Laurent Series (original) (raw)

Last Updated : 14 Nov, 2025

The Laurent series is an expansion of a complex function that includes both positive and negative powers of (z − z0). It generalizes the Taylor series, allowing representation of functions with singularities. The series is valid in an annular region around a point z0 and consists of two parts:

• Regular part: terms with non-negative powers (like the Taylor series).
• Principal part: terms with negative powers, representing the function’s behavior near singularities.

Formula for Laurent Series

Formally, for a function f(z) defined on an annulus A = {z ∈ C : r < |z - z0| < R}, the Laurent series expansion of f(z) about a point z0​ is given by:

f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n

Where,

• an are the coefficients determined by: a_n = \frac{1}{2\pi i} \int_{C} \frac{f(w)}{(w - z_0)^{n+1}} \, dw
• C is a closed contour around z0 within the region of analyticity.

The series can be split into two parts:

• **Principal Part: The terms with negative powers of

\sum_{n=-1}^{-\infty} a_n (z - z_0)^n

• **Regular Part: The terms with non-negative powers of

\sum_{n=0}^{\infty} a_n (z - z_0)^n

Convergence of Laurent Series

Convergence of the Laurent series occurs on an annulus; defined as {z: r1 < | z – z0 | < r2}.

For a Laurent series to converge, the positive and negative degree terms of the power series must converge. This convergence is uniform on compact sets within the annulus. As a result, the series defines a holomorphic function on this region.

Radius of Convergence

Radius of convergence of a power series is the distance within which the series converges to a finite value.

The convergence of a Laurent series depends on the distance from the point z0 and can be divided into three regions:

• **Interior of the Inner Radius (r 1 ): The series converges for |z - z0| < r1 only if an = 0 for all n < 0 , reducing it to a Taylor series.
• **Annulus ( r 1 < |z - z 0 | < r 2 ): The series converges in this annular region. This is the most general case for Laurent series, where both positive and negative powers of (z - z0) are present.
• **Exterior of the Outer Radius ( r 2 ): The series converges for |z - z_0| > r_2 if a_n = 0 for all n \geq 0 , turning it into a series in negative powers of (z - z_0).

Convergence Criteria

Some of the common criteria for convergence of series are:

**Cauchy-Hadamard Theorem: For a series \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n, define:

• \frac{1}{r_1} = \limsup_{n \to \infty} |a_{-n}|^{1/n}
• \frac{1}{r_2} = \limsup_{n \to \infty} |a_n|^{1/n}

The series converges in the annulus r 1 < |z - z 0 | < r 2 .

**Absolute Convergence: If the Laurent series converges at some point z1 , then it converges absolutely at every point z such that |z - z0| = |z1 - z0|.

**Uniform Convergence: The series converges uniformly on compact subsets within the annulus **r 1 < |z - z 0 | < r 2.

Laurent Series vs Taylor Series

Some of the key differences between laurent and taylor series are listed in the following table:

Applications of Laurent Series

The applications of the Laurent Series are as follows:

• **Complex analysis: Helps in studying the behavior of complex functions near singularities.
• **Residue calculus: Used for evaluating complex integrals via the residue theorem.
• **Engineering: Applied in signal processing and control theory for stability analysis.
• **Physics: Utilized in quantum mechanics and electrodynamics for potential expansions.
• **Mathematical modeling: Assists in solving differential equations and in the analysis of stability and bifurcation.

Related Articles:

• **Taylor Series
• **Maclurian Series
• **Residue Theorem
• **Power Series
• **Couchy Theorem

Solved Questions of Laurent Series

**Example 1: Find the Laurent series for f(z) = z+1/z around z0 = 0.

**Solution:

The given function can be expressed as:

f(z) = \frac{z}{z} + \frac{1}{z} = 1 + \frac{1}{z}

Hence, the Laurent series is:

f(z) = 1 + \frac{1}{z}

This series is valid in the region 0<∣z∣<∞.

**Example 2: Find the Laurent series for f(z) = z/z2 + 1 around z0 = i.

**Solution:

Using partial fractions, the function can be decomposed as:

f(z) = \frac{1}{2} \left( \frac{1}{z - i} \right) + \frac{1}{2} \left( \frac{1}{z + i} \right)

The term 1/z + i is analytic at z=i and can be expanded using a geometric series:

\frac{1}{z + i} = \frac{1}{2i} \sum_{n = 0}^{\infty} \left( -\frac{z - i}{2i} \right)^n

Therefore, the Laurent series is:

The region of convergence is 0 < ∣z−i∣ < 2.

**Example 3: Find the Laurent series for f(z) = z+1/z around 𝑧0 = v0 and determine the region of convergence.

**Solution:

The given function is:

f(z) = \frac{z + 1}{z} = 1 + \frac{1}{z}

Here, the function is already in the form of a Laurent series. We can write:

f(z) = 1 + \frac{1}{z}

The term 1 represents the analytic part with non-negative powers of z.

The term 1/𝑧 represents the principal part with negative powers of z.

Region of Convergence:

The series is valid for 0 < ∣z∣ < ∞.

**Example 4: Find the Laurent series for f(z) = z/ z2 + 1 around z0 = i. Identify the region where your answer is valid and the singular part.

**Solution:

First, use partial fractions to decompose f(z):

f(z) = \frac{z}{z^2 + 1} = \frac{1}{2} \left( \frac{1}{z - i} + \frac{1}{z + i} \right)

Expanding 1/z+i around z0=i:

\frac{1}{z + i} = \frac{1}{2i} \cdot \frac{1}{1 + \frac{z - i}{2i}} = \frac{1}{2i} \sum_{n=0}^{\infty} \left( -\frac{z - i}{2i} \right)^n

The Laurent series is:

f(z) = \frac{1}{2} \cdot \frac{1}{z - i} + \frac{1}{4i} \sum_{n=0}^{\infty} \left( -\frac{z - i}{2i} \right)^n

Singular Part:

The term 1/z − i represents the principal part.

Region of Convergence:

The series is valid for 0<∣z−i∣<2.

Practice Questions on Laurent Series

**Question 1: Determine the Laurent series for f(z)= 1/z(z−1) around z0=0.

**Question 2: Compute the Laurent series for f(z)= 1/z2 + 4 around z0=2i.

**Question 3: Find the Laurent series for f(z)= e2/z3 around z0 =0.

**Question 4: Find the Laurent series of f(z) = \frac{z}{z - 2} about z = 0 for ∣z∣ < 2and ∣z∣ > 2 .