Type of singularity of $\log z$ at $z=0$

Solution 1:

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.

Solution 2:

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 $0$

Solution 3:

$\log{z}$ 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.