<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] Your approach is correct, but without determining the order of the pole $z=0$ you are stuck because each coefficient of $f \cdot \sin$ will, in principle, be a sum with infinitely many terms. Finding the order of the pole of $f$ will allow us to kill infinitely many of these, leaving only finitely many of them, as you are going to see.
Remember that the Taylor series of $\sin$ around $z=0$ begins with $z^1$, so $0$ is a zero of order $1$ for $\sin$, therefore it is a pole of order $1$ for $\frac 1 \sin$.
For even greater clarity, remember that $z_0$ is a pole of order $n>0$ for $f$ if and only if $n$ is smallest with the property $\lim \limits _{z \to z_0} (z-z_0)^n f(z) \ne 0$. Let's check this for your $f$ and $n=1$:
$$\lim _{z \to 0} z \ \frac {z^2 + 1} {\sin z} = \lim _{z \to 0} \frac z {\sin z} (z^2 + 1) = \lim _{z \to 0} \frac z {\sin z} \lim _{z \to 0} (z^2 + 1) = 1 \cdot 1 = 1 .$$
Is $n=1$ the smallest such strictly positive number? Yes, because below it there is only $0$, and $0$ is not allowed to be the order of a pole, by definition.
Having found the order of the pole of $f$ is essential, since this allows us to restrict the lower bound for the summation index:
$$f(z) = \sum _{n = -1} ^\infty c_n z^n$$
(notice that $n$ starts now at $-1$, not at $-\infty$).
Plugging this into your own approach (and keeping only terms up to and including $z^3$, because you want only the first 4 terms) gives
$$z^2 + 1 = \left( \sum _{n =-1} ^\infty c_n z^n \right) \left( z - \frac {z^3} {3!} + \frac {z^5} {5!} - \frac {z^7} {7!} + \dots \right) = \\
\left( c_{-1} - \frac {c_{-1}} {3!} z^2 + \dots \right) + \left( c_0 z - \frac {c_0} {3!} z^3 + \dots \right) + \left( c_1 z^2 + \dots \right) + \left( c_2 z^3 + \dots \right) + \dots ,$$
so equating the coefficients of the corresponding powers of $z$ in both hands of the equation gives
$$c_{-1} = 1, \quad c_0 = 0, \quad -\frac {c_{-1}} 6 + c_1 = 1, \quad -\frac {c_0} 6 + c_2 = 0 ,$$
whence it follows that
$$c_{-1} = 1, \quad c_0 = 0, \quad c_1 = \frac 7 6, \quad c_2 = 0 ,$$
so
$$f(z) = \frac 1 z + \frac 7 6 z + \dots .$$
Notice that there is no surprise in the fact that $c_0 = c_2 = 0$, because
$$f(-z) = \frac {(-z)^2 + 1} {\sin (-z)} = \frac {z^2 + 1} {-\sin z} = -f(z) ,$$
so $f$ is odd, and the coefficients of even order of an odd function are always $0$.
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