Taylor series, Laurent series and isolated singularities

1. Convergence is part of the answer

A power series about $a$ is $\sum_{n\ge0}c_n(z-a)^n$. There is a radius $R\in[0,\infty]$ such that it converges absolutely for $|z-a|<R$ and diverges for $|z-a|>R$; boundary points need separate tests. The root test gives

$$R^{-1}=\limsup_{n\to\infty}|c_n|^{1/n},$$

with $1/0=\infty$ and $1/\infty=0$. Indeed, for $|z-a|$ below the reciprocal limsup, the roots of the terms are eventually less than a fixed number below one; above it, a subsequence of terms fails to tend to zero. On every smaller closed disk the series and all termwise derivatives converge uniformly. For the derivatives this follows because powers of $n$ are dominated by a slightly larger geometric bound. Therefore the sum is holomorphic and $c_n=f^{(n)}(a)/n!$.

C03 proves that every holomorphic function has this Taylor expansion inside any disk free of obstructions. If the function is holomorphic on $D(a,R)$, its Taylor series represents it throughout that disk. A singularity at a boundary point prevents the series from having a larger convergence disk: a larger disk would define a holomorphic continuation agreeing with the original function nearby. Not every chosen domain boundary is a true singularity, however; restricting an entire function to a small disk does not reduce its Taylor radius.

Worked example. For $f(z)=1/(1-z)$ at zero, $\sum_{n\ge0}z^n$ converges precisely for $|z|<1$. At $z=1$ it is undefined; at other points with $|z|=1$ its terms do not tend to zero. For $e^z$, the factorial coefficients give radius infinity. One must not infer a common radius from the fact that both functions are holomorphic near zero.

2. Why annuli require negative powers

Suppose $f$ is holomorphic on the annulus

$$A=\{z:r<|z-a|<R\},\qquad0\le r<R\le\infty.$$

Then it has a unique Laurent expansion

$$f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n,$$

converging absolutely at each point of $A$ and uniformly on compact subannuli. For any radius $\rho$ with $r<\rho<R$,

$$c_n=\frac1{2\pi i}\int_{|\zeta-a|=\rho}\frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta,\qquad n\in\mathbb Z.$$

Circles here are positively oriented. Cauchy theorem on an annular region, or the circle deformation of C03, makes the value independent of $\rho$.

Derivation. Fix $r<r_1<|z-a|<r_2<R$. The boundary formula for the region between these circles is

$$f(z)=\frac1{2\pi i}\int_{|\zeta-a|=r_2}\frac{f(\zeta)}{\zeta-z}\,d\zeta -\frac1{2\pi i}\int_{|\zeta-a|=r_1}\frac{f(\zeta)}{\zeta-z}\,d\zeta.$$

To justify it, remove a small disk about $z$, cut the doubly holed region along nonintersecting arcs, and apply the local-primitive cancellation argument of C02. The two sides of each cut cancel; the small circle contributes $2\pi if(z)$ by continuity as in C03. Thus no holomorphy inside the inner hole has been assumed.

In the first integral expand $1/(\zeta-z)$ in nonnegative powers of $(z-a)/(\zeta-a)$. In the second expand it as

$$-\frac1{z-a}\sum_{k\ge0}\left(\frac{\zeta-a}{z-a}\right)^k.$$

Both are uniformly convergent geometric series on the respective contours, giving positive and negative powers with the stated coefficients. For a compact subannulus, choose $r_1,r_2$ strictly beyond its boundaries; the same bounds give uniform absolute convergence. Finally integrating a uniformly convergent Laurent series termwise against $(z-a)^{-m-1}$ selects exactly its $m$th coefficient by C02's monomial circle integral. This proves uniqueness. ∎

The series is the sum of two ordinary power series, one in $z-a$ and one in $(z-a)^{-1}$; it is not an arbitrary rearrangement of a conditionally convergent doubly infinite sum.

Worked example: the same function, two annuli. Since

$$\frac1{z(z-1)}=-\frac1z+\frac1{z-1},$$

for $0<|z|<1$ we obtain

$$-z^{-1}-\sum_{n=0}^{\infty}z^n.$$

For $|z|>1$, instead write $1/[z(z-1)]=z^{-2}/(1-z^{-1})$, giving $\sum_{n=0}^{\infty}z^{-n-2}$. These different coefficient lists do not violate uniqueness: their annuli are different, separated by the singularity at $1$.

3. Classifying an isolated singularity

The point $a$ is an isolated singularity when $f$ is holomorphic on some punctured disk $0<|z-a|<R$, without a value or holomorphic definition at $a$. Its Laurent series classifies the obstruction:

• Removable: all negative coefficients vanish. The Taylor part defines a unique holomorphic extension at $a$.
• Pole of order $m\ge1$: $c_{-m}\ne0$ and all $c_n=0$ for $n<-m$. Equivalently $f(z)=g(z)/(z-a)^m$ near $a$, with $g$ holomorphic and $g(a)\ne0$.
• Essential: infinitely many negative coefficients are nonzero.

This classification concerns isolated singularities only. A branch point of $\operatorname{Log}z$ at zero is not an isolated singularity of a single-valued holomorphic function on a full punctured disk, so this trichotomy must not be applied to it without changing the setting.

Removable-singularity criterion. An isolated singularity is removable exactly when $f$ is bounded in some punctured neighbourhood. One direction follows from continuity of the extension. Conversely, if $|f|\le M$ on sufficiently small circles, the coefficient estimate gives, for every integer $k\ge1$,

$$|c_{-k}|\le M\rho^k.$$

Send $\rho\downarrow0$; all negative coefficients vanish. It also follows that if $\lim_{z\to a}f(z)$ exists finitely, that limit is the extension value.

For a pole, $|f(z)|\to\infty$ because $g(a)\ne0$. Conversely, if $|f(z)|\to\infty$, then $1/f$ is holomorphic and nonzero on a small punctured disk, tends to zero, and extends holomorphically with a zero at $a$. This extension is not identically zero, so zero isolation gives a finite positive zero order. Taking its reciprocal shows $f$ has a pole of that order.

Worked examples. The quotient $\sin z/z$ has a removable singularity at zero with extension value one. The function $(e^z-1)/z^3=z^{-2}+\tfrac12z^{-1}+\tfrac16+\cdots$ has a pole of order two, not three: numerator zeros cancel part of the apparent denominator order. The function $e^{1/z}=\sum_{n\ge0}z^{-n}/n!$ has an essential singularity at zero.

4. Essential singularities cannot omit an open disk locally

Casorati–Weierstrass theorem. If $a$ is an isolated essential singularity, then in every punctured neighbourhood of $a$, the values of $f$ are dense in $\mathbb C$.

Proof. If a punctured neighbourhood omitted an open disk about $w$, there would be a $\delta>0$ with $|f-w|\ge\delta$ there. The function $h=1/(f-w)$ would be holomorphic, bounded and nonzero on that punctured neighbourhood, hence extend holomorphically at $a$. If $h(a)\ne0$, the identity $f=w+1/h$ would give a removable extension. If $h(a)=0$, zero isolation gives a zero of finite order, because $h$ is not identically zero; then $f$ would have a pole. Both contradict essentiality. ∎

Density is not surjectivity and does not say every value is attained. For instance $e^{1/z}$ never equals zero, but its values can approach zero. Stronger value-distribution results are not needed here.

5. Exercises

P01 — Foundation. Expand $1/(2-z)$ at zero and state its radius of convergence.

P02 — Core. Find Laurent series of $1/[z(z-2)]$ on $0<|z|<2$ and on $|z|>2$.

P03 — Core. Classify the singularity at zero of $(1-\cos z)/z^4$, giving the first three nonzero Laurent terms.

P04 — Core. Classify $z\sin(1/z)$ at zero and explain why a finite limit along one ray would not settle the question.

P05 — Proof. Suppose $g$ and $h$ are holomorphic on a neighbourhood of $a$, $g$ has a zero of finite order $m\ge1$ at $a$, and $h(a)\ne0$. Determine the singularity order of $h/g$ at $a$.

P06 — Diagnosis. Why may the Laurent expansion on $|z|>1$ in §2 have no $z^{-1}$ term even though the same function has a simple pole at zero?

P07 — Proof. Prove uniqueness of a holomorphic extension across a removable isolated singularity without using the identity theorem.

P08 — Extension. Show that a function holomorphic on a punctured disk and satisfying $|f(z)|\le C|z-a|^{-\alpha}$, for a fixed real $\alpha\ge0$, has a pole of order at most $\lfloor\alpha\rfloor$ or a removable singularity.

P09 — Core. The Taylor coefficients of $f$ at zero are $c_n=3^{-n}$ for even $n$ and $0$ for odd $n$. Find the sum and its maximal convergence disk.

6. Full solutions

P01. $1/(2-z)=\tfrac12(1-z/2)^{-1}=\sum_{n\ge0}z^n/2^{n+1}$ for $|z|<2$. The pole at two prevents a larger convergence disk, so the radius is two.

P02. Partial fractions give $-1/(2z)+1/[2(z-2)]$. On the inner annulus this is $-1/(2z)-\sum_{n\ge0}z^n/2^{n+2}$. On the outer annulus use $z^{-2}(1-2/z)^{-1}=\sum_{n\ge0}2^nz^{-n-2}$. Each geometric ratio has modulus less than one on its stated annulus.

P03. Since $1-\cos z=z^2/2-z^4/24+z^6/720-\cdots$, the quotient is $\tfrac12z^{-2}-1/24+z^2/720-\cdots$. Its pole order is two and its $z^{-1}$ coefficient is zero.

P04. Expanding gives $z\sin(1/z)=1-1/(3!z^2)+1/(5!z^4)-\cdots$. Infinitely many negative powers make the isolated singularity essential. Behaviour on one ray cannot establish boundedness on a punctured disk, nor existence of a full complex limit.

P05. Write $g=(z-a)^mg_0$ with $g_0(a)\ne0$. Then $h/g=(h/g_0)/(z-a)^m$, with holomorphic numerator nonzero at $a$. The pole order is exactly $m$.

P06. A singularity classification at zero uses a punctured neighbourhood of zero with no other singularities, here $0<|z|<1$. The outer annulus excludes both zero and one and does not approach zero. Coefficients are unique on a specified annulus, not across singular circles. Its circle integrals combine the effects of both enclosed poles, which can cancel in the $z^{-1}$ coefficient.

P07. Any holomorphic extension is continuous at $a$, so its value there must equal the limit of the already fixed punctured-disk values. Two such extensions agree off $a$ by definition and at $a$ by continuity. No global uniqueness theorem is necessary.

P08. For integer $k>\alpha$, the Laurent estimate gives $|c_{-k}|\le C\rho^{k-\alpha}\to0$. Therefore all negative coefficients beyond $-\lfloor\alpha\rfloor$ vanish. The remaining finite principal part gives the stated pole bound; if it is empty the singularity is removable. In particular a bound with $0\le\alpha<1$ forces removability.

P09. The series is $\sum_{k\ge0}(z^2/9)^k=1/(1-z^2/9)$ for $|z|<3$. Its singularities at $\pm3$ make radius three maximal. The zeros among the odd coefficients do not increase that radius.

7. Summary

A Taylor series belongs to a disk; a Laurent series belongs to an annulus. Uniform contour expansion gives both existence and coefficient uniqueness. At an isolated singularity, a missing, finite or infinite negative-power part means removable, pole or essential. The coefficient of $(z-a)^{-1}$ is the residue, which the next chapter turns into a general integration method.