper half-plane ($\operatorname{Im}(z) > 0$).

Trap 2: Jordan's Lemma Sign Rules For $e^{iaz}$: if $a > 0$, close in the upper half-plane. If $a < 0$, close in the lower half-plane with a clockwise orientation factor.

Trap 3: Higher-Order Pole Formula The formula $p/q'$ applies ONLY to simple poles. Double poles require taking derivatives of $(z - z_0)^2 f(z)$.