What is Laurent’s Series?

It is an expansion of a complex function that includes both negative and positive powers of (z - a).

What is the Laurent Series and Taylor Series difference?

Laurent series can represent functions with singularities; Taylor series cannot.

Where is Laurent Series used?

In complex integration, particularly near singular points using the Residue Theorem.

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Laurent’s Series

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Laurent’s Series

Q: What are the conditions for Laurent’s Series?

The function must be analytic in an annular region around the point aa.

In complex analysis, Laurent's Series is a powerful tool for expressing complex functions, especially those with singularities. Unlike Taylor’s series, which only applies to analytic functions around a point, Laurent’s series is capable of representing functions with isolated singularities. In this blog, we’ll explore the Laurent's Series definition, formula, conditions, and a comparison with Taylor series using examples.

1.0Laurent's Series Definition

Laurent’s series is a representation of a complex function as an infinite series that includes both positive and negative powers of (z - a), where a is a point in the complex plane.

Definition:

If a function f(z) is analytic in an annular region (ring-shaped) around a point a, i.e.,

$R_{1} < ∣ z - a ∣ < R_{2}$

then f(z) can be expressed as a Laurent series: $f (z) = \sum_{n \to - \infty}^{\infty} c_{n} (z - a)^{n}$

This includes both a regular part (positive powers) and a principal part (negative powers).

2.0Laurent's Series Formula

The coefficients $c_{n}$ in the Laurent series are given by:

$c_{n} = \frac{1}{2 πi} \oint_{C} \frac{f ( ζ )}{( ζ - a ) ^{n + 1}} d ζ$

Where:

C is a positively oriented simple closed contour within the annular region.
is the integration variable.

This formula applies to any $n ϵ Z$ (positive, negative, or zero).

3.0What is the Laurent Series and Taylor Series?

Both are series expansions of complex functions, but they differ in scope and use:

Feature	Taylor Series	Laurent Series
Powers	Only non-negative integers	Both positive and negative integers
Validity	Around analytic points	Around isolated singularities
Form	$\sum_{n = 0}^{\infty} a_{n} (z - a)^{n}$	$\sum_{n = - \infty}^{\infty} a_{n} (z - a)^{n}$
Applications	Entire/analytic functions	Functions with poles or essential singularities

4.0What are the Conditions for Laurent's Series?

To express a function f(z) using Laurent’s series, the following conditions must be satisfied:

Analyticity: f(z) must be analytic in an annular region: $R_{1} < ∣ z - a ∣ < R_{2}$
Isolated Singularity: The point a can be a removable singularity, pole, or an essential singularity.
Contour Integration: The integral used in calculating the coefficients must be well-defined.

5.0Solved Example on Laurent’s Series Expansion

Example 1: Let: $f (z) = \frac{1}{z ( z - 1 )}$ . Find the Laurent series around z = 0 in the region 0 < |z| < 1.

Solution:

We write: $\frac{1}{z ( z - 1 )} = \frac{1}{z} . \frac{1}{1 - z}$

Using the geometric series: $\frac{1}{1 - z} = \sum_{n = 0}^{\infty} z^{n}$ for $∣ z ∣ < 1$

So,

$f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} z^{n} = \sum_{n = 0}^{\infty} z^{n - 1} = \sum_{n = - 1}^{\infty} z^{n}$

This is the Laurent series of f(z) in the region $0 < ∣ z ∣ < 1.$

Example 2: Find the Laurent series expansion of $f (z) = \frac{1}{z ( z + 1 )}$ valid in the region 0 < |z| < 1.

Solution:

We rewrite: $f (z) = \frac{1}{z ( z + 1 )} = \frac{1}{z} . \frac{1}{z + 1}$

Now expand $\frac{1}{z + 1}$ as a geometric series:

$\frac{1}{z + 1} = \frac{1}{1 + z} = \sum_{n = 0}^{\infty} (- a)^{n} z^{n}$ for $∣ z ∣ < 1$

So:

$f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} (- 1)^{n} z^{n} = \sum_{n = 0}^{\infty} (- 1)^{n} z^{n - 1}$

Hence, the Laurent series is: $f (z) = \sum_{n = 0}^{\infty} (- 1)^{n} z^{n - 1}$

Example 3: Find the Laurent series of $f (z) = \frac{1}{z ( z - 1 )}$ valid in $1 < ∣ z ∣ < \infty$

Solution:

Rewrite:

$f (z) = \frac{1}{z ( z - 1 )} = \frac{1}{z} . \frac{1}{z - \frac{1}{z}}$

Now expand:

$\frac{1}{1 - \frac{1}{z}} = \sum_{n = 0}^{\infty} \frac{1}{z ^{n}} = \sum_{n = 0}^{\infty} z^{- n}$ for $∣ z ∣ > 1$

Then:

$f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} z^{- n} = \sum_{n = 0}^{\infty} z^{- n - 1} = \sum_{n = 1}^{\infty} z^{- n}$

Example 4: Find the Laurent expansion of $f (z) = \frac{1}{z ^{2} - 1}$ in the region $0 < ∣ z ∣ < 10 < ∣ z ∣ < 1$

Solution:

We write:

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

Now expand each term around z = 0, for |z| < 1:

Expand $\frac{1}{z - 1}$ as: $\frac{1}{z - 1} = - \frac{1}{1 - z} = - \sim_{n = 0}^{\infty} z^{n}$
Expand $\frac{1}{z + 1} = \sum_{n = 0}^{\infty} (- 1)^{n} z^{n}$

So:

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

Thus,

$f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n} only even powers survive$

Example 5: Expand $f (z) = e^{\frac{1}{z}}$ as a Laurent series about z = 0 (valid in $0 < ∣ z ∣ < \infty$ )

Solution:

We use the exponential series:

$f (z) = e^{\frac{1}{z}} = \sum_{n = 0}^{\infty} \frac{1}{n !} . \frac{1}{z ^{n}} = \sum_{n = 0}^{\infty} \frac{1}{n !} z^{- n}$

This is a purely principal part, indicating an essential singularity at z = 0.

Example 6: Find the Laurent expansion of $f (z) = \frac{1}{( z - 1 ) ^{2} ( z - 2 )}$ valid in 1 < |z| < 2

Solution:

This requires partial fractions: $f (z) = \frac{A}{z - 2} + \frac{B}{z - 1} + \frac{C}{( z - 1 ) ^{2}}$

Use algebra to solve for A, B, C.

After decomposition, you expand:

$\frac{1}{z - 2}$ in powers of $\frac{1}{z}$ (for outer region)
$\frac{1}{z - 1}, \frac{1}{( z - 1 ) ^{2}}$ in powers of $\frac{z}{1}$ (for inner region)

6.0Practice Questions on Laurent Series

Q1. Find the Laurent series of $f (z) = \frac{1}{z ^{2} - 1}$ in the annulus $1 < ∣ z ∣ < \infty$

Q2. Find the Laurent series for $f (z) = \frac{z}{z ^{2} - 4}$ about z = 0, valid in the region |z| < 2.

Q3. Find the Laurent expansion of $f (z) = s in (\frac{1}{z})$ around z = 0, up to the term .

Q4. Determine the Laurent expansion for $f (z) = \frac{1}{z - 2}$ in the region |z| > 2. (Hint: Express in terms of 1/z)

Q5. For the function $f (z) = \frac{e ^{z}}{z ^{3}}$ find the first three non-zero terms of the Laurent expansion at z = 0 and identify the type of singularity.

Q6. Find the residue of $f (z) = \frac{1}{( z - 1 ) ^{2} ( z + 2 )}$ at z = 1 using Laurent series.

Q7. Find the Laurent expansion of $f (z) = \frac{l n ( 1 + z )}{z}$ about z = 0 in the region 0 < |z| < 1.

Q8. Use the Laurent coefficient formula: $c_{n} = \frac{1}{2 πi} \oint_{c} \frac{f ( z )}{( z - a ) ^{n + 1}} d z$ to find the coefficient $c_{- 1}$ of the function $f (z) = \frac{1}{z ( z - 2 )}$ in the annulus 0 < |z| < 2.

Q9. Expand $f (z) = \frac{1}{z ( z - 1 )}$ in both regions:

a) 0 < |z| < 1

b) |z| > 1 and compare the series.

Q10. Using the Laurent expansion, evaluate the integral: $\oint_{∣ z ∣ = 1} \frac{e ^{x}}{z ^{2}} d z$

7.0Applications of Laurent Series

Residue Calculation for contour integrals.
Classifying singularities (removable, pole, or essential).
Solving complex integrals using the Residue Theorem.
Representing functions with isolated singularities.

Also Read

Arithmetic Progression	Geometric Progression	Weighted Arithmetic Mean
Arithmetic Mean	Harmonic Mean	Arithmetic Progression Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Laurent’s Series

In complex analysis, Laurent's Series is a powerful tool for expressing complex functions, especially those with singularities. Unlike Taylor’s series, which only applies to analytic functions around a point, Laurent’s series is capable of representing functions with isolated singularities. In this blog, we’ll explore the Laurent's Series definition, formula, conditions, and a comparison with Taylor series using examples.

1.0Laurent's Series Definition

Laurent’s series is a representation of a complex function as an infinite series that includes both positive and negative powers of (z - a), where a is a point in the complex plane.

Definition:

If a function f(z) is analytic in an annular region (ring-shaped) around a point a, i.e.,

$R_{1} < ∣ z - a ∣ < R_{2}$

then f(z) can be expressed as a Laurent series: $f (z) = \sum_{n \to - \infty}^{\infty} c_{n} (z - a)^{n}$

This includes both a regular part (positive powers) and a principal part (negative powers).

2.0Laurent's Series Formula

The coefficients $c_{n}$ in the Laurent series are given by:

$c_{n} = \frac{1}{2 πi} \oint_{C} \frac{f ( ζ )}{( ζ - a ) ^{n + 1}} d ζ$

Where:

C is a positively oriented simple closed contour within the annular region.
is the integration variable.

This formula applies to any $n ϵ Z$ (positive, negative, or zero).

3.0What is the Laurent Series and Taylor Series?

Both are series expansions of complex functions, but they differ in scope and use:

Feature	Taylor Series	Laurent Series
Powers	Only non-negative integers	Both positive and negative integers
Validity	Around analytic points	Around isolated singularities
Form	$\sum_{n = 0}^{\infty} a_{n} (z - a)^{n}$	$\sum_{n = - \infty}^{\infty} a_{n} (z - a)^{n}$
Applications	Entire/analytic functions	Functions with poles or essential singularities

4.0What are the Conditions for Laurent's Series?

To express a function f(z) using Laurent’s series, the following conditions must be satisfied:

Analyticity: f(z) must be analytic in an annular region: $R_{1} < ∣ z - a ∣ < R_{2}$
Isolated Singularity: The point a can be a removable singularity, pole, or an essential singularity.
Contour Integration: The integral used in calculating the coefficients must be well-defined.

5.0Solved Example on Laurent’s Series Expansion

Example 1: Let: $f (z) = \frac{1}{z ( z - 1 )}$ . Find the Laurent series around z = 0 in the region 0 < |z| < 1.

Solution:

We write: $\frac{1}{z ( z - 1 )} = \frac{1}{z} . \frac{1}{1 - z}$

Using the geometric series: $\frac{1}{1 - z} = \sum_{n = 0}^{\infty} z^{n}$ for $∣ z ∣ < 1$

So,

$f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} z^{n} = \sum_{n = 0}^{\infty} z^{n - 1} = \sum_{n = - 1}^{\infty} z^{n}$

This is the Laurent series of f(z) in the region $0 < ∣ z ∣ < 1.$

Example 2: Find the Laurent series expansion of $f (z) = \frac{1}{z ( z + 1 )}$ valid in the region 0 < |z| < 1.

Solution:

We rewrite: $f (z) = \frac{1}{z ( z + 1 )} = \frac{1}{z} . \frac{1}{z + 1}$

Now expand $\frac{1}{z + 1}$ as a geometric series:

$\frac{1}{z + 1} = \frac{1}{1 + z} = \sum_{n = 0}^{\infty} (- a)^{n} z^{n}$ for $∣ z ∣ < 1$

So:

$f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} (- 1)^{n} z^{n} = \sum_{n = 0}^{\infty} (- 1)^{n} z^{n - 1}$

Hence, the Laurent series is: $f (z) = \sum_{n = 0}^{\infty} (- 1)^{n} z^{n - 1}$

Example 3: Find the Laurent series of $f (z) = \frac{1}{z ( z - 1 )}$ valid in $1 < ∣ z ∣ < \infty$

Solution:

Rewrite:

$f (z) = \frac{1}{z ( z - 1 )} = \frac{1}{z} . \frac{1}{z - \frac{1}{z}}$

Now expand:

$\frac{1}{1 - \frac{1}{z}} = \sum_{n = 0}^{\infty} \frac{1}{z ^{n}} = \sum_{n = 0}^{\infty} z^{- n}$ for $∣ z ∣ > 1$

Then:

$f (z) = \frac{1}{z} \sum_{n = 0}^{\infty} z^{- n} = \sum_{n = 0}^{\infty} z^{- n - 1} = \sum_{n = 1}^{\infty} z^{- n}$

Example 4: Find the Laurent expansion of $f (z) = \frac{1}{z ^{2} - 1}$ in the region $0 < ∣ z ∣ < 10 < ∣ z ∣ < 1$

Solution:

We write:

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

Now expand each term around z = 0, for |z| < 1:

Expand $\frac{1}{z - 1}$ as: $\frac{1}{z - 1} = - \frac{1}{1 - z} = - \sim_{n = 0}^{\infty} z^{n}$
Expand $\frac{1}{z + 1} = \sum_{n = 0}^{\infty} (- 1)^{n} z^{n}$

So:

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

Thus,

$f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n} only even powers survive$

Example 5: Expand $f (z) = e^{\frac{1}{z}}$ as a Laurent series about z = 0 (valid in $0 < ∣ z ∣ < \infty$ )

Solution:

We use the exponential series:

$f (z) = e^{\frac{1}{z}} = \sum_{n = 0}^{\infty} \frac{1}{n !} . \frac{1}{z ^{n}} = \sum_{n = 0}^{\infty} \frac{1}{n !} z^{- n}$

This is a purely principal part, indicating an essential singularity at z = 0.

Example 6: Find the Laurent expansion of $f (z) = \frac{1}{( z - 1 ) ^{2} ( z - 2 )}$ valid in 1 < |z| < 2

Solution:

This requires partial fractions: $f (z) = \frac{A}{z - 2} + \frac{B}{z - 1} + \frac{C}{( z - 1 ) ^{2}}$

Use algebra to solve for A, B, C.

After decomposition, you expand:

$\frac{1}{z - 2}$ in powers of $\frac{1}{z}$ (for outer region)
$\frac{1}{z - 1}, \frac{1}{( z - 1 ) ^{2}}$ in powers of $\frac{z}{1}$ (for inner region)