Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$ 
Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$ in $\{|z|>1\}$. 

Ok, so for $|z|>1 \iff |\frac{1}{z}|<1$ I can write
$\frac{1}{z-1}=\frac{1}{z}\cdot \frac{1}{1-\frac{1}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac{1}{z})^n=\frac{1}{z}\sum_{n<0}z^n$
Which is the Laureant Series for $\frac{1}{z-1}$.
But I don't know what to do with the $e^z$ multiplying. How should I proceed?

Looks like to me we can just do this, 
\begin{align}
\frac{e^z}{z-1} &= \frac{1}{z-1}\sum_{n \geq 0} \frac{e(z-1)^n}{n!}.\\
&=\frac{e}{z-1}\left(1 + (z-1) + \frac{1}{2!}(z-1)^2 +  \frac{1}{3!}(z-1)^2 +\dots \right) \\
&=e\left(\frac{1}{z-1} + 1 + \frac{1}{2!}(z-1) +  \frac{1}{3!}(z-1)^2 +\dots \right)
\end{align}
So it suffices to find the sum of coefficient of $z$ in $(z-1)^n$, that is
\begin{align}
\sum_{n \geq 1} \frac{1}{(n+1)!} \binom{n}{n-1} (-1)^{n -1} &=\sum_{n \geq 1} \frac{n}{(n+1)!} (-1)^{n-1}\\
&=\sum_{n \geq 0} \frac{n+1}{(n+2)!} (-1)^{n}\\
&=\sum_{n \geq 0} \frac{1}{(n+2)n!} (-1)^{n}\\
&=\frac{e - 2}{e}.
\end{align}
Therefore the coefficient of $z$ is $e-2.$
