Why does the branch cut for log(1+z), cutting away the negative axis, start at -1? 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 2πi term, on (-1,0]? Any comments are welcome. Thanks,

Messiah Sutton

Messiah Sutton

Answered question

2022-11-17

Why does the branch cut for log(1+z), cutting away the negative axis, start at -1?
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 π i term, on (-1,0]?
Any comments are welcome.
Thanks,

Answer & Explanation

martinmommy26nv8

martinmommy26nv8

Beginner2022-11-18Added 16 answers

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

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