Why does the branch cut for log(1+z), cutting away the negative axis, start at -1?

Question

I am really confused about this concept.

Is it because from -1 to 0, the inputs (1+z) are just viewed as inputs that aren't actually from the negative axis at all, and so log(1+z) is well-defined and continuous on (-1,0]? I don't see the single-valuedness on (-1,0], though. Shouldn't log(1+z) jump by a $2\pi i$ term, on (-1,0]?

Any comments are welcome.

Thanks,

Answer 1

Let $w=1+z$. Now, the branch cut for $\log w$ starts at $w=0$. So, the branch cut for $\log(1+z)$ starts where $1+z=0$. In other words, where $z=-1$.