How to construct this Laurent series? How do I construct the following Laurent series (clipped off Wolfram Alpha)?
I know that the numerator can be written as $-1+\frac{\pi}2 z-...$
Alternatively (without the Laurent series), how can I see that the pole at $1$ is of order $2$?

 A: A related problem.
Hint: Here is a start
$$ \frac{\sin\left(\frac{(z-1)\pi}{2}+\frac{\pi}{2}\right)-1}{(z-1)^4}(1+(z-1))^{-5}=\frac{\cos\left(\frac{(z-1)\pi}{2}\right)-1}{(z-1)^4}(1+(z-1))^{-5} .$$
Now, just expand the series and collect terms of powers of $z-1$.
A: By l'Hospital, check that
$$\lim_{z\to 1}f(z)\,$$
isn't finite, so you have a pole. Now check that also
$$\lim_{z\to 1}(z-1)f(z)\,$$
isn't finite, but
$$\lim_{z\to 1}(z-1)^2f(z)\,$$
is finite , so the pole has order two.
Added on request: Let us write:
$$\sin\frac{\pi}{2}z=\sin\left(\frac{\pi}{2}(z-1)+\frac{\pi}{2}\right)=\cos\frac{\pi}{2}(z-1)=1-\frac{\pi^2}{8}(z-1)^2+\frac{\pi^4}{24\cdot 16}(z-1)^4-\ldots$$
$$\frac{1}{z^5}=\frac{1}{(1+(z-1))^5}=\left(1-(z-1)+(z-1)^2-(z-1)^3+\ldots\right)^5=1-5(z-1)+\ldots$$
The first expansion above is true for any $\,z\,$, the second though only for $\,|z-1|<1\,$ , which we can assume for $\,z\,$ pretty close to $\,1\,$ , thus
$$\frac{\sin\frac{\pi}{2}z-1}{z^5(z-1)^4}=\frac{1}{(z-1)^4}\left(-\frac{\pi^2}{8}(z-1)^2-\ldots\right)\left(1-5(z-1)+\ldots\right)=$$
$$=\frac{1}{(z-1)^4}\left(-\frac{\pi^2}{8}(z-1)^2+\frac{5\pi^2}{8}(z-1)^3+\ldots\right)=-\frac{\pi^2}{8(z-1)^2}+\frac{5\pi^2}{8(z-1)}+\ldots$$
The advantage in the above is that you see $\,z=1\,$ is a double pole and also you can see the residue!
