This is a three part question. 1) Compute a Laurent series for 1/Log(z) centered at z=1 where Log(z) is the principal branch of the log function. 1) Compute a Laurent series for $1/Log(z)$ centered at $z=1$ where $Log(z)$ is the principal branch of the log function.
2) Determine the annular region in which the Laurent expansion is valid.
3) Use this Laurent expansion to compute $Res_{s=1}\ 1/Log(z)$ and $\int_{|z-1|=1/2}1/Log(z)\ dz $
Here is what I have done so far.
For part $1$ I did the following. The function $Logz$ has a simple zero at $z=1$ and is undefined at $z=0$.
-It follows that the residue of $f$ at $1$ is $lim_{z−>1} (z−1)/(Logz)$
-Here $z−1/Logz\  =\  z−1/(Logz−Log1)$ ==> $lim_{z−>1} (z−1)/(Logz−Log1)$ ==> l'Hopital's rule, then take the limit to end up with 1. However, I am unsure if what I have done is correct, and I am not sure how to tackle part 2.
 A: Here's one way to go. We can write
$$
\log z=\log(1+(z-1))=\sum_{n=0}^\infty (-1)^n\frac{(z-1)^{n+1}}{n+1}.
$$
Therefore, 
\begin{align}
\frac{1}{\log z}&=\frac{1}{(z-1)\left[\sum_{n=0}^\infty (-1)^n\frac{(z-1)^{n}}{n+1}\right]} \\
&=\frac{1}{(z-1)\left[1+\sum_{n=1}^\infty (-1)^n\frac{(z-1)^{n}}{n+1}\right]} \\
&=\frac{1}{z-1}\frac{1}{1-\sum_{n=1}^\infty (-1)^{n+1}\frac{(z-1)^{n}}{n+1}}
\end{align}
A geometric series arises from the right term.
EDIT: To follow up, you aren't going to get a nice formula for the entire Laurent series. My method allows (if you multiply term by term or apply long division) for you to compute some early terms. The most important term is the residue, which is $1$.
For the radius, the function $f(z)=\frac{1}{\log z}$ is analytic everywhere in its domain. About the point $z=1$, the nearest singularity is at $z=0$, which is a distance of $1$. Hence the Laurent series is valid for $0<|z-1|<1$.
A: $1/\ln z=1/((z-1)-(z-1)^2/2+(z-1)^3/3-\mathcal O((z-1)^4))$.
For the Laurent series, write $g(z)\ln z=1$.
Get $g(z)=1/(z-1)+1/2-1/6(z-1)+\mathcal O((z-1)^2)$.
The annulus of convergence is $0\lt|z-1|\lt1$.
Thus the residue is $1$.
