Question

Find the Laurent series for the following functions about the indicated points; hence find the residue of the function at the point. (Be sure you have the Laurent series which converges near the point.) $$\frac{1}{z^2-5z+6},z=2$$

Short answer

The residue at z=2 is -1.

Step-by-step solution

Factor the Denominator

First, factor the denominator of the given function: $$\frac{1}{z^2-5z+6}$$ This can be factored as: $$\frac{1}{(z-2)(z-3)}$$

Decompose into Partial Fractions

Decompose the fraction into partial fractions: $$\frac{1}{(z-2)(z-3)}=\frac{A}{z-2}+\frac{B}{z-3}$$ Solve for A and B: $$1=A(z-3)+B(z-2)$$ By substituting suitable values for z, we find: For $z=2$: $$1=A(2-3)+B(2-2)\Rightarrow A=-1$$ For $z=3$: $$1=A(3-3)+B(3-2)\Rightarrow B=1$$

Write the Partial Fractions

Write the function in terms of the partial fractions: $$\frac{1}{(z-2)(z-3)}=\frac{-1}{z-2}+\frac{1}{z-3}$$

Find the Laurent Series About z=2

To express $\frac{1}{z-3}$ as a Laurent series that converges near $z=2$, rewrite it as: $$\frac{1}{z-3}=\frac{1}{(z-2)-1}=-\frac{1}{z-2}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{-1}{z-2}\right)}^n$$ Thus, the Laurent series becomes: $$\frac{1}{(z-2)(z-3)}=-\frac{1}{z-2}+\frac{1}{z-2}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{z-2}\right)}^n$$ $$=-\frac{1}{z-2}+\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{z-2}\right)}^{n+1}$$

Identifying the Residue

The residue at $z=2$ is the coefficient of $\frac{1}{z-2}$ in the Laurent series, which is: $$-1$$

Key concepts

residue

In complex analysis, the residue of a function at a particular point is a key concept when dealing with Laurent series. A residue gives us crucial information about the behavior of a function near a singularity.
When you expand a function into a Laurent series, it takes the form: ote{f(z) = \frac{a_{-1}}{(z - z_0)} + \frac{a_{-2}}{(z - z_0)^2} + ... + a_0 + a_1(z - z_0) + ... } The coefficient $a_{-1}$ is called the residue of the function at $z=z_0$.
In our solution, we determined that the residue at $z=2$ is $-1$, derived from the term $-\frac{1}{z-2}$ in the Laurent series.
Understanding residues helps in evaluating complex integrals and in describing the nature of singularities.

partial fractions

Partial fractions are a way to break down more complex rational functions into simpler ones that are easier to integrate or expand.
For the function $\frac{1}{z^2-5z+6}$, we factored the denominator to $(z-2)(z-3)$.
We then decomposed this into: $\frac{A}{z-2}+\frac{B}{z-3}$.
Solving for $A$ and $B$ involved substituting specific values for $z$ to form simple equations. This yielded:

• $A=-1$ when $z=2$
• $B=1$ when $z=3$

Thus, we wrote $\frac{1}{(z-2)(z-3)}$ as $\frac{-1}{z-2}+\frac{1}{z-3}$. These simpler fractions pave the way to finding the Laurent series.

complex analysis

Complex analysis, the study of functions that operate on complex numbers, provides essential tools for many areas of mathematics and engineering. It deals with complex functions and their properties, including convergence, differentiation, and integration.
One critical component of complex analysis is the Laurent series, an expansion that includes both positive and negative powers of $z$. This is particularly useful for functions with singularities. In our exercise, we utilized the Laurent series to understand the behavior of $\frac{1}{z^2-5z+6}$ around $z=2$.
We started with partial fraction decomposition and then proceeded by rewriting $\frac{1}{z-3}$ as a series expansion around $z=2$. This approach, rooted in complex analysis, gave us insight into the function's behavior near that point, enabling us to identify the residue at the singularity.

convergence

Convergence is a fundamental concept when dealing with series in mathematics. A series is said to converge if its sum approaches a finite limit as the number of terms increases.
For Laurent series, convergence determines how close or far from the singularity the series accurately represents the function.
In our solution, we focused on finding the Laurent series that converges around $z=2$.
We started by expressing the function in terms of partial fractions and then expanded $\frac{1}{z-3}$ as a series around $z=2$.
This involved the geometric series expansion, ensuring that the series converges in the annulus where $|z-2|<1$.
Proper convergence is crucial to ensuring accurate representations, allowing us to effectively use Laurent series to analyze complex functions.