Laurent Series: Annuli
students, imagine trying to describe a complex function not just near one point, but in a whole ring-shaped region around that point. 🌟 That ring is called an annulus, and it is one of the most important settings for Laurent series. In this lesson, you will learn what an annulus is, why it matters, and how it helps us expand complex functions when ordinary power series are not enough.
What is an annulus?
In complex analysis, an annulus is the set of points between two circles with the same center. If the center is $z_0$, an annulus can be written as
$$r<|z-z_0|<R$$
where $0\le r<R\le \infty$.
This means the points are at a distance greater than $r$ and less than $R$ from the center $z_0$. The set looks like a ring or a donut-shaped region. 🍩
For example, if we take
$$1<|z|<3,$$
then the annulus is the region between the circles of radius $1$ and $3$ centered at the origin. The point $z=0$ is not included, and neither are the boundary circles $|z|=1$ and $|z|=3$.
Why does this matter? Because many complex functions cannot be written as a single power series on a disk around a point if there is a singularity inside that disk. But they may still be expandable on an annulus around the singularity.
Why annuli matter in Laurent series
A Laurent series is like a power series with extra terms that allow negative powers of $z-z_0$. Around a point $z_0$, it has the form
$$\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n.$$
This is useful when a function is analytic in a ring-shaped region rather than in a full disk.
A key idea is this: a Laurent series does not usually converge everywhere. It converges on an annulus centered at $z_0$.
For example, suppose a function has one singularity at $z_0=0$. Then it may be analytic for all $z$ such that
$$1<|z|<4.$$
On that annulus, the function might have a Laurent series that includes both positive and negative powers. The negative powers are needed because the function may behave badly near the center, but still be well-behaved in the ring region itself.
Inner and outer radii
Every annulus has two radii:
- the inner radius $r$
- the outer radius $R$
The annulus is
$$r<|z-z_0|<R.$$
If $r=0$, the region becomes a punctured disk:
$$0<|z-z_0|<R.$$
If $R=\infty$, the region extends outward forever:
$$r<|z-z_0|<\infty.$$
The exact annulus of convergence for a Laurent series depends on where the function fails to be analytic. The series converges in the largest annulus centered at $z_0$ that avoids singularities.
Connection between singularities and annuli
To understand Laurent series, students, it helps to think about singularities. A singularity is a point where a function is not analytic. If the nearest singularity is too close, a power series centered at $z_0$ may fail on a disk. But a Laurent series can still work on a surrounding annulus.
Suppose a function is analytic on the region
$$1<|z|<2.$$
Then any Laurent series about $z_0=0$ that represents the function must converge only in that annulus. It cannot usually converge at $|z|=1$ or $|z|=2$ if those are boundaries of analyticity.
This helps explain a central fact: the shape of the domain determines the kind of series we use. A disk gives a Taylor series. An annulus gives a Laurent series. 📘
Example 1: Expanding a simple function on an annulus
Consider
$$f(z)=\frac{1}{z(1-z)}.$$
This function has singularities at $z=0$ and $z=1$. If we want a Laurent series centered at $z_0=0$, we must choose a region where the function is analytic except at the center itself. One valid annulus is
$$0<|z|<1.$$
Now rewrite the function:
$$f(z)=\frac{1}{z}\cdot\frac{1}{1-z}.$$
For $|z|<1$, we can use the geometric series
$$\frac{1}{1-z}=\sum_{n=0}^{\infty} z^n.$$
So
$$f(z)=\frac{1}{z}\sum_{n=0}^{\infty} z^n=\sum_{n=0}^{\infty} z^{n-1}.$$
This becomes
$$f(z)=\frac{1}{z}+1+z+z^2+\cdots.$$
This is a Laurent series because it contains the negative power $z^{-1}$. The series converges on the annulus
$$0<|z|<1.$$
Notice the region is not a full disk because the function is not analytic at $z=0$.
Example 2: Different annuli give different expansions
Now consider
$$g(z)=\frac{1}{z(1-z)}$$
again, but centered at $z_0=0$ and on the region
$$|z|>1.$$
This is also a valid annulus, since the function is analytic for large $|z|$ away from $z=0$ and $z=1$.
We rewrite the function in a different way:
$$g(z)=\frac{1}{z(1-z)}=-\frac{1}{z}\cdot\frac{1}{z-1}.$$
For $|z|>1$,
$$\frac{1}{z-1}=\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}.$$
Using the geometric series in $\frac{1}{z}$,
$$\frac{1}{1-\frac{1}{z}}=\sum_{n=0}^{\infty} z^{-n}$$
for $|z|>1$. Thus
$$g(z)=-\frac{1}{z^2}\sum_{n=0}^{\infty} z^{-n}=-\sum_{n=0}^{\infty} z^{-n-2}.$$
So on the annulus
$$|z|>1,$$
the Laurent expansion is a different one:
$$g(z)=-\frac{1}{z^2}-\frac{1}{z^3}-\frac{1}{z^4}-\cdots.$$
This example shows an important idea: the same function can have different Laurent series in different annuli. The region matters just as much as the function.
How to think about annuli in practice
When solving problems, students, it helps to follow these steps:
- Find the center $z_0$.
- Locate singularities of the function.
- Choose a region where the function is analytic.
- Describe the annulus using inequalities like $r<|z-z_0|<R$.
- Expand the function using algebraic rewriting and known series.
The most important part is deciding which inequalities define the valid region. For example, if a denominator contains $1-z$, then one expansion may require $|z|<1$, while another may require $|z|>1$. These are different annuli and lead to different series.
Annuli and the broader Laurent series picture
Annuli are not just a side topic. They are the natural home of Laurent series. A Laurent series exists because analytic functions can be represented in regions that surround singularities without including them.
The full Laurent series has two parts:
$$\sum_{n=0}^{\infty} a_n (z-z_0)^n$$
and
$$\sum_{n=1}^{\infty} a_{-n}(z-z_0)^{-n}.$$
The first part looks like a Taylor series. The second part is called the principal part. The annulus tells us where both parts together converge.
If a function is analytic in a disk, then the negative-power terms are all zero, and the Laurent series becomes a Taylor series. If there is a singularity at the center, then the negative-power terms may appear. That is exactly why annuli are so useful: they allow us to describe functions near singularities in a precise and manageable way.
Conclusion
Annuli are ring-shaped regions described by inequalities like $r<|z-z_0|<R$. In complex analysis, they are essential because Laurent series typically converge on annuli rather than on full disks. The size and location of the annulus depend on the singularities of the function, and different annuli can give different series expansions for the same function. By understanding annuli, you gain the foundation needed to work with Laurent series, principal parts, and expansion methods in complex analysis. 🌟
Study Notes
- An annulus is the region $r<|z-z_0|<R$ between two circles with the same center $z_0$.
- Annuli are shaped like rings and may be finite, punctured, or extend outward to infinity.
- Laurent series often converge on annuli, not on full disks.
- A Laurent series has terms with both positive and negative powers:
$$\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n.$$
- The principal part is the negative-power part of the Laurent series.
- Singularities determine where an annulus of convergence can exist.
- The same function can have different Laurent expansions on different annuli.
- To find a Laurent expansion, identify the center, singularities, valid region, and then rewrite the function for a known series expansion.
- If the negative-power terms vanish, the Laurent series reduces to a Taylor series.
- Annuli connect directly to the broader study of Laurent series and function behavior near singularities.