Say we want to calculate the Laurent series of $\tfrac{1}{z^2(z-1)}$ about $z_0=0.$ Now I know that one way to do it is to say that $f(z)=\tfrac{1}{z^2}(\tfrac{1}{z-1})$ and appy the geometric series expansion to the brackets term. But I wanted to try and do it a different way :

First we split f into partial fractions and compute the Laurent series separately.Now consider the Laurent expansion of $\tfrac{1}{z^2}$

We know that $0$ is a pole of order 2 which implies that $\forall n>2,a_{-n}=0$. Therefore $\tfrac{1}{z^2}$ haa series expansion $\tfrac{a_{-2}}{(z-z_0)^2}+\tfrac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+...$

Now to calculate the the $a_{-2}$ coefficient I applied the following trick

$a_n=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{f(z)}{(z-z_0)^{n+1} }dz=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+3}}dx=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{z^{n+3}}dz$

$a_{n-2}=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+1} }dz=\tfrac{f^n(z_0)}{n!}$ But as n $f(z)=1$ this implies that $n$ must be zero and so $a_{-2}=1$

Now when I tried to use the same trick on $a_{-1}$ It doesn't work because now we can't use Cauchy's formula. Also when I tried to use u substitution by letting $u=z-z_0$ it returns that $a_{-1}=-\tfrac{1}{z}$ but I know this is not rue as I know from the method that I mentioned in the first paragraph that $a_{-1}=1$ So does anyone have any suggestions on how I can find $a_{-1}$ continuing with the method I'm trying to use ?

Note : originally this had a typo that said expansion about 1 , it should always have read expansion about zero.


You're expanding around $z=1$. In a neighborhood of that point, $\frac1{z^2}$ is an analytic function. There won't be any negative-degree terms coming from that part - only the pole at $1$ can contribute anything of negative degree to the expansion.

Although - a Laurent series requires more than just a point to specify it. It requires a band. In this case, two possibilities - do you want the nearby series for $0<|z-1|<1$ (negative-degree terms from $\frac1{z-1}$, positive degree terms from $\frac1z$), or the distant series for $1<|z-1|$ (all negative-degree terms)?

  • $\begingroup$ That was a typo actually . It was meant to be an expansion around 0. sorry about that ... $\endgroup$ – excalibirr Jan 13 at 16:50

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