Taylor and Laurent Series

From Local Taylor Expansions to Series on Annuli

The Identity Theorem relied on the local Taylor expansion of a holomorphic function: its coefficients determine whether a zero is isolated or whether the function vanishes near that point. Here we examine how those expansions arise from Cauchy’s integral formula and how the same idea produces a more general expansion when the domain is an annulus rather than a disk.

A Taylor series uses nonnegative powers of $z-a$ and describes a function near the center $a$. A Laurent series also permits negative powers. Those extra terms allow an expansion on a region that excludes the center. The domain matters: even one function can have different series around the same center on different annuli.

Taylor Series from Cauchy’s Integral Formula

Suppose $f$ is holomorphic on the disk $D(a,R)=\{z:|z-a|<R\}$. Choose a radius $r$ with $0<r<R$. Cauchy’s Integral Formula applies on the circle $|\zeta-a|=r$. For $|z-a|<r$, its kernel has the geometric expansion

\frac{1}{\zeta-z} =\frac{1}{\zeta-a}\frac{1}{1-\frac{z-a}{\zeta-a}} =\sum_{n=0}^{\infty}\frac{(z-a)^n}{(\zeta-a)^{n+1}}.

The geometric series converges uniformly as $\zeta$ runs over this circle, provided $z$ stays in a smaller closed disk $|z-a|\leq s<r$. Thus it can be integrated term by term in Cauchy’s Integral Formula. This gives coefficients expressed as integrals around the circle.

Theorem (Taylor Expansion from Cauchy’s Integral Formula): Let $f$ be holomorphic on $D(a,R)$, where $R>0$. For each $n\geq0$, set $c_n=f^{(n)}(a)/n!$. Then, for every $|z-a|<R$, $f$ is represented by the Taylor series $\sum_{n=0}^{\infty}c_n(z-a)^n$. For every $r$ with $0<r<R$, the coefficients also satisfy $$ c_n=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta. $$

Proof. First fix $r$ with $0<r<R$, and let $|z-a|<r$. By Cauchy’s Integral Formula,

f(z)=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-z}\,d\zeta.

For fixed $z$, the geometric expansion of $1/(\zeta-z)$ converges uniformly on the circle because $|z-a|/|\zeta-a|=|z-a|/r<1$. Substitution into the integral and term-by-term integration therefore give

f(z)=\sum_{n=0}^{\infty} \left(\frac{1}{2\pi i}\int_{|\zeta-a|=r} \frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta\right)(z-a)^n.

The power-series coefficients in this representation must be the Taylor coefficients, by the Uniqueness of Power Series Coefficients established earlier in the course. The Differentiation of a Power Series theorem identifies those coefficients as $f^{(n)}(a)/n!$. This proves the formula for points within radius $r$. Since $r$ can be chosen larger than any specified $|z-a|<R$, the representation holds throughout $D(a,R)$. The same uniqueness result shows that the integral formula gives the same coefficient for every permitted radius $r$. $\square$

The theorem provides more than a formal expansion: every point of the disk lies within the region of convergence. It also gives a useful estimate for how quickly Taylor polynomials approximate $f$. If $M_r=\max_{|\zeta-a|=r}|f(\zeta)|$, then for $|z-a|=s<r$, the geometric-series remainder in the integral formula yields

\left|f(z)-\sum_{n=0}^{N}c_n(z-a)^n\right| \leq \frac{M_r r}{r-s}\left(\frac{s}{r}\right)^{N+1}.

Indeed, the omitted part of the kernel is bounded by $(s/r)^{N+1}/(r-s)$ on the circle, whose length is $2\pi r$. The displayed estimate follows after multiplying by $M_r/(2\pi)$. Since $s/r<1$, the bound tends to zero as $N$ increases.

Worked Example: Expanding a Function About a Shifted Center

Let $f(z)=1/(3-z)$ and expand about $a=1$. Write $z=1+(z-1)$. Then

\frac{1}{3-z} =\frac{1}{2-(z-1)} =\frac{1}{2}\frac{1}{1-\frac{z-1}{2}} =\sum_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+1}}, \qquad |z-1|<2.

The last equality is the geometric series, valid when $|(z-1)/2|<1$. For example, at $z=0$, the series gives $\sum_{n=0}^{\infty}(-1)^n/2^{n+1}=1/3$, agreeing with $f(0)=1/3$. The restriction on the center-relative distance is essential: the series is asserted on the disk of radius $2$, not at its boundary.

Laurent Series on an Annulus

Now let $f$ be holomorphic on an annulus centered at $a$, with inner radius $r_0$ and outer radius $R$, where $0\leq r_0<R\leq\infty$. Thus $f$ is defined whenever $r_0<|z-a|<R$. Choose two circles in the annulus, with radii $\rho$ and $\sigma$, so that

r_0<\rho<|z-a|<\sigma<R.

Cauchy’s theorem applied to the region between the two circles, with a small disk around $z$ removed, gives the annular form of Cauchy’s formula:

f(z)=\frac{1}{2\pi i}\int_{|\zeta-a|=\sigma} \frac{f(\zeta)}{\zeta-z}\,d\zeta -\frac{1}{2\pi i}\int_{|\zeta-a|=\rho} \frac{f(\zeta)}{\zeta-z}\,d\zeta.

Here both circles are traversed counterclockwise. The minus sign reflects the clockwise orientation of the inner boundary of the annular region. Expanding the outer integral uses $|z-a|<\sigma$, while expanding the inner integral uses $\rho<|z-a|$. These two geometric expansions produce, respectively, nonnegative and negative powers.

Theorem (Laurent Expansion on an Annulus): If $f$ is holomorphic on $r_0<|z-a|<R$, then it has a Laurent expansion $$ f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n $$ throughout that annulus. The series converges uniformly on every compact subannulus. For any $\tau$ with $r_0<\tau<R$, its coefficients are $$ c_n=\frac{1}{2\pi i}\int_{|\zeta-a|=\tau} \frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta, \qquad n\in\mathbb{Z}. $$ The coefficients are unique.

Proof. Fix $z$ in the annulus and choose $\rho,\sigma$ as above. On the outer circle, the geometric expansion is

\frac{1}{\zeta-z} =\sum_{n=0}^{\infty} \frac{(z-a)^n}{(\zeta-a)^{n+1}}, \qquad |\zeta-a|=\sigma.

On the inner circle, the appropriate expansion is

\frac{1}{\zeta-z} =-\sum_{k=0}^{\infty}\frac{(\zeta-a)^k}{(z-a)^{k+1}}, \qquad |\zeta-a|=\rho.

Each series converges uniformly on its circle. Substitute them into the annular formula and integrate term by term. The outer integral gives the terms with $n\geq0$, with coefficients

c_n=\frac{1}{2\pi i}\int_{|\zeta-a|=\sigma} \frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta \qquad (n\geq0).

The negative of the inner integral gives the terms $(z-a)^{-k-1}$, with coefficients

c_{-k-1}=\frac{1}{2\pi i}\int_{|\zeta-a|=\rho} f(\zeta)(\zeta-a)^k\,d\zeta \qquad (k\geq0).

These are precisely the stated coefficient integrals for the corresponding integer indices. The integral defining any $c_n$ is unchanged when its circle is moved to another radius within the annulus: the integrand $f(\zeta)/(\zeta-a)^{n+1}$ is holomorphic between the circles, so Cauchy’s theorem makes the difference of the two circle integrals zero.

To check convergence locally uniformly, take any compact subannulus $K$ within the given annulus. Choose $\rho$ smaller than every value of $|z-a|$ on $K$ and $\sigma$ larger than every such value, with both radii still in the annulus. On $K$, the ratios in the two geometric expansions are bounded above by constants strictly less than $1$. Their tails therefore converge uniformly, and so does the Laurent series obtained by integration.

Finally, suppose a Laurent series represents $f$ on the annulus. On any circle $|z-a|=\tau$ within it, the series converges uniformly. Integrating term by term against $(z-a)^{-m-1}$ isolates $2\pi i$ times the coefficient of $(z-a)^m$: the integral of $(z-a)^j(z-a)^{-m-1}$ is zero when $j\neq m$, and is $2\pi i$ when $j=m$. Thus every coefficient must equal the stated contour integral. This proves uniqueness and completes the proof. $\square$

Worked Example: Two Laurent Expansions on Different Annuli

Consider $f(z)=1/(z(z-2))$, centered at $0$. Its domain excludes $0$ and $2$, so the annuli $0<|z|<2$ and $|z|>2$ are both available. Partial fractions give

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

For $0<|z|<2$, expand the second term in nonnegative powers:

\frac{1}{2(z-2)} =-\frac{1}{4}\frac{1}{1-z/2} =-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+2}}.

Consequently,

\frac{1}{z(z-2)} =-\frac{1}{2z}-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+2}}, \qquad 0<|z|<2.

For $|z|>2$, instead expand in negative powers:

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

The $n=0$ term in this last sum is $1/(2z)$, which cancels the term $-1/(2z)$ in the partial-fraction formula. Therefore

\frac{1}{z(z-2)} =\sum_{n=1}^{\infty}\frac{2^{n-1}}{z^{n+1}}, \qquad |z|>2.

The two expansions do not conflict. They represent the same function on different annuli, and each geometric series has its own convergence condition. At $z=1$, the first expansion gives $-1/2-\sum_{n=0}^{\infty}1/2^{n+2}=-1/2-1/2=-1$, which agrees with $1/(1(1-2))=-1$.

Reading the Expansion Region Correctly

For Taylor series, the allowed terms have powers $0,1,2,\ldots$, and the expansion is centered on a disk. A Laurent series allows powers $\ldots,-2,-1,0,1,2,\ldots$, making it suitable for an annulus that may exclude its center. The negative-power terms are not optional notation: their presence is what permits the series to represent functions that are not defined at $a$.

A useful check is to ask which geometric series is being used. Rewriting a denominator in terms of $(z-a)/(\zeta-a)$ requires the first quantity to have smaller modulus; rewriting it in terms of $(\zeta-a)/(z-a)$ requires the reverse. Confusing these conditions can lead to a formally plausible expansion that converges in the wrong region.

The coefficient integral also explains why Laurent coefficients are determined by the function on the annulus, not by an arbitrary choice of circle within it. The next tutorial will examine what the behavior of the negative-power terms can tell us about excluded points.

Check Your Understanding

Use the integral formulas and geometric expansions to answer the following questions.

Why can the kernel in Cauchy’s Integral Formula be expanded uniformly on a circle when the point $z$ lies strictly inside that circle?
For a Laurent expansion on an annulus, which circle integral produces the nonnegative powers, and which produces the negative powers?
Expand $1/(4-z)$ about $a=0$ and state the disk on which the geometric expansion is valid.
Why can the same function have different Laurent expansions on two different annuli centered at the same point?
How does integrating a Laurent series around a circle isolate one specified coefficient?

What You'll Learn