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)?


  • $\begingroup$ Where is (the real) square root $\sqrt{x+1}$ defined? Certainly for some negative numbers. $\endgroup$
    – mrf
    Oct 22, 2015 at 5:50
  • $\begingroup$ Hi @mrf, but if I naively cut away all of the negative axis, then circling around, say, -.5, also makes the function jump by an argument of 2pi there, too, making the function multivalued there, and -.5 would also be considered a branch point of the square root (with argument x+1). Where am I going wrong with this thinking? Thanks, $\endgroup$
    – user282725
    Oct 22, 2015 at 6:11

1 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 $z \notin C-\{s\}$.


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