Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function  $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$? 
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}} 
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots )\\
1& = \Big ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big ) \cdot \Big ( z^3 \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big )\\
\end{align*}
After some more thought...
Is it true to say that because f(z) has a pole of order 4 at $z=0$ that our $b_n$'s only go out to the 4th term? Meaning there are no $b_5$, $b_6$, etc like how wrote previously. That is,
$\frac{1}{z^3 \sin{(z)}} 
= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big )\\$
followed by 
\begin{align*}
1 
&= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots\Big )\\
\end{align*}
Which then when multiplying out $b_1z^2$ with each term from sin(z)'s Laurent expansion will never yield a $\frac{1}{z}$ term, concluding that the coefficient $b_1 = 0$?
 A: The function is even so the residue at $z = 0$ is zero, since the residue is the coefficient of ${1 \over z}$ in the Laurent expansion.
A: It's much simpler than what you do, using asymptotic analysis:
\begin{align}
\frac1{\sin z}&=\frac 1{z-\cfrac{z^3}6+\cfrac{z^5}{120}+O(z^7)}\\
&=\frac1z\cdot\frac1{1-\biggl(\underbrace{\cfrac{z^2}6-\cfrac{z^4}{120}+O(z^6)}_{=\,u}\biggr)} \\
&=\frac1z\biggl[1+
\cfrac{z^2}6-\cfrac{z^4}{120}+\biggl( \cfrac{z^2}6-\cfrac{z^4}{120}\biggr)^{\!\!2}+O(z^6)\biggr] \\
&=\frac1z\biggl[1+
\cfrac{z^2}6-\cfrac{z^4}{120}+ \cfrac{z^4}{36}+O(z^6)\biggr] \\
&=1+ \cfrac{z^2}6+\frac{7z^4}{360}+O(z^6)\\
\text{so that }\qquad
\frac1{z^3}\frac1{\sin z}&=\frac1{z^4}\biggl[1+
\cfrac{z^2}6+ \cfrac{7z^4}{360}+O(z^6)\biggr] \\
&=\frac1{z^4}+\frac1{6z^2}+ \cfrac{7}{360}+O(z^2).
\end{align}
Finally, $\;\operatorname{Res}(f,0)=0$.
In this case, it could have been anticipated: the function $\dfrac1{z^3\sin z}$ is even, and therefore, its Laurent  expansion around $0$ has only terms of even degree, so $a_{-1}=0$.
A: Hint. The function 
$$
z \mapsto \frac{1}{\sin(z)}-\frac{1}{z}
$$ is regular and odd near $0$ thus the residue of 
$$
f(z) = \frac{1}{z^3 \sin(z)}
$$is equal to zero.
A: I came up with the solution my professor was looking for and figured I would share it here for both my own clarification and for anyone else in the future who wants to find the solution via a Laurent expansion.
For our function $f(z) = \frac{1}{z^3 \sin{(z)}}$, we can solve for the residue of $f$ at $z = 0$ by doing the following:
We know that this is a pole of order 4, as $z^3$ has order 3 at $z=0$ and $\sin{(z)}$ has a pole of order 1 (simple pole) at $z=0$. So our general Laurent series for $$\frac{1}{z^3 \sin{(z)}} = \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + \dots$$ We want to solve for $b_1$, which is our residue.
We can do some algebra and some known Laurent expansions to get
\begin{align*}
\frac{1}{z^3 \sin{(z)}} &= \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + \dots\\
1 &= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + \dots \Big ) \cdot \Big (z^3 \sin{(z}) \Big )\\
1 &= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + \dots \Big ) \cdot \Big (\sin{(z}) \Big )\\
1 &= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + \dots \Big ) \cdot \Big (z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big)\\
\end{align*}
Gathering all the constant terms from the left and right sides of the equation, we have
\begin{align*}
1 &= \frac{b_4}{z} \cdot z\\
& = b_4
\end{align*}
So $b_4 = 1$.
Similarly, gathering all the z terms from the left and right sides, we have
\begin{align*}
0z &= b_3 \cdot z\\
0 &= b_3z
\end{align*}
So $b_3 = 0$.
Remembering that $b_4 = 1$ and gathering all the $z^2$ terms from the left and right sides, we have
\begin{align*}
0z^2 &= \frac{b_4}{z} \cdot \Big (-\frac{z^3}{3!} \Big) + b_2z \cdot z\\
0 &= -\frac{b_4}{3!}z^2 +b_2z^2\\
0 &= \Big (-\frac{1}{3!} + b_2 \Big )z^2\\
\implies  b_2&= \frac{1}{3!}
\end{align*}
So $b_2 = \frac{1}{6}$.
Remembering that $b_3 = 0$ and gathering all the $z^3$ terms, we have
\begin{align*}
0z^3 &= -\frac{b_3}{3!}z^3 + b_1z^3\\
\implies 0 &= -\frac{0}{3!} + b_1\\
\implies b_1 &= 0
\end{align*}
Thus the residue of $f(x) = \frac{1}{z^3 \sin{(z)}} = b_1 = 0$
A: $\displaystyle z \mapsto -z$ does not change 
$\displaystyle {1 \over z^{3}\sin\left(z\right)}.\quad$ So,
$\displaystyle {1 \over z^{3}\sin\left(z\right)}
\,\,\,\stackrel{\mathrm{as}\ z\ \to\ 0}{\sim}\,\,\, {b_{4} \over z^{4}}
+ {b_{2} \over z^{2}} + {\color{red}{\large 0} \over z} + a_{0}$ 
