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Find Laurent Series Calculator – Calculator

Find Laurent Series Calculator






Laurent Series Calculator – Find Expansions Around Singularities


Laurent Series Calculator

Calculate Laurent Series for f(z) = 1/((z-a)(z-b)) around z=0

This calculator finds the Laurent series expansion for the function f(z) = 1/((z-a)(z-b)) centered at z0=0 for different regions of convergence defined by ‘a’ and ‘b’.







Series will appear here…

Details:

Region: –
Coeff of 1/(z-a): –
Coeff of 1/(z-b): –
Table of coefficients an
n an (Real) an (Imag)
Enter values and calculate

Chart of |an| vs n

The function is f(z) = (1/(a-b)) * [1/(z-a) – 1/(z-b)]. The series for 1/(z-c) around z=0 depends on whether |z| < |c| or |z| > |c|.

What is a Laurent Series Calculator?

A Laurent series calculator is a tool designed to find the Laurent series expansion of a complex function around a point, particularly within an annulus of convergence. A Laurent series is a representation of a complex function f(z) as a series of powers of (z-z0), including both non-negative and negative powers. It’s an extension of the Taylor series, applicable even when the function has singularities at or near the center z0, as long as it’s analytic in an annulus around z0. Our Laurent series calculator specifically handles the function f(z) = 1/((z-a)(z-b)) around z=0.

This type of calculator is used by students, engineers, and mathematicians working with complex analysis, especially when analyzing the behavior of functions near singularities, calculating residues, and evaluating complex integrals.

Common misconceptions include thinking a Laurent series is the same as a Taylor series (Taylor series only have non-negative powers and require the function to be analytic at the center) or that every function has a unique Laurent series around a point (it depends on the annulus of convergence).

Laurent Series Formula and Mathematical Explanation

A Laurent series of a function f(z) about a point z0, convergent in an annulus R1 < |z-z0| < R2, is given by:

f(z) = ∑n=-∞ an (z-z0)n

where the coefficients an are calculated using the integral formula:

an = (1 / (2πi)) ∮C f(w) / (w-z0)n+1 dw

Here, C is any simple closed contour within the annulus R1 < |z-z0| < R2, oriented counter-clockwise and encircling z0.

Our Laurent series calculator deals with f(z) = 1/((z-a)(z-b)) centered at z0=0. We first use partial fraction decomposition:

f(z) = 1/((z-a)(z-b)) = (1/(a-b)) * [1/(z-a) – 1/(z-b)]

Then, we expand 1/(z-a) and 1/(z-b) around z=0 using the geometric series formula 1/(1-u) = ∑ un, adapting it for |z|<|c| or |z|>|c|:

  • If |z| < |c|: 1/(z-c) = -1/c * 1/(1-z/c) = -1/c ∑n=0 (z/c)n
  • If |z| > |c|: 1/(z-c) = 1/z * 1/(1-c/z) = 1/z ∑n=0 (c/z)n = ∑n=0 cn z-n-1

The regions of convergence around z=0 are determined by the magnitudes of ‘a’ and ‘b’. Assuming |a| < |b| (and a, b non-zero), the regions are |z| < |a|, |a| < |z| < |b|, and |z| > |b|.

Variables in Laurent Series Calculation
Variable Meaning Unit Typical Range
f(z) The complex function Complex number Varies
z The complex variable Complex number Varies
z0 Center of the expansion (0 in our calculator) Complex number 0
a, b Locations of singularities of f(z) Complex numbers (real in our calc) Non-zero real numbers
an Laurent series coefficients Complex numbers Varies
n Index of the series terms Integer -∞ to ∞
R1, R2 Inner and outer radii of the annulus of convergence Real, non-negative 0 ≤ R1 < R2 ≤ ∞

Practical Examples (Real-World Use Cases)

Example 1: |z| < |a|

Let f(z) = 1/((z-1)(z-2)), so a=1, b=2. Center z0=0. Consider the region |z| < 1 (min(|a|,|b|)=1). f(z) = (1/(1-2)) * [1/(z-1) - 1/(z-2)] = -1/(z-1) + 1/(z-2). In |z|<1: -1/(z-1) = 1/(1-z) = 1 + z + z2 + …
1/(z-2) = -1/2 * 1/(1-z/2) = -1/2 (1 + z/2 + (z/2)2 + …)
So, f(z) = (1 + z + z2 + …) – 1/2 (1 + z/2 + z2/4 + …) = 1/2 + 3/4 z + 7/8 z2 + …
Using the Laurent series calculator with a=1, b=2, region |z|<1, you'd get these coefficients.

Example 2: |a| < |z| < |b|

Let f(z) = 1/((z-1)(z-2)), a=1, b=2, z0=0. Consider the region 1 < |z| < 2. -1/(z-1) = -1/z * 1/(1-1/z) = -1/z (1 + 1/z + 1/z2 + …) = -z-1 – z-2 – z-3 – …
1/(z-2) = -1/2 (1 + z/2 + (z/2)2 + …)
So, f(z) = (-z-1 – z-2 – …) – 1/2 (1 + z/2 + z2/4 + …) = … – z-2 – z-1 – 1/2 – z/4 – z2/8 – …
The Laurent series calculator will show terms for n < 0 and n ≥ 0.

How to Use This Laurent Series Calculator

  1. Enter ‘a’ and ‘b’: Input the values for the singularities ‘a’ and ‘b’ of the function f(z) = 1/((z-a)(z-b)). Ensure a ≠ b, a ≠ 0, b ≠ 0.
  2. Enter Number of Terms (N): Specify how many terms you want to see, from n=-N to n=N.
  3. Select Region: Based on the values of |a| and |b|, the calculator will offer possible regions of convergence around z=0. Select the desired region.
  4. Calculate: Click “Calculate Series” or observe the real-time update.
  5. View Results: The primary result shows the series expansion. Intermediate results show the region, partial fraction coefficients, and a table of an coefficients. A chart visualizes |an|.
  6. Interpret: The table and series show the principal part (negative powers of z) and the analytic part (non-negative powers of z) of the Laurent series in the chosen region.

Understanding the results helps in analyzing the function’s behavior near z=0 and its singularities, and is crucial for residue calculation.

Key Factors That Affect Laurent Series Results

  • Function f(z): The form of the function dictates the complexity and nature of the series. Our Laurent series calculator is specific to f(z)=1/((z-a)(z-b)).
  • Center of Expansion (z0): The point around which the series is expanded (fixed at 0 here). Changing z0 changes the series and regions.
  • Location of Singularities (a, b): The values of ‘a’ and ‘b’ determine the boundaries of the annuli of convergence.
  • Region of Convergence: The chosen annulus dictates which form of the geometric series expansion is used for each term from the partial fractions, thus defining the coefficients an.
  • Number of Terms: How many terms are calculated and displayed. A Laurent series generally has infinitely many terms.
  • Analyticity: The function must be analytic within the annulus for the Laurent series to be valid there. Find more about this with our complex analysis calculator.

Frequently Asked Questions (FAQ)

What is the difference between a Taylor series and a Laurent series?
A Taylor series is a special case of a Laurent series where all coefficients an for n < 0 are zero. Taylor series represent functions analytic at the center z0, while Laurent series can represent functions analytic in an annulus around z0, even if z0 is a singularity. See our Taylor series calculator for comparison.
What is the principal part of a Laurent series?
The principal part consists of the terms with negative powers of (z-z0): ∑n=-∞-1 an (z-z0)n. It characterizes the nature of the singularity at z0 (if R1=0).
What is the analytic part of a Laurent series?
The analytic part (or regular part) consists of the terms with non-negative powers: ∑n=0 an (z-z0)n. This part behaves like a Taylor series.
Why is the region of convergence important for a Laurent series calculator?
The coefficients an and the form of the series depend on the annulus of convergence. A function can have different Laurent series around the same point z0 but valid in different annuli. Our Laurent series calculator helps visualize this.
What is a singularity in complex analysis?
A singularity of a complex function is a point where the function is not analytic (not differentiable). The Laurent series is very useful for classifying singularities (poles, essential singularities, removable singularities). Explore more with our complex analysis basics guide.
Can this calculator handle functions other than 1/((z-a)(z-b))?
No, this specific Laurent series calculator is designed only for f(z) = 1/((z-a)(z-b)) centered at z=0. Finding Laurent series for general functions automatically is much more complex.
What if a=b or a=0 or b=0?
If a=b, the function becomes 1/(z-a)2, requiring a different approach (differentiation of geometric series). If a=0 or b=0, the center z=0 is a singularity of a different order, and the regions change. Our calculator requires a ≠ b, a ≠ 0, b ≠ 0.
How are the coefficients an calculated here without integration?
For functions like 1/((z-a)(z-b)), we use partial fraction decomposition and known geometric series expansions, which avoids direct complex integration for each an when the regions are simple annuli centered at z=0. Check out our partial fraction decomposition tool.

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