1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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)?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.