Laurent series of a function with $\cos^2(z)$

Question

I am having problems finding the Laurent series of

$$f(z) = \frac{z^2-\dfrac{\pi^2}{4}}{\cos^2(z)}$$ at $z=\dfrac{\pi}{2}$.

Answer 1

Putting $z=w+\pi/2$, you can rewrite this as: $$ \frac{z^2-\frac{\pi^2}{4}}{\cos^2(z)}=w(w+\pi)\csc^2(w)=g(w) $$ Notice that $\frac{\mathrm{d}}{\mathrm{d}w}(-\cot(w))=\csc^2(w)$, so we have: $$ \csc^2(w)=\frac{\mathrm{d}}{\mathrm{d}w}\sum_{n=0}^\infty\frac{(-1)^{n+1}2^{2n}B_{2n}}{(2n)!}w^{2n-1}=\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n-1)2^{2n}B_{2n}}{(2n)!}w^{2n-2} $$ Therefore: $$ g(w)=\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n-1)2^{2n}B_{2n}}{(2n)!}\left(\pi w^{2n-1}+w^{2n}\right) $$ This isn't completely explicitly in the form $\sum a_nw^n$, but it's still easy to read off all the expansion coefficients. To get back to $f$, just plug in $w=z-\pi/2$.