Understanding the Laurent series of a function is crucial for unlocking a powerful tool in mathematical analysis. It provides a way to represent a function as an infinite sum of complex exponentials, revealing its behavior near specific points in the complex plane. Unlike other series expansions, the Laurent series is particularly adept at handling singularities, allowing for a deeper exploration of functions with complex singularities.
To embark on the journey of determining the Laurent series of a function, we must first define an isolated singularity. An isolated singularity occurs at a point in the complex plane where the function fails to be analytic, but its behavior near that point can be described by a Laurent series. By analyzing the function’s behavior around the singularity, we can identify its order and principal part, which are essential components for constructing the Laurent series.
Furthermore, the coefficients of the Laurent series are determined through a process of contour integration. By integrating the function around a carefully chosen contour that encircles the singularity, we can extract the coefficients of the individual terms in the series. This approach provides a systematic way to represent the function in a form that captures its behavior near both regular and singular points, offering a comprehensive understanding of its analytical properties.
Identifying Isolated Singularities
Before we can determine the Laurent series of a function, we need to locate its isolated singularities. These are points where the function is not defined or has a removable discontinuity. To identify isolated singularities, we can examine the denominator of the function.
- If the denominator has a factor of $(z-a)^n$ with $n>0$, then $z=a$ is an isolated singularity.
- If the denominator has a factor of $(z-a)^n$ with $n<0$, then $z=a$ is not an isolated singularity.
- If the denominator has a factor of $e^{az}-1$ or $e^{az}+1$ with $a\neq 0$, then $z=0$ is an isolated singularity.
Example
Consider the function $f(z)=\frac{1}{z^2-1}$. The denominator can be factored as $(z-1)(z+1)$. Both $z=1$ and $z=-1$ are isolated singularities because the denominator has factors of $(z-1)^1$ and $(z+1)^1$.
Poles of Positive and Negative Order
In complex analysis, a pole of a function is a point in the complex plane where the function is not defined due to an infinite discontinuity. Poles can be of two types: positive order and negative order.
Poles of Positive Order
A pole of positive order occurs when the denominator of a rational function has a factor of the form $(z – a)^n$, where $n$ is a positive integer. The order of the pole is $n$. At a pole of order $n$, the function has the following Laurent series expansion:
“`
f(z) = \frac{a_{-n}}{(z – a)^n} + \frac{a_{-n+1}}{(z – a)^{n-1}} + \cdots + \frac{a_{-1}}{z – a} + a_0 + a_1(z – a) + a_2(z – a)^2 + \cdots
“`
where $a_k$ are complex coefficients.
Example
Consider the function $f(z) = \frac{1}{(z – 2)^3}$. This function has a pole of order 3 at $z = 2$. The Laurent series expansion of $f(z)$ around $z = 2$ is:
“`
f(z) = \frac{1}{(z – 2)^3} + \frac{1}{(z – 2)^2} + \frac{1}{z – 2} + 1 + (z – 2) + (z – 2)^2 + \cdots
“`
Poles of Negative Order
A pole of negative order occurs when the denominator of a rational function has a factor of the form $(z – a)^{-n}$, where $n$ is a positive integer. The order of the pole is $-n$. At a pole of order $-n$, the function has the following Laurent series expansion:
“`
f(z) = a_{-n} + a_{-n+1}(z – a) + a_{-n+2}(z – a)^2 + \cdots + a_{-1}(z – a)^{n-1} + \frac{a_0}{(z – a)^n} + \frac{a_1}{(z – a)^{n+1}} + \cdots
“`
where $a_k$ are complex coefficients.
Example
Consider the function $f(z) = \frac{z}{(z – 1)^2}$. This function has a pole of order $-2$ at $z = 1$. The Laurent series expansion of $f(z)$ around $z = 1$ is:
“`
f(z) = 1 + (z – 1) + (z – 1)^2 + \frac{1}{z – 1} + \frac{1}{(z – 1)^2} + \frac{1}{(z – 1)^3} + \cdots
“`
Laurent Series Expansion for Poles
A pole is a point in the complex plane where the function has a removable singularity. This means that the function can be made continuous at the pole by removing the singularity. Laurent series expansion for poles can be used to find the residues of the function at the pole, which are important in many applications such as finding the zeros of a function.
To find the Laurent series expansion for a pole, we first need to find the order of the pole. The order of the pole is the largest integer n such that (z – a)^n f(z) is analytic at z = a. Once we know the order of the pole, we can use the following formula to find the Laurent series expansion for the function:
$$\sum_{n=-\infty}^{\infty} c_n (z – a)^n$$
Where
$$c_n = \frac{1}{2\pi i} \int_{C} \frac{f(z)}{(z – a)^{n+1}} dz$$
and C is a circle centered at z = a with radius r such that C does not enclose any other singularities of f(z).
The following table shows the Laurent series expansion for some common types of poles:
| Pole Type | Laurent Series Expansion |
|---|---|
| Simple pole | $$\frac{a_{-1}}{z – a} + \sum_{n=0}^{\infty} c_n (z – a)^n$$ |
| Double pole | $$\frac{a_{-2}}{(z – a)^2} + \frac{a_{-1}}{z – a} + \sum_{n=0}^{\infty} c_n (z – a)^n$$ |
| Triple pole | $$\frac{a_{-3}}{(z – a)^3} + \frac{a_{-2}}{(z – a)^2} + \frac{a_{-1}}{z – a} + \sum_{n=0}^{\infty} c_n (z – a)^n$$ |
Principal Part of the Laurent Series
The principal part of a Laurent series, also known as the singular part, is the portion that contains the negative powers of \(z-a\). In general, the principal part of a Laurent series for a function \(f(z)\) centered at \(z=a\) takes the form:
“`
\sum_{n=1}^{\infty} \frac{a_{-n}}{(z-a)^n}
“`
Where \(a_{-n}\) are the coefficients of the negative powers of \(z-a\). The principal part of the Laurent series represents the contributions from the isolated singularity at \(z=a\). Depending on the nature of the singularity, the principal part may have a finite number of terms or an infinite number of terms.
Pole of Order \(m\):
If \(f(z)\) has a pole of order \(m\) at \(z=a\), then the principal part of its Laurent series contains exactly \(m\) terms:
“`
\frac{a_{-1}}{z-a} + \frac{a_{-2}}{(z-a)^2} + \cdots + \frac{a_{-m}}{(z-a)^m}
“`
where \(a_{-1}, a_{-2}, \cdots, a_{-m}\) are non-zero constants.
Essential Singularity:
If \(f(z)\) has an essential singularity at \(z=a\), then the principal part of its Laurent series contains an infinite number of negative terms:
“`
\sum_{n=1}^{\infty} \frac{a_{-n}}{(z-a)^n}
“`
where at least one of the coefficients \(a_{-n}\) is non-zero for all \(n\).
Removable Singularity:
If \(f(z)\) has a removable singularity at \(z=a\), then the principal part of its Laurent series is simply \(0\), indicating that there are no negative power terms in the series:
“`
0
“`
Expansion of Meromorphic Functions
A meromorphic function is a function that is holomorphic except for a set of isolated singularities.
Laurent series can be used to expand meromorphic functions around their singularities.
The Laurent series of a meromorphic function \(f(z)\) around a singularity \(z_0\) has the following form:
$$\sum_{n=-\infty}^\infty a_n(z-z_0)^n$$
where \(a_n\) are constants.
The principal part of the Laurent series is the sum of the terms with negative powers of \((z-z_0)\).
The order of the singularity is the degree of the pole of the principal part.
For example, the Laurent series of the function \(f(z) = \frac{1}{z-1}\) around the singularity \(z=1\) is
$$\sum_{n=-\infty}^\infty (-1)^n(z-1)^n = \frac{1}{z-1} – 1 + (z-1) – (z-1)^2 + …$$
The principal part of this series is \(\frac{1}{z-1}\), and the order of the singularity is 1.
The following table summarizes the steps for expanding a meromorphic function around a singularity:
| Step | Description |
|---|---|
| 1 | Find the residues of the function at the singularity. |
| 2 | Write the principal part of the Laurent series as a sum of terms with negative powers of \((z-z_0)\). |
| 3 | Find the Laurent series of the function by adding the principal part to a regular function. |
Determination of Laurent Series at Infinity
To determine the Laurent series of a function at infinity, we follow these steps:
1. Simplify the Function into a Rational Form
First, we simplify the function into a rational form where the denominator is linear in the variable. This involves dividing the numerator by the denominator using polynomial long division.
2. Factor the Denominator
Next, we factor the denominator of the rational function into linear factors.
3. Create a Principal Part for Poles
For each linear factor (z – a) in the denominator, we create a principal part of the form \( \frac{A}{z – a} \), where A is a constant.
4. Find the Coefficients A
To find the constants A, we use the method of residues. This involves evaluating the integral of the function \( f(z) \over z – a \) around the circle centred at \( z = 0 \) with radius \( R \), and then taking the limit as \( R \to \infty \).
5. Find the Principal Part for Infinity
We create a principal part of the form \( \sum_{n=0}^\infty a_n z^n \), where the \( a_n \) are constants.
6. Combine Principal Parts to Form Laurent Series
Finally, we combine the principal parts to form the Laurent series of the function:
$$ f(z) = \left(\sum_{n=1}^\infty \frac{A_n}{z – a_n}\right) + \left(\sum_{n=0}^\infty a_n z^n\right) $$
Where the first term represents the principal part for poles, and the second term represents the principal part for infinity.
Laurent Series for Rational Functions
A rational function is a function that can be expressed as the quotient of two polynomials. In other words, it is a function of the form
$$f(z) = \frac{p(z)}{q(z)},$$
where
The Laurent series for a rational function can be determined by using the following steps:
1. Factor the denominator into linear factors.
If the denominator of the rational function can be factored into linear factors, then the Laurent series can be written as a sum of partial fractions.
2. Find the residues of the rational function.
The residue of a rational function at a singularity is the coefficient of the
3. Write the Laurent series for the rational function.
The Laurent series for a rational function is a sum of the partial fractions and the residues of the function.
For example, consider the rational function
$$f(z) = \frac{1}{z^2-1}.$$
The denominator of this function can be factored into the linear factors
$$z^2-1 = (z-1)(z+1).$$
The residues of the function at the singularities
Therefore, the Laurent series for the function is
$$f(z) = \frac{1}{2(z-1)} – \frac{1}{2(z+1)}.$$
This series converges for all
Cauchy’s Integral Formula and Laurent Series
Cauchy’s Integral Formula
One magical formula that assists in determining the Laurent series for a function is Cauchy’s Integral Formula:
\(f(z) = \frac{1}{2\pi i} \int\limits_\gamma \frac{f(w)}{w-z} dw\)
In this formula, \(f(w)\) is the function we seek to decipher, \(w\) is a complex variable traveling along a contour \(\gamma\), and \(z\) is a specific point within or outside the contour.
Laurent Series for a Function Within a Circle
When the complex function \(f(z)\) is analytic inside and on a positively oriented circle centered at the origin with radius \(R>0\), then it has an associated Laurent series valid for \(0<|z|
\(f(z) = \sum\limits_{n=-\infty}^\infty a_n z^n\) = \(…+\frac{a_{-2}}{z^2} + \frac{a_{-1}}{z} + a_0 + a_1 z + a_2 z^2 + …\)
Here, the coefficients \(a_n\) are given by:
\(a_n = \frac{1}{2\pi i} \int\limits_{|z|=R} \frac{f(z)}{z^{n+1}} dz\)
Laurent Series for a Function Outside a Circle
In the case where the function \(f(z)\) is analytic outside and on a positively oriented circle centered at the origin with radius \(R>0\), its Laurent series converges for \( |z| > R \):
\(f(z) = \sum\limits_{n=-\infty}^\infty a_n z^n\) = \(…+ a_{-2} z^2 + a_{-1} z + a_0 + \frac{a_1}{z} + \frac{a_2}{z^2} + …\)
The coefficients \(a_n\) are still calculated using the same formula as before:
\(a_n = \frac{1}{2\pi i} \int\limits_{|z|=R} \frac{f(z)}{z^{n+1}} dz\)
Laurent Series for a Function with an Essential Singularity
When the function \(f(z)\) has an essential singularity at the origin, its Laurent series consists of infinitely many nonzero terms in both the positive and negative powers of \(z\):
\(f(z) = \sum\limits_{n=-\infty}^\infty a_n z^n\) = \(… + a_{-2} z^2 + a_{-1} z + a_0 + a_1 z + a_2 z^2 + …\)
The coefficients \(a_n\) are still obtained using the same formula:
\(a_n = \frac{1}{2\pi i} \int\limits_{|z|=R} \frac{f(z)}{z^{n+1}} dz\)
Laurent Series Representation
The Laurent series of a function
f(z)
in a domain
\Omega
specifies the function as a sum of its Taylor series and a principal part, which contains the negative powers of
z-z_0
. The Laurent series for a function that is analytic in an annular region is given by:
$$f(z)=\sum_{n=-\infty}^{\infty}c_n(z-z_0)^n$$
where
c_n
are the Laurent coefficients.
Application to Complex Integration
The Laurent series representation allows us to evaluate complex integrals using the residue theorem. The residue of a function at a singularity
z_0
is the coefficient of the
(z-z_0)^{-1}
term in its Laurent series. The residue theorem states that the integral of a function around a closed contour
C
enclosing a singularity
z_0
is equal to
2\pi i
times the residue of the function at
z_0
.
Calculating Residues
To calculate the residue of a function at a pole
z_0
, we can use the following formula:
$$\operatorname{Res}[f(z), z_0] = \lim_{z \to z_0} (z – z_0) f(z)$$
Cauchy’s Integral Formula
Cauchy’s integral formula is a powerful tool for evaluating complex integrals. It states that if
f(z)
is analytic in a domain containing a closed contour
C
and
z_0
is inside
C
, then the integral of
f(z)
around
C
is equal to
2\pi i
times the value of
f(z)
at
z_0
.
$$ \oint_C f(z) dz = 2\pi i f(z_0) $$
Example
Consider the integral:
$$ I = \oint_C \frac{1}{z^2 + 1} dz $$
where
C
is the unit circle centered at the origin. The function
f(z) = \frac{1}{z^2 + 1}
has two poles at
z = \pm i
. The residue of
f(z)
at
z = i
is:
$$ \operatorname{Res}[f(z), i] = \lim_{z \to i} (z – i) \frac{1}{z^2 + 1} = \frac{1}{2i} $$
Using Cauchy’s integral formula, we can evaluate the integral as:
$$ I = 2\pi i \operatorname{Res}[f(z), i] = 2\pi i \cdot \frac{1}{2i} = \pi $$
Convergence and Error Estimation
The Laurent series of a function f(z) converges uniformly in an annulus r1 < |z| < r2 if and only if f(z) is continuous on the boundary of the annulus. If f(z) is continuous on the closed annulus [r1, r2], then the Laurent series converges uniformly to f(z) on the open annulus r1 < |z| < r2.
The error in approximating f(z) by its Laurent series with n terms is given by the remainder term:
Remainder Term
Rn(z) = f(z) – Sn(z)
where Sn(z) is the nth partial sum of the Laurent series.
The remainder term can be estimated using the following formula:
Error Estimation
|Rn(z)| ≤ M/(r- |z|)n+1
where M = max |f(z)| on the boundary of the annulus.
| Condition | Convergence |
|---|---|
| f(z) is continuous on the boundary of the annulus r1 < |z| < r2 | Laurent series converges uniformly in the annulus |
| f(z) is continuous on the closed annulus [r1, r2] | Laurent series converges uniformly to f(z) in the open annulus r1 < |z| < r2 |
| |f(z)| ≤ M on the boundary of the annulus r1 < |z| < r2 | Error in approximating f(z) by its Laurent series with n terms is bounded by M/(r- |z|)n+1 |
How to Determine the Laurent Series of a Function
The Laurent series of a function $f(z)$ is a representation of the function as a sum of powers of \(z-a\), where \(a\) is a singular point of the function. The series is valid in an annular region around the singular point, and it can be used to evaluate the function at points in that region.
To determine the Laurent series of a function, you can use the following steps:
1. Find the isolated singular points of the function.
2. For each singular point, find the order of the pole or zero.
3. Write the Laurent series in the form
$$f(z) = \sum_{n=-\infty}^{\infty} c_n (z-a)^n$$
where \(c_n\) are the coefficients of the series.
The coefficients \(c_n\) can be found using the following formulas:
$$c_n = \frac{1}{2\pi i} \int_{C} \frac{f(z)}{(z-a)^{n+1}} dz$$
where \(C\) is a circle around the singular point.
People Also Ask About How to Determine the Laurent Series of a Function
What is the Laurent series?
The Laurent series is a representation of a function as a sum of powers of \(z-a\), where \(a\) is a singular point of the function. The series is valid in an annular region around the singular point, and it can be used to evaluate the function at points in that region.
How do I find the Laurent series of a function?
To find the Laurent series of a function, you can use the following steps:
- Find the isolated singular points of the function.
- For each singular point, find the order of the pole or zero.
- Write the Laurent series in the form
$$f(z) = \sum_{n=-\infty}^{\infty} c_n (z-a)^n$$
where \(c_n\) are the coefficients of the series.
What are the coefficients of the Laurent series?
The coefficients of the Laurent series are given by the following formulas:
$$c_n = \frac{1}{2\pi i} \int_{C} \frac{f(z)}{(z-a)^{n+1}} dz$$
where \(C\) is a circle around the singular point.
