Chapter 5 • Theory & Derivations

Complex Series Expansions: Taylor, Laurent & Singularity Classification

Taylor series representation, Laurent series on annular domains, classification of isolated singularities (removable, poles, essential), Casorati-Weierstrass theorem, and Picard's Great Theorem.

§5.1 Taylor Series Expansion of Holomorphic Functions & Analytic Radii of Convergence

1. Taylor's Expansion Theorem

Theorem 5.1 (Taylor's Theorem): Let $f(z)$ be holomorphic on an open disk $D(z_0, R)$. Then $f(z)$ has a unique power series representation: $$f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n$$ converging absolutely for all $|z - z_0| < R$, where the coefficients $a_n$ are given by: $$a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i} \oint_\gamma \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}\,d\zeta$$ The radius of convergence $R$ equals the exact distance from $z_0$ to the nearest singularity of $f(z)$.

§5.2 Laurent Series Expansion on Annular Domains r < |z - z0| < R

1. Laurent's Theorem

When a function $f(z)$ is holomorphic in an annular region $A = \{z \in \mathbb{C} : r < |z - z_0| < R\}$ ($0 \le r < R \le \infty$), it can be represented by a series containing both positive and negative powers of $(z - z_0)$:

Theorem 5.2 (Laurent's Theorem): For any $z \in A$: $$f(z) = \sum_{n=-\infty}^\infty c_n (z - z_0)^n = \underbrace{\sum_{n=1}^\infty \frac{b_n}{(z - z_0)^n}}_{\text{Principal Part}} + \underbrace{\sum_{n=0}^\infty a_n (z - z_0)^n}_{\text{Analytic Part}}$$ where the coefficients are given by: $$c_n = \frac{1}{2\pi i} \oint_\gamma \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}\,d\zeta, \quad n \in \mathbb{Z}$$ along any simple closed counter-clockwise contour $\gamma$ in the annulus.

§5.3 Classification of Isolated Singularities: Removable, Poles & Essential Singularities

1. Types of Isolated Singularities

Let $z_0$ be an isolated singularity of $f(z)$ (meaning $f$ is holomorphic on punctured disk $0 < |z - z_0| < R$). The nature of the singularity is determined by the principal part $\sum_{n=1}^\infty \frac{b_n}{(z - z_0)^n}$ of its Laurent series:

Classification:

Removable Singularity: The principal part vanishes entirely ($b_n = 0$ for all $n \ge 1$). $\lim_{z \to z_0} f(z)$ exists and is finite. (Riemann's Removable Singularity Theorem: bounded near $z_0 \implies$ removable).
Pole of Order $m \ge 1$: The principal part terminates with finite non-zero terms: $$\frac{b_m}{(z - z_0)^m} + \dots + \frac{b_1}{z - z_0}, \quad b_m \ne 0$$ In this case, $\lim_{z \to z_0} |f(z)| = \infty$. If $m = 1$, it is a simple pole.
Essential Singularity: The principal part contains infinitely many non-zero terms. The limit $\lim_{z \to z_0} f(z)$ does not exist (not even as $\infty$). Example: $e^{1/z} = \sum_{n=0}^\infty \frac{1}{n! z^n}$ at $z = 0$.

§5.4 Behavior Near Essential Singularities: Casorati-Weierstrass & Picard Theorems

1. The Casorati-Weierstrass Theorem

Theorem 5.3 (Casorati-Weierstrass Theorem): Let $z_0$ be an isolated essential singularity of $f(z)$. Then for any complex number $w \in \mathbb{C}$, any $\epsilon > 0$, and any $\delta > 0$, there exists a point $z \in D(z_0, \delta) \setminus \{z_0\}$ such that: $$|f(z) - w| < \epsilon$$ In other words, the image of any punctured neighborhood of an essential singularity is dense in the entire complex plane $\mathbb{C}$!

2. Picard's Great Theorem

Theorem 5.4 (Picard's Great Theorem): In any punctured neighborhood of an isolated essential singularity, $f(z)$ assumes every complex value infinitely many times, with at most one possible exception! (For example, $e^{1/z}$ assumes every complex value except $0$).

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 • Foundational Example 5.1: Laurent Series in Annular Domain 1 < |z| < 2

Find the Laurent series expansion of $f(z) = \frac{1}{(z - 1)(z - 2)}$ in the annular region $1 < |z| < 2$.

Step 1: Partial Fraction Decomposition

\frac{1}{(z - 1)(z - 2)} = \frac{A}{z - 1} + \frac{B}{z - 2}

$1 = A(z - 2) + B(z - 1)$. At $z = 1$: $1 = -A \implies A = -1$. At $z = 2$: $1 = B \implies B = 1$.

f(z) = -\frac{1}{z - 1} + \frac{1}{z - 2}

Step 2: Expand $-\frac{1}{z - 1}$ for $|z| > 1$ (Principal Part)
Since $|z| > 1 \implies \left|\frac{1}{z}\right| < 1$:

-\frac{1}{z - 1} = -\frac{1}{z(1 - 1/z)} = -\frac{1}{z} \sum_{n=0}^\infty \frac{1}{z^n} = -\sum_{n=0}^\infty \frac{1}{z^{n+1}} = -\sum_{m=1}^\infty \frac{1}{z^m}

Step 3: Expand $\frac{1}{z - 2}$ for $|z| < 2$ (Analytic Part)
Since $|z| < 2 \implies \left|\frac{z}{2}\right| < 1$:

\frac{1}{z - 2} = -\frac{1}{2(1 - z/2)} = -\frac{1}{2} \sum_{n=0}^\infty \left(\frac{z}{2}\right)^n = -\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}

Step 4: Combine both series

f(z) = -\sum_{m=1}^\infty \frac{1}{z^m} - \sum_{n=0}^\infty \frac{z^n}{2^{n+1}} = \dots - \frac{1}{z^2} - \frac{1}{z} - \frac{1}{2} - \frac{z}{4} - \frac{z^2}{8} - \dots

Final Answer & Physical Insight

$f(z) = -\sum_{m=1}^\infty \frac{1}{z^m} - \sum_{n=0}^\infty \frac{z^n}{2^{n+1}}$

Tier 2 • Intermediate Exam Example 5.2: Rouché's Theorem: Roots of Polynomial in Concentric Annuli

Use Rouché's theorem to prove that all five roots of $z^5 + 3z + 1 = 0$ lie in $|z| < 2$, and exactly one root lies in $|z| < 1$.

Part 1: Disk $|z| < 2$
Let $f(z) = z^5$ and $g(z) = 3z + 1$. On the circle $C_2: |z| = 2$:

$$|f(z)| = |z|^5 = 2^5 = 32$$

|g(z)| = |3z + 1| \le 3|z| + 1 = 3(2) + 1 = 7

Since $|g(z)| = 7 < 32 = |f(z)|$ on $|z| = 2$, by Rouché's Theorem, $f(z) + g(z) = z^5 + 3z + 1$ has the same number of roots inside $|z| < 2$ as $f(z) = z^5$. Since $z^5$ has 5 roots (at $z = 0$), all 5 roots of $z^5 + 3z + 1 = 0$ lie inside $|z| < 2$.

Part 2: Disk $|z| < 1$
Now choose $f_1(z) = 3z$ and $g_1(z) = z^5 + 1$. On the circle $C_1: |z| = 1$:

$$|f_1(z)| = 3|z| = 3(1) = 3$$

|g_1(z)| = |z^5 + 1| \le |z|^5 + 1 = 1 + 1 = 2

Since $|g_1(z)| = 2 < 3 = |f_1(z)|$ on $|z| = 1$, by Rouché's Theorem, $f_1(z) + g_1(z) = z^5 + 3z + 1$ has the same number of roots inside $|z| < 1$ as $f_1(z) = 3z$. Since $3z = 0$ has exactly 1 root (at $z = 0$), $z^5 + 3z + 1 = 0$ has exactly 1 root inside $|z| < 1$.

(The remaining 4 roots lie in the annular region $1 \le |z| < 2$).

Final Answer & Physical Insight

Proved: All 5 roots lie in $|z| < 2$; exactly 1 root in $|z| < 1$; 4 roots in $1 \le |z| < 2$.

Tier 3 • Honors Challenge Example 5.3: Mellin-Type Branch Cut Integral via Keyhole Contour

Evaluate $\int_0^\infty \frac{x^{a-1}}{1 + x}\,dx = \frac{\pi}{\sin(\pi a)}$ for $0 < a < 1$ using a keyhole contour around the positive real axis.

Step 1: Complex extension and branch cut
Consider $f(z) = \frac{z^{a-1}}{1 + z}$. We choose the branch cut along $[0, \infty)$ with $0 \le \arg z < 2\pi$. Poles: simple pole at $z = -1 = e^{i\pi}$.

Step 2: Residue at $z = -1$

\text{Res}(f, -1) = \lim_{z \to -1} (z + 1) \frac{z^{a-1}}{1 + z} = (-1)^{a-1} = (e^{i\pi})^{a-1} = e^{i\pi(a - 1)} = -e^{i\pi a}

Step 3: Keyhole contour integration
Let the keyhole contour consist of:

Upper edge $L_1$: $z = x + i0^+ \implies z^{a-1} = x^{a-1}$.
Large circle $C_R$: $|f(z)| \le \frac{R^{a-1}}{R - 1} \implies \int_{C_R} \to 0$ as $R \to \infty$ since $a < 1$.
Lower edge $L_2$: $z = x - i0^+ \implies z = x e^{2\pi i} \implies z^{a-1} = x^{a-1} e^{2\pi i (a - 1)} = x^{a-1} e^{2\pi i a}$.
Small circle $C_\epsilon$: $|f(z)| \le \frac{\epsilon^{a-1}}{1 - \epsilon} \implies \int_{C_\epsilon} \to 0$ as $\epsilon \to 0$ since $a > 0$.

Step 4: Combine the edges

\int_0^\infty \frac{x^{a-1}}{1 + x}\,dx - e^{2\pi i a} \int_0^\infty \frac{x^{a-1}}{1 + x}\,dx = 2\pi i\, \text{Res}(f, -1)

(1 - e^{2\pi i a}) \int_0^\infty \frac{x^{a-1}}{1 + x}\,dx = 2\pi i (-e^{i\pi a})

Divide both sides by $-e^{i\pi a}$:

(e^{i\pi a} - e^{-i\pi a}) \int_0^\infty \frac{x^{a-1}}{1 + x}\,dx = 2\pi i

Since $e^{i\pi a} - e^{-i\pi a} = 2i\sin(\pi a)$:

2i\sin(\pi a) \int_0^\infty \frac{x^{a-1}}{1 + x}\,dx = 2\pi i \implies \int_0^\infty \frac{x^{a-1}}{1 + x}\,dx = \frac{\pi}{\sin(\pi a)}

Final Answer & Physical Insight

$\int_0^\infty \frac{x^{a-1}}{1 + x}\,dx = \frac{\pi}{\sin(\pi a)}$