How to construct this Laurent series?

Solution 1:

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 $z-1$.

Solution 2:

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!