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Definition(Branch): A branch of a multiple-valued function $f$ is any single-valued function $F$ that is analytic in some domain at each point $z$ of which the value $F(z)$ is one of the possible values of $f.$

This is the definition of branch of multi valued functions.

Clearly $$Log(z)=ln(r)+i\theta (r>0,-\pi<\theta<\pi)$$ is a branch of the multi valued function $log(z).$

Now my question is according to the above definition $$f(z)=ln(r)+i\theta (r>0,0<\theta<\pi)$$ is also a branch of the multi valued function $log(z)?$ I am thinking so because $f$ is also single valued analytic(in upper half domain) assuming exactly one of various possible values of $Log.$ Am i right? Please suggest me. Thanks in advance.

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    $\begingroup$ I think "some domain" is intended to imply an open set. So if you said $0 < \theta < \pi$ then it would be a "branch" of $\log z$.. $\endgroup$ – GEdgar Oct 23 '15 at 12:30
  • $\begingroup$ but is is not usually consider as branch...i don't know why. $\endgroup$ – neelkanth Oct 23 '15 at 12:46
  • $\begingroup$ Definition(Branch): A branch of a multiple-valued function $f$ is $\color{teal}{any}$ single-valued function $F$ that is $\underline {analytic}$ in some domain at each point $z$ $\color{teal}{of \ which}$ the value $F(z)$ is one of the possible values of $f.$ here I don't understand the word analytic This is the definition of branch of multi valued functions. Why does $$Log(z)=ln(r)+i\theta (r>0,-\pi<\theta<\pi)$$ is a branch of the multi valued function $log(z)?$ Is $ln(r)+i\theta $ a possible value of $log(z)$? $\endgroup$ – IggyPass Oct 23 '15 at 13:09
  • $\begingroup$ yes $Log(z)$ is one of the infinite values of $log(z).$ $\endgroup$ – neelkanth Oct 23 '15 at 13:11
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Yes, you're correct. It fulfills all the requirements of the definition.

The definition does not put any requirements of the domain in which the branch is defined. For example this becomes handy if you analytically extend the real analytical function $\ln(1+x) = \sum x^n/n$, it would converge in a unit disc around $0$ which makes the McLaurin series a branch of $\ln(1+x)$

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  • $\begingroup$ for this whole real line will be our branch cut. Am i right? $\endgroup$ – neelkanth Oct 23 '15 at 13:06
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    $\begingroup$ @neela No my example is a branch of $ln(1+z)$ with only the unit disc as domain. The point is that one is allowed to exclude much more than minimally required which sometimes is practical. $\endgroup$ – skyking Oct 23 '15 at 13:39

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