Taylor and Laurent series, singularities, residues

Recall the one integral everything rests on, from Chapter 7: $\displaystyle\oint_{|z|=r} z^n\,dz = 0$ for every integer $n \ne -1$, and $= 2\pi i$ when $n = -1$. Now integrate a Laurent series $\sum a_n (z-z_0)^n$ term by term around a small loop. Every term dies — except the single term $a_{-1}(z-z_0)^{-1}$, which contributes $a_{-1}\cdot 2\pi i$. So

The entire loop integral collapses to one coefficient. That coefficient $a_{-1}$ has a name — the residue — and that single sentence is Cauchy's residue theorem in embryo. Everything else in this chapter is machinery for finding $a_{-1}$ without writing out the whole series.

A point where an analytic $q$ has a zero of order $m$ is one where $q(z) = (z-z_0)^m\,(\text{something nonzero})$. Flip the function over: $\dfrac{1}{q}$ then has a pole of order $m$ there. More generally a quotient $\dfrac{p}{q}$ has, at a zero of order $m$ of $q$, a pole of order at most $m$ — "at most," because the numerator $p$ may vanish there too and cancel some of the factors. That "at most" is not a footnote: it is the exact wire you will trip in practice problem P3, where a denominator $(z-1)^2$ hides a singularity that is only a simple pole.

If $f$ is analytic throughout the disk $|z - z_0| < R$, then for every $z$ in that disk

The radius of convergence is the distance from $z_0$ to the nearest singularity of $f$. The series is exactly as large as the biggest singularity-free disk around $z_0$ — no larger, because a power series cannot converge across a point where the function blows up.

If $f$ is analytic on an annulus $r < |z - z_0| < R$ (a ring, possibly with $r = 0$ or $R = \infty$), then throughout that ring

The negative-power terms are the principal part. A function can have several different Laurent series about the same $z_0$ — one for each ring on which it is analytic. The ring you are working in decides which series is correct.

An exam never asks for "the Laurent series." It asks for the series valid in $1 < |z| < 2$, or in $0 < |z| < 1$. Read the ring off the question, and write your ring down before you expand — the next worked example shows two different answers for one function, and the only thing that tells them apart is the inequality.

Let $z_0$ be an isolated singularity of $f$, with Laurent series $\sum a_n (z-z_0)^n$ valid in a punctured disk $0 < |z - z_0| < R$. Look only at the principal part (the negative powers):

• Removable — no negative powers at all. $f$ extends to an analytic function; $\lim_{z\to z_0} f(z)$ exists and is finite.
• Pole of order $m$ — the lowest power is $(z-z_0)^{-m}$ ($a_{-m} \ne 0$, and $a_{-k} = 0$ for all $k > m$). $m = 1$ is a simple pole.
• Essential — infinitely many negative powers.

One closing line, because it is reassuring and occasionally examined: even at an essential singularity the residue is just $a_{-1}$ read from the series. For $e^{1/z} = 1 + \dfrac1z + \dfrac{1}{2!\,z^2} + \cdots$, the coefficient of $\tfrac1z$ is $1$, so $\operatorname{Res}_{0} e^{1/z} = 1$ and $\displaystyle\oint_{|z|=1} e^{1/z}\,dz = 2\pi i$. No residue formula applies to an essential singularity — but none is needed. The series is the formula.

If $z_0$ is a simple pole of $f$,

Multiply away the single bad factor, then evaluate at $z_0$.

Kill the pole with $(z-z_0)^m$, differentiate $m-1$ times, divide by $(m-1)!$, evaluate. For $m = 1$ this is the simple-pole rule (no derivative, $0! = 1$).

If $f = \dfrac{p}{q}$ with $p, q$ analytic, $p(z_0) \ne 0$, and $q$ has a simple zero at $z_0$ (so $q(z_0) = 0$, $q'(z_0) \ne 0$), then

No factoring of $q$ needed — differentiate the denominator and evaluate. This is the fastest route for things like $\dfrac{2z+1}{z^2 - z - 2}$ (you will use it in P5).

Let $C$ be a positively oriented simple closed contour, and let $f$ be analytic on and inside $C$ except at finitely many isolated singularities $z_1, \dots, z_k$ inside $C$. Then

Key idea: expand $f$ in its Laurent series at each enclosed singularity and integrate term-by-term — every power dies except $a_{-1}$, since $\oint (z - z_0)^n\,dz = 0$ for all $n \ne -1$.

This is the single formula that replaces Chapter 8's split-and-apply ritual: locate the poles inside, add their residues, multiply by $2\pi i$. Step 0 — locate every pole and keep only the enclosed ones — is as decisive here as it was for Cauchy's integral formula.

For any $\displaystyle\int_0^{2\pi} R(\cos\theta, \sin\theta)\,d\theta$ with $R$ a rational function, set $z = e^{i\theta}$. As $\theta$ runs $0 \to 2\pi$, $z$ traces the unit circle once counter-clockwise, and

The integral becomes $\displaystyle\oint_{|z|=1}(\text{rational in } z)\,dz$ — then it is just Step 0 and the residue theorem.

• The value jumps. Slide $R$ across a pole's radius and the integral snaps to a new value — it does not drift. The integral counts only the poles the contour encloses; the size of the contour is otherwise irrelevant.
• Cancellation is real. For $f = \dfrac{1}{(z-1)(z-3)}$ with both poles enclosed, the residues $-\tfrac12$ and $+\tfrac12$ add to $0$: enclosing a pole does not guarantee a nonzero integral.
• No pole inside $\Rightarrow 0$. Shrink $R$ below every pole and the integral is $0$ — Cauchy–Goursat, the same answer Chapter 7 gave.
• On the circle is off-limits. When $R$ equals a pole's modulus the contour passes through a singularity and the integral is undefined — the readout refuses to answer. (For $f = \dfrac{3z-1}{z(z-1)}$, the pole at $z = 0$ sits inside for every $R$ on this slider, so only the crossing at $z = 1$ shows the jump.)

In standard Indian engineering math exams, and likely in MCC201A unless your class notes differ, this chapter wears four recognisable costumes — each one you have now seen worked. The phrasing on the left maps to the worked example on the right:

• "Expand $f(z)$ valid in $\langle$region$\rangle$" → worked example 3B. Write the inequality down first; it decides the series. (Practice: P1, P2.)
• "Discuss / state the nature of the singularities of $f$" → worked example 3C. Factor the numerator before trusting the denominator. (Practice: P3.)
• "Evaluate $\oint_C f\,dz$ using Cauchy's residue theorem" → worked example 3E. Step 0, residues of the enclosed poles, times $2\pi i$. (Practice: P5.)
• "Evaluate $\int_0^{2\pi} \dots\,d\theta$" → worked example 3F. Substitute $z = e^{i\theta}$ and run the template. (Practice: P6.)

1. Step 0 — locate. Find every singularity and mark which lie inside $C$.
2. No singularity inside? The integral is $0$ (Cauchy–Goursat). Stop.
3. Classify each enclosed pole. Simple → $\lim (z-z_0)f$ or $p(z_0)/q'(z_0)$. Order $m$ → the derivative formula. Essential → read $a_{-1}$ off the series.
4. Add the enclosed residues, multiply by $2\pi i$.

You may see these two names on a paper; this course does not drill them, but they should not be alien. Both count zeros and poles by integrating, and both build directly on the residue idea:

• The argument principle: if $f$ is analytic inside and on a positively oriented simple closed contour $C$ except for poles, and has no zero on $C$, then $\dfrac{1}{2\pi i}\displaystyle\oint_C \dfrac{f'(z)}{f(z)}\,dz = Z - P$, the number of zeros minus the number of poles inside $C$ (each counted with its order).
• Rouché's theorem: if $f$ and $g$ are analytic inside and on $C$ and $|g(z)| < |f(z)|$ everywhere on $C$, then $f$ and $f + g$ have the same number of zeros inside $C$ (counted with order). It is the standard tool for "how many roots lie in this disk."

Recognition only — no proof, no worked evaluation here. If your class notes assign them, treat this as the vocabulary and the statements, and come back for the method.

Re-do two of P1–P6 from a blank page tomorrow — not re-read, re-do. Expanding in a region and locating-then-residue are muscle memory worth more than another hour of reading now.

This closes Unit 1. The arc ran from "what is a complex number" to "evaluate real integrals that ordinary methods make genuinely painful" — analyticity, the Cauchy–Riemann test, conformal and Möbius maps, contour integration, Cauchy's formula, and now series and residues. What comes next is Unit 2: double integrals — a genuine change of subject, back to real multivariable calculus. Take the win here first.