<s>[INST] Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$ How does one find the first four terms of the Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$? My approach was this:
$(z^2+1) = f(z)\sin(z) = \left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\sin(z)=\left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\left(z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040} \right)$.
How do we proceed further?
 [/INST] For example, very close to zero we have
$$\frac{z^2+1}{\sin z}=\frac{z^2+1}{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots}=\frac{z^2+1}{z\left(1-\frac{z^2}6+\frac{z^4}{120}-\ldots\right)}\stackrel{\text{Devel. of geom. series}}=$$
$$=\left(z+\frac1z\right)\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)=\frac1z+\frac76z+\frac7{36}z^3+\ldots$$
With the above you already have the pole at $\;z=0\;$ is a simple one and its residue is $\;1\;$ .
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