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.) $$\frac{\sin \pi z}{4z^2-1},z=\frac{1}{2}$$

Short answer

The Laurent series combines partial fraction and geometric series. The residue is $\frac{1}{4}$.

Step-by-step solution

Identify the singularity

First, determine the singularity point of the function. The point of interest given in the problem is at $z=\frac{1}{2}$. Identify the zeros of the denominator, which determines the singularities: $$4z^2-1=0(2z-1)(2z+1)=0z=\frac{1}{2}\text{ or }z=-\frac{1}{2}.$$ So, $z=\frac{1}{2}$ is indeed a singularity.

Rewrite the function for series expansion

Rewrite the function suitable for series expansion around the singularity $z=\frac{1}{2}$. Express it as: $$\frac{\sin (\pi z)}{4z^2-1}.$$

Decompose the function

Decompose the function into partial fractions to simplify the Laurent series expansion: $$\frac{1}{4z^2-1}=\frac{1}{4(z-\frac{1}{2})(z+\frac{1}{2})}$$. Use partial fraction decomposition: $$\frac{1}{4(z-\frac{1}{2})(z+\frac{1}{2})}=\frac{A}{z-\frac{1}{2}}+\frac{B}{z+\frac{1}{2}}.$$

Solve for the coefficients

Solve for the coefficients A and B by equating and solving the equations: $$1=A(z+\frac{1}{2})+B(z-\frac{1}{2})$$. Inputting suitable values (substitute $z=\frac{1}{2},-\frac{1}{2}$) leads to:$$A=\frac{1}{4}\text{and}B=\frac{-1}{4}.$$ So: $$\frac{1}{4(z-\frac{1}{2})(z+\frac{1}{2})}=\frac{1/4}{z-\frac{1}{2}}-\frac{1/4}{z+\frac{1}{2}}.$$

Expand into Laurent series

Express $\frac{1}{z+\frac{1}{2}}$ as a Laurent series around $z=\frac{1}{2}$ using geometric series expansion:$$\frac{1}{z+\frac{1}{2}}=\frac{1}{z-\frac{1}{2}+1}=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(z-\frac{1}{2}{)}^n.$$ Thus,$$\frac{1}{4(z-\frac{1}{2})(z+\frac{1}{2})}=\frac{1}{4}\cdot \frac{1}{z-\frac{1}{2}}-\frac{1}{4}\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(z-\frac{1}{2}{)}^n.$$

Combine into final Laurent series

Combine the series with the sine function: $$\frac{\sin (\pi z)}{4(z-\frac{1}{2})(z+\frac{1}{2})}=\sin (\pi z)(\frac{1}{4(z-\frac{1}{2})}-\frac{1}{4}\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(z-\frac{1}{2}{)}^n).$$ The first term contributes to singular part and higher terms contribute to the regular part of Laurent series.

Find the residue

The residue at $z=\frac{1}{2}$ is the coefficient of $\frac{1}{z-\frac{1}{2}}$ in the Laurent series: $$\text{Residue}=\frac{\sin (\pi \cdot \frac{1}{2})}{4}=\frac{1}{4}.$$

Key concepts

Singularity

In complex analysis, a singularity of a function is a point where the function is not defined or not analytic. For the given function $\frac{\text{sin}\text{pi z}}{4z^2-1}$, the singularity is determined by the zeros of the denominator. By solving $$4z^2-1=0$$ we find that the function has singularities at $z=\frac{1}{2}$ and $z=-\frac{1}{2}$ since these values make the denominator equal to zero. Singularity points are important because they play a key role in understanding the behavior of the function near those points. For instance, they determine the Laurent series' convergence region around the given point.

Partial Fraction Decomposition

Partial fraction decomposition breaks down a complex rational function into simpler fractions. This method is useful for integrating or expanding functions into series. For $\frac{1}{4z^2-1}$, partial fraction decomposition helps us decompose it into:

$$\frac{1}{4(z-\frac{1}{2})(z+\frac{1}{2})}=\frac{A}{z-\frac{1}{2}}+\frac{B}{z+\frac{1}{2}}$$
By solving for coefficients $A$ and $B$, we substitute values and solve equations to find:

$$A=\frac{1}{4},B=-\frac{1}{4}.$$
This simplifies the given function into:

$$\frac{1/4}{z-\frac{1}{2}}-\frac{1/4}{z+\frac{1}{2}}$$
which is easier to expand into a Laurent series.

Geometric Series Expansion

Geometric series expansion is a method to write rational functions as infinite series. This is particularly helpful when dealing with complex functions. The term $\frac{1}{z+\frac{1}{2}}$ can be written as a geometric series:

$$\frac{1}{z+\frac{1}{2}}=\frac{1}{z-\frac{1}{2}+1}=\text{thinspace thinspace sum\_{n=0}^{thinspace thinspaceinfty} (-1)^n (z - frac{1}{2})^n}$$
This transforms our expression into an infinite series around $z=\frac{1}{2}$. The terms have diminishing influence as $n$ grows, simplifying the analysis around the singularity point.

Residue

The residue of a function at a singularity is a key concept in complex analysis, especially for evaluating integrals via the residue theorem. The residue is simply the coefficient of $\frac{1}{z-z_0}$ in the Laurent series expansion of the function around $z_0$. For this function, the Laurent series at $z=\frac{1}{2}$ combines:

$$\frac{\text{sin}\text{thinspacepi z}}{4(z-\frac{1}{2})(z+\frac{1}{2})}=\text{sin}\text{thinspacepi z}(\frac{1}{4(z-\frac{1}{2})}-\frac{1}{4}\text{thinspacesum\_{n=0}^{thinspacethinspaceinfty}}(-1{)}^n(z-\frac{1}{2}{)}^n)$$
The coefficient of $\frac{1}{z-\frac{1}{2}}$ is simply:
$$\frac{\text{sin}\text{thinspacepi thinspacefrac{1}{2}}}{4}=\frac{1}{4}$$
Thus, the residue at $z=\frac{1}{2}$ is $\frac{1}{4}$.