Branch points and natural maximal domain of log(1+z)/(1-z)

Bruce Sherman

Bruce Sherman

Answered question


Branch points and natural maximal domain of log 1 + z 1 z
So I have a function log 1 + z 1 z and I'm supposed to find its branch points, natural maximal domain.
So far I converted z to x + i y but no real good. Can't check holomorphicity using this. I know that log is holomorphic after a branch cut. How do I proceed?
I also conclude that 1 + z 1 z is holomorphic as it doesn't contain z ¯

Answer & Explanation

Houston Ellis

Houston Ellis

Beginner2022-10-04Added 4 answers

We can't allow 0 or in log, so the points z = ± 1 are out.
Also: the complex logarithm is multi-valued, acquiring a multiple of 2 π i as its argument moves along a closed curve separating 0 from . If you don't want this to happen, you must make sure that your domain has no closed curves separating 1 from 1There is no canonical way of making this happen. Cutting out the line segment from −1 to 1 is one natural choice; there is another one with cuts along the real axis (I leave it for you to find it), and plenty of others, which are less nice.


Beginner2022-10-05Added 2 answers

We can define
(1) log ( 1 + z 1 z ) = π 2 i + i z ( 1 w + 1 1 w 1 ) d w
Since the residues at + 1 and 1 cancel, the integral along any closed path that circles them both an equal number of times will be 0. The smallest branch cut which forces any closed path to circle both an equal number of times is the segment [ 1 , + 1 ]. Thus the largest natural domain would be C [ 1 , + 1 ]

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