= Solution
On the stated branch of the <principal complex logarithm>, the only denominator zero is $z=1$, and $\log z=(z-1)+O((z-1)^2)$. Therefore
$$
\boxed{z=1\text{ is a simple pole, with residue }1.}
$$
Zero is a logarithmic <branch point>, not a removable isolated singularity, even though $1/\log z$ tends to zero as $z\to0$ within the cut plane. Continuation around zero changes $\log z$ by $2\pi i$. The nonpositive real axis is the chosen <branch cut>; its nonzero points are not intrinsically branch points and admit local analytic continuation. There are no other finite singularities inside the specified branch domain.
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