I have some serious problems with Laurent series. I don't know how to manipulate even the easiest series because I don't understand where to do what.

I know what a Laurent series is and i know how to calculate the radius of convergence and use them and everything. It so frustrated that I don't get how to do the right manipulation.

Look at this one


with singularities at $z=-1$ and $z=0$,

I want to find the Laurent series for the neighborhoods $0<|z|<1$ and $1<|z|$

why is the manipulation for the first one $\frac{1}{z}\frac{1}{1-(-z)}$ and $\frac{1}{z^2}\frac{1}{1-(-\frac{1}{z})}$ for the second one?

As soon as I get the right manipulation stated above I can do everything else by my own but simply don't understand it, and this is a really easy one :(


2 Answers 2


Remember that the expansion

$$\frac1{1-z} = \sum_{n=0}^{\infty} z^n$$

is only valid when $|z| \lt 1$. So each region defined will require manipulation in order to get something that may be expanded likewise. For instance,

$$\frac1{z+z^2} = \frac1{z} \frac1{1+z} $$

so that when $|z| \lt 1$, we may expand the $1/(1+z)$ term as above. However, when $|z| \gt 1$, $1/|z| \lt 1$ and we must factor out the $z$ from the $1/(1+z)$ term in order to be able to expand as above:

$$\frac1{z+z^2} = \frac1{z} \frac1{z} \frac1{1+\frac1{z}} = \frac1{z^2} \frac1{1+\frac1{z}}$$

Now we can expand the one over the binomial piece.

  • $\begingroup$ So if we expand around another point $z=-1$ we need to make sure that $|z+1|<1$ if we wanna do it like the first expression you wrote first. That would be in the neighborhood of radius one around the point $z=-1$, right? And outside this we need to manipulate it so we have it on the latter form? $\endgroup$
    – user269620
    Oct 10, 2015 at 13:43
  • $\begingroup$ Really helpful answer - 1/(1-z) is only valid for |z|<1. I understood the derivations from Taylor series to Laurent series, but I just couldn't understand why they were writing the denominator in different forms, thinking about being inside/outside the curves of the annulus... this is the missing link in my text. Thanks!! $\endgroup$
    – amelia
    Apr 8, 2016 at 15:23

The Laurent series around $z = a$ should look like

$$\sum_{n \ge -m} c_n (z - a)^n = c_{-m} (z - a)^{-m} + c_{-m + 1} (z - a)^{-m + 1} + \dotsb$$

This you can manufacture as

$$\frac{f(z)}{(z - a)^m}$$

where $f(z)$ has a "regular" power series.


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