Finding the laurent series for $e^{\sin(1/z)}$ I began with the expansion of $e^z$, which is  \begin{eqnarray}  e^z\ = \sum_{n=0}^\infty \frac{z^n}{(n)!}  \end{eqnarray}
Then I found the expansion for sin(1/z), which is  \begin{eqnarray} \sin\left(\frac{1}{z}\right) = \sum_{n=0}^\infty (-1)^n \frac{z^{-2n-1}}{(2n+1)!}  \end{eqnarray}
But now I don't know how to go further?
Should I just substitute the first four values for $\sin\left(\frac 1 z \right)$ into the formula for $e^z$?

Substitute $\sin( \frac{1}{z})=\frac{1}{z}-\frac{1}{6z^3}+\cdots$ into the series for $e^z$ ... we have 
\begin{eqnarray*}
e^{\sin( \frac{1}{z})}=1+(\frac{1}{z}-\frac{1}{6z^3}+\cdots)+\frac{1}{2}(\frac{1}{z}-\frac{1}{6z^3}+\cdots)^2+\frac{1}{6}(\frac{1}{z}-\frac{1}{6z^3}+\cdots)^3 +\frac{1}{4!}(\frac{1}{z}+\cdots)^3 \cdots
\end{eqnarray*}
Expanding & collecting terms ... we have
\begin{eqnarray*}
e^{\sin( \frac{1}{z})}=1+\frac{1}{z}+\frac{1}{2}\frac{1}{z^2}+\frac{1}{z^3}(-\frac{1}{6}+\frac{1}{6})+\frac{1}{z^4}(\frac{1}{4!}-\frac{2}{2 \times 6} )+\cdots
\end{eqnarray*}
So the first $4$ non zero terms are 
\begin{eqnarray*}
e^{\sin( \frac{1}{z})}=1+\frac{1}{z}+\frac{1}{2}\frac{1}{z^2}-\frac{1}{8}\frac{1}{z^4}+\cdots
\end{eqnarray*}
