Find the laurent series expansion of $f(z)$ at $z=0$

Question

Let $$f(z)={1\over \sin z}-{1\over z}+{2z \over z^2-\pi^2}$$

Find the $z^{-3},z^{-2},..., z^3$ term of the laurent series expansion for $f(z)$ at $z=0$

I found $${1\over \sin z}-{1\over z}=\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots$$

Also $${2z \over z^2-\pi^2}={1\over z- \pi}+{1\over z+ \pi}$$

Is it true that $${1\over z- \pi}=-{1\over \pi}({1\over 1- z/\pi})=-{1\over \pi}\sum_{k=0}^\infty ({z\over \pi})^k$$ at $z=0$?

Answer 1

Since we're interested in $\;|z|\;$ close to zero, we can try the geometric series:

$$\frac1{\sin z}-\frac1z-\frac2{\pi^2}\frac z{1-\left(\frac z\pi\right)^2}=\frac1{z\left(1-\frac{z^2}6+\mathcal O(z^4)\right)}-\frac1z-\frac2{\pi^2}z\left(1+\frac{z^2}{\pi^2}+\mathcal O(z^4)\right)=$$$${}$$

$$=\frac1z\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)-\frac1z-\frac2{\pi^2}z-\frac2{\pi^4}z^3+\mathcal O(z^5)$$

You can see the principal part is zero...as expected, since zero is a removable singularity:

$$\lim_{z\to0}\left(\frac1{\sin z}-\frac1z\right)=0$$

and the last summand is analytic at zero.