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 $b_1 = 0$?

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.
