The Central Idea
Near a point \(z_0\), an analytic function has a Taylor expansion
\[ \boxed{ f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n. } \]Near an isolated singularity, the more general Laurent expansion is
\[ \boxed{ f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n. } \]The negative-power terms determine the singularity type.
Power Series Review
A power series centered at \(z_0\) has the form
\[ \sum_{n=0}^{\infty}a_n(z-z_0)^n. \]It converges inside some disk
\[ |z-z_0|The number \(R\) is the radius of convergence.
Taylor Series for Analytic Functions
If \(f\) is analytic near \(z_0\), then
\[ \boxed{ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n. } \]The coefficients are
\[ \boxed{ a_n=\frac{f^{(n)}(z_0)}{n!}. } \]Coefficients from Cauchy's Integral Formula
The Taylor coefficients can also be written as contour integrals:
\[ \boxed{ a_n = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta. } \]This directly connects local series coefficients with boundary data.
Example: Exponential Series
The exponential function has
\[ \boxed{ e^z = 1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots. } \]This series converges for every complex \(z\).
Example: Geometric Series
For
\[ |z|<1, \]we have
\[ \boxed{ \frac1{1-z} = 1+z+z^2+z^3+\cdots. } \]The nearest singularity to the center \(0\) is at \(z=1\), so the radius of convergence is \(1\).
Using the Geometric Series as a Template
Many complex series expansions come from rewriting an expression into the form
\[ \frac1{1-w} \]or
\[ \frac1{1+w}. \]Then use
\[ \frac1{1-w} = \sum_{n=0}^{\infty}w^n \]whenever
\[ |w|<1. \]Example: Expand About a Nonzero Center
Expand
\[ \frac1{z} \]about
\[ z_0=1. \]Write
\[ z=1+(z-1). \]Then
\[ \frac1z = \frac1{1+(z-1)}. \]Using
\[ \frac1{1+w} = \sum_{n=0}^{\infty}(-1)^n w^n, \]we obtain
\[ \boxed{ \frac1z = \sum_{n=0}^{\infty} (-1)^n(z-1)^n, \qquad |z-1|<1. } \]Why Taylor Series Are Not Enough
A Taylor series uses only nonnegative powers:
\[ (z-z_0)^0, (z-z_0)^1, (z-z_0)^2, \ldots \]But a function such as
\[ \frac1{z-z_0} \]cannot be represented by a Taylor series at \(z_0\), because it is not analytic there.
To describe such functions locally, we allow negative powers.
Laurent Series
A Laurent series centered at \(z_0\) has the form
\[ \boxed{ \sum_{n=-\infty}^{\infty} a_n(z-z_0)^n. } \]It may be separated into two parts:
\[ \sum_{n=0}^{\infty}a_n(z-z_0)^n \]and
\[ \sum_{n=1}^{\infty} \frac{a_{-n}}{(z-z_0)^n}. \]The Principal Part
The negative-power portion
\[ \boxed{ \sum_{n=1}^{\infty} \frac{a_{-n}}{(z-z_0)^n} } \]is called the principal part of the Laurent series.
The principal part carries the information about the singularity at \(z_0\).
The Regular Part
The nonnegative-power portion
\[ \boxed{ \sum_{n=0}^{\infty}a_n(z-z_0)^n } \]behaves like an ordinary Taylor series and is analytic at \(z_0\).
Laurent Series Converge on Annuli
Unlike a Taylor series, a Laurent series may converge on a ring-shaped region:
\[ \boxed{ r<|z-z_0|Different annuli can produce different Laurent expansions of the same function.
Example: Different Expansions in Different Regions
Consider
\[ f(z)=\frac1{z(1-z)}. \]The singularities are at
\[ z=0 \]and
\[ z=1. \]Around \(z_0=0\), the natural regions are
\[ 0<|z|<1 \]and
\[ |z|>1. \]Expansion for \(0<|z|<1\)
Write
\[ \frac1{z(1-z)} = \frac1z\cdot\frac1{1-z}. \]For \(|z|<1\),
\[ \frac1{1-z} = \sum_{n=0}^{\infty}z^n. \]Therefore,
\[ \boxed{ \frac1{z(1-z)} = \frac1z+1+z+z^2+\cdots, \qquad 0<|z|<1. } \]Expansion for \(|z|>1\)
Rewrite
\[ \frac1{1-z} = -\frac1z\cdot\frac1{1-\frac1z}. \]For \(|z|>1\),
\[ \frac1{1-\frac1z} = \sum_{n=0}^{\infty}\frac1{z^n}. \]Therefore,
\[ \frac1{z(1-z)} = -\frac1{z^2} \left( 1+\frac1z+\frac1{z^2}+\cdots \right). \]Hence,
\[ \boxed{ \frac1{z(1-z)} = -\frac1{z^2} -\frac1{z^3} -\frac1{z^4} -\cdots, \qquad |z|>1. } \]Isolated Singularities
A point \(z_0\) is an isolated singularity of \(f\) if \(f\) is not analytic at \(z_0\) but is analytic throughout some punctured disk
\[ \boxed{ 0<|z-z_0|The Three Main Types of Isolated Singularities
An isolated singularity is classified as:
- a removable singularity;
- a pole;
- an essential singularity.
The Laurent principal part tells us which type occurs.
Removable Singularities
A singularity at \(z_0\) is removable when the Laurent series contains no negative powers.
Thus the principal part is zero:
\[ \boxed{ a_{-1}=a_{-2}=a_{-3}=\cdots=0. } \]The function can then be redefined at \(z_0\) to become analytic there.
Example: A Removable Singularity
Consider
\[ f(z)=\frac{\sin z}{z}, \qquad z\neq0. \]Since
\[ \sin z = z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots, \]dividing by \(z\) gives
\[ \frac{\sin z}{z} = 1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots. \]There are no negative powers.
Therefore \(z=0\) is removable, and defining
\[ \boxed{f(0)=1} \]makes the function analytic at the origin.
A Limit Test for Removable Singularities
If
\[ \lim_{z\to z_0}f(z) \]exists and is finite, then the singularity is removable.
Define
\[ f(z_0) = \lim_{z\to z_0}f(z). \]The extended function becomes analytic under the removable-singularity theorem.
Poles
A singularity at \(z_0\) is a pole of order \(m\) when the principal part has finitely many negative-power terms and the most negative power is
\[ \frac{a_{-m}}{(z-z_0)^m}, \qquad a_{-m}\neq0. \]Thus the Laurent series has the form
\[ \boxed{ \frac{a_{-m}}{(z-z_0)^m} +\cdots+ \frac{a_{-1}}{z-z_0} +a_0+a_1(z-z_0)+\cdots. } \]Simple Poles
A pole of order \(1\) is called a simple pole.
Near a simple pole,
\[ f(z) = \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0) +\cdots. \]The coefficient \(a_{-1}\) will later be called the residue.
Example: A Pole of Order Three
Consider
\[ f(z)=\frac{e^z}{z^3}. \]Since
\[ e^z = 1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots, \]dividing by \(z^3\) gives
\[ \frac{e^z}{z^3} = \frac1{z^3} + \frac1{z^2} + \frac1{2z} + \frac1{3!} + \frac{z}{4!} +\cdots. \]The most negative power is \(z^{-3}\).
Therefore,
\[ \boxed{ z=0\text{ is a pole of order }3. } \]A Factor Test for Poles
If
\[ f(z)=\frac{g(z)}{(z-z_0)^m} \]where \(g\) is analytic near \(z_0\) and
\[ g(z_0)\neq0, \]then
\[ \boxed{ z_0\text{ is a pole of order }m. } \]A Limit Test for the Order of a Pole
If
\[ \lim_{z\to z_0} (z-z_0)^m f(z) \]exists and is nonzero and finite, while smaller powers do not remove the singularity, then \(z_0\) is a pole of order \(m\).
Essential Singularities
A singularity at \(z_0\) is essential when the Laurent principal part contains infinitely many nonzero negative-power terms.
Symbolically,
\[ \boxed{ \sum_{n=1}^{\infty} \frac{a_{-n}}{(z-z_0)^n} } \]has infinitely many nonzero coefficients.
Example: \(e^{1/z}\)
Expand:
\[ e^{1/z} = 1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots. \]There are infinitely many negative powers.
Therefore,
\[ \boxed{ z=0\text{ is an essential singularity of }e^{1/z}. } \]Example: \(\sin(1/z)\)
Using the sine series,
\[ \sin\left(\frac1z\right) = \frac1z - \frac1{3!z^3} + \frac1{5!z^5} -\cdots. \]Again, infinitely many negative powers occur.
Therefore,
\[ \boxed{ z=0\text{ is an essential singularity}. } \]Laurent-Series Classification
At an isolated singularity \(z_0\):
- Removable: no negative-power terms.
- Pole: finitely many negative-power terms.
- Essential: infinitely many negative-power terms.
Behavior Near a Pole
If \(z_0\) is a pole, then
\[ |f(z)|\to\infty \]as
\[ z\to z_0. \]Thus poles behave like controlled blow-ups.
Behavior Near an Essential Singularity
Essential singularities behave far more irregularly than poles.
The Casorati–Weierstrass theorem states that near an essential singularity, the values of the function come arbitrarily close to every complex number.
Picard's theorem strengthens this dramatically.
A Preview of Picard's Theorem
Near an essential singularity, a function takes every complex value, with at most one exception, infinitely often in every punctured neighborhood.
This illustrates how different essential singularities are from removable singularities and poles.
Nonisolated Singularities
Not every singularity is isolated.
A point is a nonisolated singularity when every neighborhood contains other singular points.
Laurent-series classification around a single punctured disk does not apply in the same way.
Branch Points
Functions such as
\[ \log z \]and
\[ z^{1/2} \]exhibit branch behavior around certain points.
These are not ordinary isolated singularities of a single-valued analytic function on a punctured disk.
They require branch choices and branch cuts.
Laurent Coefficients
If \(f\) is analytic on an annulus around \(z_0\), then the Laurent coefficients satisfy
\[ \boxed{ a_n = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta, } \]where \(C\) is any positively oriented circle lying in the annulus.
The Coefficient of \((z-z_0)^{-1}\)
In the Laurent series
\[ f(z) = \cdots + \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0+\cdots, \]the coefficient
\[ \boxed{a_{-1}} \]is especially important.
It is called the residue of \(f\) at \(z_0\).
Why the Residue Matters
Around a small positively oriented circle centered at \(z_0\),
\[ \oint_C(z-z_0)^n\,dz \]vanishes for every integer \(n\neq-1\).
Only the term
\[ \frac{a_{-1}}{z-z_0} \]contributes.
Therefore,
\[ \boxed{ \oint_C f(z)\,dz = 2\pi i\,a_{-1}. } \]This is the local mechanism behind the residue theorem.
Example: Read a Residue from a Laurent Series
Consider
\[ \frac{e^z}{z^3} = \frac1{z^3} + \frac1{z^2} + \frac1{2z} + \frac1{3!} +\cdots. \]The coefficient of \(1/z\) is
\[ \boxed{\frac12}. \]Therefore the residue at \(z=0\) is \(1/2\).
Partial Fractions and Laurent Structure
Rational functions are often easiest to analyze by partial fractions.
For example,
\[ \frac1{z(z-1)} = -\frac1z+\frac1{z-1}. \]Each term makes one singularity explicit.
Functions with Multiple Singularities
A function may have several isolated singularities.
Each one is analyzed locally with a Laurent series centered at that point.
Around different centers, the same function can have different local expansions and different principal parts.
The Nearest Singularity Controls a Taylor Radius
If \(f\) is analytic at \(z_0\), its Taylor series can continue until the nearest singularity is encountered.
Thus the radius of convergence is determined by the distance from \(z_0\) to the nearest obstruction to analyticity.
Example: Radius from Singularities
Consider
\[ f(z)=\frac1{z^2+4}. \]The singularities satisfy
\[ z^2+4=0, \]so
\[ z=\pm2i. \]A Taylor series centered at \(0\) therefore has radius
\[ \boxed{R=2}. \]Common Errors
- Using a Taylor series at a singular point: use a Laurent series instead.
- Ignoring the region of convergence: the same function can have different Laurent series on different annuli.
- Calling every blow-up a simple pole: determine the actual order.
- Confusing a removable singularity with a zero: a removable singularity concerns the definition or analyticity of the function at the point.
- Missing infinitely many negative powers: that signals an essential singularity.
- Confusing branch points with isolated singularities: they require separate branch analysis.
A Reliable Procedure for Finding a Laurent Expansion
- Choose the center \(z_0\).
- Locate nearby singularities.
- Identify the annulus of interest.
- Rewrite the function using geometric-series forms when possible.
- Expand each factor in the form valid for that annulus.
- Combine the series carefully.
- State the region of convergence explicitly.
A Reliable Procedure for Classifying an Isolated Singularity
- Confirm that the singularity is isolated.
- Find or infer the Laurent expansion near the point.
- Inspect the principal part.
- If there are no negative powers, the singularity is removable.
- If there are finitely many negative powers, it is a pole.
- If there are infinitely many negative powers, it is essential.
A Reliable Procedure for Finding Pole Order
- Factor the denominator or isolate powers of \(z-z_0\).
- Cancel common factors only when justified.
- Write \[ f(z)=\frac{g(z)}{(z-z_0)^m} \] with \(g(z_0)\neq0\).
- Conclude that the pole has order \(m\).
Worked Example: Classify \(\frac{1-\cos z}{z^2}\) at \(0\)
Use
\[ \cos z = 1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots. \]Then
\[ 1-\cos z = \frac{z^2}{2!} - \frac{z^4}{4!} +\cdots. \]Divide by \(z^2\):
\[ \frac{1-\cos z}{z^2} = \frac12 - \frac{z^2}{4!} +\cdots. \]There are no negative powers.
Therefore,
\[ \boxed{ z=0\text{ is removable}. } \]Worked Example: Classify \(\frac{1}{(z-2)^4}\)
The Laurent series is already visible:
\[ \boxed{ \frac{1}{(z-2)^4}. } \]The principal part contains one term, and the highest negative power is \((z-2)^{-4}\).
Therefore,
\[ \boxed{ z=2\text{ is a pole of order }4. } \]Worked Example: Classify \(e^{1/(z-1)}\)
Expand:
\[ e^{1/(z-1)} = 1 + \frac1{z-1} + \frac1{2!(z-1)^2} + \frac1{3!(z-1)^3} +\cdots. \]Infinitely many negative powers occur.
Therefore,
\[ \boxed{ z=1\text{ is an essential singularity}. } \]Practice Problems
Practice 1
Classify the singularity at \(z=0\) of
\[ \frac{e^z-1}{z}. \]Practice 2
Classify the singularity at \(z=0\) of
\[ \frac1{z^5}. \]Practice 3
Classify the singularity at \(z=0\) of
\[ e^{1/z}. \]Practice 4
Find the Laurent expansion of
\[ \frac1{z(1-z)} \]for \(0<|z|<1\).
Practice 5
Find the Laurent expansion of
\[ \frac1{z(1-z)} \]for \(|z|>1\).
Practice 6
Find the order of the pole of
\[ \frac{z+1}{(z-3)^2} \]at \(z=3\).
Practice 7
Classify the singularity at \(z=0\) of
\[ \frac{\sin z}{z}. \]Practice 8
Find the radius of convergence of the Taylor series about \(0\) for
\[ \frac1{z^2+9}. \]Practice 9
Identify the residue at \(z=0\) from
\[ \frac{e^z}{z^2}. \]Practice 10
What feature of a Laurent series distinguishes an essential singularity?
Practice Answers
Practice 1
Since
\[ e^z-1 = z+\frac{z^2}{2!}+\cdots, \]we get
\[ \frac{e^z-1}{z} = 1+\frac z{2!}+\cdots. \]Thus
\[ \boxed{\text{removable}}. \]Practice 2
\[ \boxed{\text{pole of order }5}. \]Practice 3
\[ \boxed{\text{essential singularity}}. \]Practice 4
\[ \boxed{ \frac1z+1+z+z^2+\cdots, \qquad 0<|z|<1. } \]Practice 5
\[ \boxed{ -\frac1{z^2} -\frac1{z^3} -\frac1{z^4} -\cdots, \qquad |z|>1. } \]Practice 6
The numerator is nonzero at \(z=3\), while the denominator contains \((z-3)^2\).
\[ \boxed{\text{pole of order }2}. \]Practice 7
\[ \boxed{\text{removable}}. \]Practice 8
The singularities are at
\[ z=\pm3i. \]Thus
\[ \boxed{R=3}. \]Practice 9
Expand:
\[ \frac{e^z}{z^2} = \frac1{z^2} + \frac1z + \frac1{2!} +\cdots. \]The coefficient of \(1/z\) is
\[ \boxed{1}. \]Practice 10
The principal part contains infinitely many nonzero negative-power terms.
Checking Strategies
- State the center and region of convergence for every local series.
- Locate nearby singularities before deciding which expansion is valid.
- Use geometric-series templates whenever possible.
- Inspect the principal part to classify an isolated singularity.
- For a suspected pole, factor the function into an analytic nonzero part times a negative power.
- Verify removable singularities by checking whether the finite limit exists.
- Watch specifically for the coefficient of \((z-z_0)^{-1}\), because it will control residues.
Connections to Earlier Mathematics
- Cauchy's integral formula: contour integrals determine series coefficients.
- Power series: Taylor expansions describe analytic points.
- Complex integration: the \(1/(z-z_0)\) Laurent term is the only one that survives around a closed circle.
- Algebra: partial fractions expose rational singularities.
- Topology: annuli reflect the geometry of excluded singular points.
- Residues: the coefficient \(a_{-1}\) becomes the key local invariant for contour integration.
What to Remember
Taylor series use nonnegative powers:
\[ \boxed{ f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n. } \]Laurent series allow both positive and negative powers:
\[ \boxed{ f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n. } \]At an isolated singularity:
- no negative powers \(\Rightarrow\) removable;
- finitely many negative powers \(\Rightarrow\) pole;
- infinitely many negative powers \(\Rightarrow\) essential.
The coefficient
\[ \boxed{a_{-1}} \]is the residue.
The essential structural idea is: the Laurent principal part is a complete local fingerprint of an isolated singularity.
Where We Go Next
Laurent series isolate the coefficient that matters most for contour integration: the coefficient of
\[ \frac1{z-z_0}. \]The next lesson, Residues, develops efficient ways to compute that coefficient and uses the residue theorem to evaluate contour integrals from singularity data rather than direct parametrization.