Why does the branch cut for the principal branch of log(z+1) start at z=-1?

Question

If I cut away the negative real axis to make log(z+1) single-valued, why does the branch cut start at z=-1 and not at the origin z=0?

Why does the shift in argument from log(z) to log(z+1) make it ... single-valued on the real interval (-1,0)?

Thanks,

Answer 1

In general if a cut $C$ makes $f$ single valued, then $C-\{s\} = \{c-s| c \in C\}$ will make $z \mapsto f(z+s)$ single valued.

This is because $z+s \notin C$ iff $z \notin C-\{s\}$.