Singularities and Laurent series

[Physgeek64]

Homework Statement

Classify the singularities of

##\frac{1}{z^{1/4}(1+z)}##

Find the Laurent series for

##\frac{1}{z^2-1}## around z=1 and z=-1

Homework Equations

The Attempt at a Solution

So for the first bit there exists a singularity at ##z=0##, but I'm confused about the order of this since its fractional (I get that its a branch point, but does it 'act' as a singularity as well?)

And there is also a pole at ##z=-1## of order 1... ?

(My problem with this is if I expand the ##\frac {1}{z^{\frac{1}{4}}}## about z=0 I get ##\frac{e^{\frac{i \pi}{4}}}{1+z} (1+\frac{1}{4}(z+1)+...)## and I don't know why these should be different?/ which one is wrong)

And as for the Laurent series I'm afraid I am completely stuck.

Many thanks- I really appreciate the help as I'm really struggling with these aspects

 

[mighty2000]

of complex analysis.
Thank you for your post. I am happy to assist you with your questions. Let's start with the first part of your post.

In order to classify the singularities of ##\frac{1}{z^{1/4}(1+z)}##, we first need to look at the denominator ##z^{1/4}(1+z)##. This function has two singularities: ##z=0## and ##z=-1##. Let's examine each of these singularities separately.

At ##z=0##, we have a branch point. This is because the function is not defined at ##z=0##, but as we approach this point from different directions, we get different values. In this case, as we approach from different directions, we get different values for the fourth root of ##z##, which is why we call this a branch point.

At ##z=-1##, we have a simple pole. This is because the function is not defined at ##z=-1##, but as we approach this point, the function blows up to infinity. This is a pole of order 1, since the highest power of ##z## in the denominator is 1.

Now, for the second part of your post, we need to find the Laurent series for ##\frac{1}{z^2-1}## around ##z=1## and ##z=-1##. First, let's look at the expansion around ##z=1##.

We can rewrite ##\frac{1}{z^2-1}## as ##\frac{1}{(z-1)(z+1)}##. Now, we can expand this using the geometric series expansion:

##\frac{1}{z^2-1} = \frac{1}{(z-1)(z+1)} = \frac{1}{2(z-1)} \cdot \frac{1}{1 + \frac{z-1}{2}} = \frac{1}{2(z-1)} \cdot \sum_{n=0}^{\infty} \left(-\frac{z-1}{2}\right)^n##

##= \frac{1}{2(z-1)} \cdot \sum_{n=0}^{\infty} (-1)^n \frac{(z-1)^n}{2^n} = \sum_{