I have never taken complex analysis, but I am preparing for a GRE for this week end and I am trying to learn a bit about Laurent Series.

So far what I get is that the Laurent Series are of form $$ \sum_{n=-\infty}^\infty a_n (z-z_0)^n = \sum_{i=1}^\infty {a_{-i}(z-z_0)^{-i}} + \sum_{j=0}^\infty {a_j(z-z_0)^i}$$

where $a_i$ is the usual Taylor coefficients and $a_{-j}$ is given by $${1\over{2\pi i}} \int_c {f(z)dz \over {(z-z_0)^{-j+1}}}$$.

I have absolutely no idea how this works, but I saw that in practice, we just manipulate the Taylor Series to get the Laurent Series some how.

For example, the Taylor expansion of $$ \frac{1}{1-z} =1+z+z^2+...$$ for $|z|<1$.

So this is what I would really like to understand. Supposedly in order to avoid the singularity at $z = 1$, now $1 \over 1-z$ must be expanded in a Laurent Series in the region $1<|z|<\infty$.

To do so the series will be manipulated as such

$$ \frac{1}{1-z} = \frac{-1}{z} \cdot \frac{1}{\left(1- \frac{1}{z} \right)} $$

Since $1<|z|$, $|1/z|<1$, so the Taylor expansion gives us $$ -\frac{1}{z} - \frac{1}{z^2} - \cdots $$ for $1<|z|$.

I only have a couple of example problems in the GRE practice book, and I failed to understand all of them. I recon that I am not getting the motivation of when to use the Taylor Series and when the Laurent.

I would also like to claim that since they both represent $1 \over {1-z}$, why do they look different and HOW are they different ?

My book also did not generalize how to manipulate the Taylor Series to make it into a Laurent series, so can someone guide me to where I could learn this a little bit more with concrete examples and details or explain this to me?

I know I am asking a lot, but mathematics means the life to me and I want to do as good as possible on the GRE.

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    $\begingroup$ Taylor for holomorphic, Laurent in presence of isolated singularities. They are different because in the practice, you must have a concept of infinite summation where negative powers of the variable appear $\endgroup$ Apr 13, 2013 at 8:46
  • $\begingroup$ Thanks! I will start by looking for what holomorphic means. $\endgroup$
    – hyg17
    Apr 14, 2013 at 4:38
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    $\begingroup$ The GRE has complex series now? $\endgroup$
    – Jwan622
    Nov 23, 2020 at 6:42

1 Answer 1


Well, the taylor series only works when your function is holomorphic, the laurent series works still for isolated singularities.

They both represent the function, but one only converges when $|z|>1$ and the other only converges when $|z|<1$.

When $f$ is holomorphic, the taylor series and the laurent series are the same, and with Cauchy's theorem you can see that. If you want to be as good as possible, you have to calculate those things on your own, that's how you learn the most.

If you want some exercises, come chat, or search google, and if you can't solve them, ask a new question.

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    $\begingroup$ Thanks! I will look up what holomorphic means and probably go on chat and ask for some more examples I can work on. $\endgroup$
    – hyg17
    Apr 14, 2013 at 4:39
  • $\begingroup$ Down vote from me. What do you mean by, "only work for your function is holomophic"? With $\vert z \vert >/< 1$ I guess you refer to his example. What's about other points $\vert z \vert = 1$ but $z≠1$? Those aren't singularities. This answer doesn't seem complete to me. $\endgroup$ Mar 29, 2017 at 20:30
  • $\begingroup$ @FelixCrazzolara already got the answer? Would you like to explain it to me? $\endgroup$
    – Unknown123
    Dec 30, 2020 at 9:32
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    $\begingroup$ @FelixCrazzolara For points |z|=1 the series doesn't converge absolute necessarily and in those cases the order in which you sum it up does matter so it won't have a relationship with the function. I don't think you can say anything general in those cases, but if you know something i would like to hear it! $\endgroup$ Mar 4, 2021 at 17:06

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