Laurent Series

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,

• a_n 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 (r₁): The series converges for |z - z₀| < r₁ only if a_n = 0 for all n < 0 , reducing it to a Taylor series.

• Annulus ( r₁ < |z - z₀| < r₂ ): 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 ( r₂ ): 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 r₁ < |z - z₀| < r₂ .

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 r₁ < |z - z₀| < r₂.

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 .