genral question about centre of Laurent series

Question

I am wondering about the following

suppose we wanted the Laurent series of $$g(z)=\frac{z-\sin z}{z^6}$$

using standard method for taylor series of sine ,

I would obtain

$$g(z)=z^{-3}/3!-z^{-1}/5!+z/7!-...$$

but now if I wanted the Laurent series of g about $z=\pi/2$

would I simply do

$$g(z)=(z-\pi/2)^{-3}/3!-(z-\pi/2)^{-1}/5!+(z-\pi/2)/7!-...$$

Or is this not a legitimate thing to do?

If not, then how would I go about such things?

But I am confused for a few reasons, the first is that there does not appear to be a singulairty at $z=\pi/2$ so would the laurent series not just be equal to the taylor series? in which case we wouldnt expect any negative term

Thanks

Answer 1

It is like you say in the last paragraph. Around a regular point (or removable singularity) the Laurent series coincides with the Taylor series; there are no negative powers.

As for the first part of the question, consider $f(z)=1/z$. The Laurent series around $z_0=0$ is $1/z$. But the Laurent expansion around $z_0=1$ is not $1/(z-1)$. It is $$ \frac1z=\frac{1}{1+(z-1)}=\sum_{n=0}^\infty(-1)^n(z-1)^n,\quad |z-1|<1. $$