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:

$c_n=\frac{1}{2\pi i}{\oint }_C\frac{f(\zeta )}{(\zeta -a{)}^{n+1}}d\zeta$

Where:

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

This formula applies to any $nϵZ$ (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:

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:

1. Analyticity: f(z) must be analytic in an annular region: $R_1<|z-a|<R_2$
2. Isolated Singularity: The point a can be a removable singularity, pole, or an essential singularity.
3. 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:

1. Expand $\frac{1}{z-1}$ as: $\frac{1}{z-1}=-\frac{1}{1-z}=-{\sim }_{n=0}^{\infty }{z}^n$
2. 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,

$f(z)={\sum }_{n=0}^{\infty }{a}_n{z}^n\text{only even powers survive}$ 

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 $f(z)=sin(\frac{1}{z})$ 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 $f(z)=\frac{ln(1+z)}{z}$ about z = 0 in the region 0 < |z| < 1.

Q8. Use the Laurent coefficient formula: $c_n=\frac{1}{2\pi i}{\oint }_c\frac{f(z)}{(z-a{)}^{n+1}}dz$ 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:  ${\oint }_{|z|=1}\frac{e^x}{z^2}dz$

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.

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