Let $f(z)=\frac{1}{(z-1)^2(z+1)^2}$. While trying to expand this function into the Laurent series, convergent in $P(0,1,2):=\lbrace z\in\mathbb{C}:1<|z|<2\rbrace$, a few questions popped into my mind.

  1. We can write $f(z)=\frac{1}{4}\left(\frac{1}{(z-1)^2}+\frac{1}{(z+1)^2}\right)$. Both functions inside parentheses are complex derivatives of functions which have immediate Laurent series expansion: $\frac{1}{1-z}$ and $-\frac{1}{z+1}$. Now, can we differentiate the obtained series term by term to get the desired expansion of $f$? If so, is it because the Laurent series is convergent almost uniformly?
  2. Could someone verify that the Laurent series of $f$ is convergent in $P(0,1,\infty)$?
  • $\begingroup$ Here is a related problem. $\endgroup$ Commented Feb 5, 2013 at 10:25
  • $\begingroup$ I just realized I wrote something which is not true ($f$ is not the sum of the two functions above). But maybe I won't edit the post, since this mistake led me to a question I wanted to ask anyway. $\endgroup$
    – czachur
    Commented Feb 5, 2013 at 11:56

2 Answers 2



$$z\in P(0,1,2)\Longleftrightarrow \frac{1}{2}<\frac{1}{|z|}<1\Longleftrightarrow z\in P\left(0,\frac{1}{2},1\right)$$

so using the well-known developments


we get



Can you take it from here?

Added: A different approach:



Check this thoroughly, please.

  • $\begingroup$ I got here at my first attempt, but having to deal with squares made me change the strategy. I assume I need to get rid of them in some way, right? $\endgroup$
    – czachur
    Commented Feb 5, 2013 at 12:15
  • $\begingroup$ No if what you want is only some few terms of the Laurent expansion, otherwise you'll have to look for some patter, which I'm not sure it exists. $\endgroup$
    – DonAntonio
    Commented Feb 5, 2013 at 12:29
  • $\begingroup$ @czachur, please do check the added part to my answer. This may work nice. $\endgroup$
    – DonAntonio
    Commented Feb 5, 2013 at 12:41

Yes, it's ok to differentiate a Laurent series termwise, since it converges locally uniformly on its annulus of convergence (and the termwise differentiated series is a new Laurent series with the same annulus of convergence).

2) Yes, since the only poles of $f$ are at $z = \pm 1$. (I.e., $f$ is holomorphic on every annulus $1 < |z| < r$ for $r > 1$, and on that annulus you will get a convergent Laurent series.)


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