What Type of Singularity is z=1 in the Function (z^2-1)/(z-1)^2?

[csnsc14320]

Homework Statement

Say whether the indicated point is regular, an essential singularity, or a pole, and if a pole of what order it is.

$$\frac{z^2-1}{(z-1)^2}, z = 1$$

Homework Equations

The Attempt at a Solution

Right now I'm just sort of stuck on how to put this into a laurent series - I can't seem to expand the denominator in a series about 1 because I keep getting infinity :(

any hints or suggestions?

 

[Dick]

You don't have to do a full Laurent series. Just factor the numerator and denominator and simplify.

 

[csnsc14320]

You don't have to do a full Laurent series. Just factor the numerator and denominator and simplify.

i found that it reduces to $$\frac{z+1}{z-1}$$ but again don't I have to expand this and i have the same problem?

 

[Dick]

I don't believe it reduces to (z+1)/(z-1). Did you enter the problem wrong? If not, try that again and show you did it.

 

[csnsc14320]

I don't believe reduces to (z+1)/(z-1). Did you enter the problem wrong? If not, try that again and show you did it.

oops! it should read

$$\frac{z^2-1}{(z-1)^2}$$

 

[Dick]

oops! it should read

$$\frac{z^2-1}{(z-1)^2}$$

Ok, now I believe you. Now that has the form f(z)/(z-1) where f(z)=z+1. f(1) is not zero. You can classify the singularity just using that. It's also pretty easy to show the full Laurent series if you don't believe me. Write it as (2-(1-z))/(1-z).