Type of singularity of $\log z$ at $z=0$ What type of singularity is $z=0$ for $\log z$ (any branch)?
What is the Laurent series for $\log z$ centered at 0, if exist? If the Laurent series has the form $\sum_{k=-\infty}^{\infty} a_kx^k$, then certainly among $a_{-1},a_{-2},...,a_{-j},...$, at least one is nonzero (or otherwise $\log z$ would be analytic at $0$). Since $\lim_{z\to 0}z\log z=0$, we must have $a_{-j}=0$ for all $j>0$, a contradiction. Hence the Laurent series centered at $0$ cannot exist.
Is the singularity at $0$ a pole? If so, what is its order? Thanks.
 A: The singularity is not a pole, since Logz (or at least its real part) does not blow up to $\infty$ as you approach 0. Maybe the best name is that $0$ is a branch point for Logz;  and it is a branch point of infinite index. The branch point tells you that if you wind around the unit circle a point once, i.e., if you wind around by a value of $2\pi$, you do not ever return to your initial value  , i.e., if you consider the values of {$Log(z+2n\pi)$} for $n=1,2,3,...$, these are all different values. Compare and contrast this with the case for
the n-th root $z^{1/n}$ . Here $0$ is also a branch point for $z^{1/n}$ , but this time it is of index n, because $e^{i\theta/n}=e^{i\theta+2n\pi/n}$ , i.e., you return to the original value of your function after looping n times.
EDIT: As pointed out in the comments, I was wrong in my claim that |Logz| does not go to $\infty$ as $z\rightarrow 0$ ; it does, since lnx does blow up near $0$
A: The singularity of $\log z$ is not an isolated singularity, so the usual classification into, pole, essential, or removable does not apply. In particular, there is no Laurent expansion about 0 and you cannot apply residue theory. 
In this case the singularity is known as a branch point, and it is the typical example.
A: $\log{z}$ is viewed as a branch point due to its multivaluedness.  That is, $\log{z}$ is only determined to within an integer multiple of $i 2 \pi$.  $\log{z}$ is unique within a single branch; that is, as long as a contour along which $\log{z}$ is uniquely defined does not cross a branch cut that has been defined.  That $\log{z}$ blows up as $z \to 0$ is beside the point; you simply do not include branch points such as those for which the argument of the log function is zero because of the nonuniqueness of the log near there.  
