Laurent Series Expansion of $\frac{1}{z^2-1}$

In summary, the conversation discusses calculating the Laurent series expansion for the function 1/(z^2-1) at three different points (z=1, z=-1, z=0). The singularity is classified as removable for z=0 and as poles for z=1 and z=-1. The region of convergence is determined to be -1 < |z| < 1. The conversation also includes a discussion on how to express the function as a power series in terms of (z-1) by manipulating the denominator.
  • #1
gtfitzpatrick
379
0

Homework Statement



Calculate the laurent series expansion about he points specified, classify the singularity and sate the region of convergence for.

[itex]\frac{1}{z^2 - 1}[/itex] at (i) z=1 (ii) z=-1 (iii)z=0

Homework Equations





The Attempt at a Solution



[itex]\frac{1}{z^2 - 1} = \frac{1}{2(z-1)} - \frac{1}{2(z+1)}[/itex]

i observe that f(z) is analytic in {z≠ -1,1} so is anyalytic in -1 < [itex]\left|z\right| < 1[/itex]

@z=1 [itex]\frac{-1}{2}\frac{1}{1-(-z)} = \frac{-1}{2}\sum (-z)^n[/itex]

similarly

@z=-1 [itex]\frac{1}{2}\frac{1}{(z-1)} = \frac{-1}{2}\frac{1}{1-z)} = \frac{-1}{2}\sum (z)^n[/itex]

and

@z=0 is it just the sum of both of them? singularities are at 1, -1 and region of convergence -1 < [itex]\left|z\right| < 1[/itex]

am i anywhere near?
 
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  • #2
gtfitzpatrick said:

Homework Statement



Calculate the laurent series expansion about he points specified, classify the singularity and sate the region of convergence for.

[itex]\frac{1}{z^2 - 1}[/itex] at (i) z=1 (ii) z=-1 (iii)z=0

Homework Equations


The Attempt at a Solution



[itex]\frac{1}{z^2 - 1} = \frac{1}{2(z-1)} - \frac{1}{2(z+1)}[/itex]

i observe that f(z) is analytic in {z≠ -1,1} so is anyalytic in -1 < [itex]\left|z\right| < 1[/itex]

@z=1 [itex]\frac{-1}{2}\frac{1}{1-(-z)} = \frac{-1}{2}\sum (-z)^n[/itex]

similarly

@z=-1 [itex]\frac{1}{2}\frac{1}{(z-1)} = \frac{-1}{2}\frac{1}{1-z)} = \frac{-1}{2}\sum (z)^n[/itex]

and

@z=0 is it just the sum of both of them? singularities are at 1, -1 and region of convergence -1 < [itex]\left|z\right| < 1[/itex]

am i anywhere near?

No. When you expand around a point z0, then the series should be in powers of (z-z0). So for the first one, the first partial term is already in powers of z-1. However for the second term:

[tex]-1/2 \frac{1}{z+1}[/tex]

that's not. But you can do so by forcing a factor of z-1 in the denominaor by:

[tex]\frac{1}{z+1}=\frac{1}{z-1+2}=\frac{1}{2+z-1}|=\frac{1}{2(1+\frac{z-1}{2})}[/tex]

which you can then easily express that as a power series in terms of (z-1) right? Same dif with the other ones.
 
  • #3
thanks a million jackmell
 

1. What is a Laurent series expansion?

A Laurent series expansion is a representation of a complex function as an infinite sum of powers of z, including negative and fractional powers. It is similar to a Taylor series expansion, but it also takes into account the singularities of the function.

2. Why is the Laurent series expansion important?

The Laurent series expansion is important because it allows us to represent a complex function in a simpler form, making it easier to analyze and manipulate. It also helps us to understand the behavior of the function near its singularities.

3. How do you find the Laurent series expansion of a function?

To find the Laurent series expansion of a function, we first determine the singularities of the function and the region in which it is valid. Then, we use the coefficients of the series to represent the function in terms of powers of z and its singularities. These coefficients can be found by differentiating and integrating the function, and using the Cauchy integral formula.

4. What is the Laurent series expansion of $\frac{1}{z^2-1}$?

The Laurent series expansion of $\frac{1}{z^2-1}$ is $\frac{1}{2z}+\frac{1}{4z^2}+\frac{1}{8z^3}+...$ for $|z|>1$, and $-\frac{1}{2z}-\frac{1}{4z^2}-\frac{1}{8z^3}-...$ for $|z|<1$. This can be obtained by using the geometric series formula and expressing the function as $\frac{1}{z-1}-\frac{1}{z+1}$.

5. What is the region of convergence for the Laurent series expansion of $\frac{1}{z^2-1}$?

The region of convergence for the Laurent series expansion of $\frac{1}{z^2-1}$ is the annulus $1<|z|<\infty$. This means that the series will converge for all values of z that lie within this annulus. Outside of this region, the series will diverge.

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