Branch of logarithm which is real when z>0 I am familiar with the complex logarithm and its branches, but still this confuses me. I read this in a textbook: "For complex z!=0,log(z) denotes that branch of the logarithm which is real when z>0.What does this mean?

Amiya Melendez 2022-10-13 Answered
Branch of logarithm which is real when z>0
I am familiar with the complex logarithm and its branches, but still this confuses me. I read this in a textbook:
"For complex z 0 , l o g ( z ) denotes that branch of the logarithm which is real when z > 0
What does this mean?
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Answers (2)

Carly Yang
Answered 2022-10-14 Author has 19 answers
It doesn't mean anything unless you specify the region G you want to define the complex logarithm on. The statement seems to try to do this with G = C { 0 }, but that is impossible: In general, the complex logarithm is only defined up to 2 k π, k Z , and if your G winds around 0, you cannot make the function continuous. Instead, G could be any simply connected subset of C { 0 }, and the usual choice is to remove the nonpositive reals from C , i.e. G = C { , 0 ]. With this (or any other simply connected G that contains the positive reals), it is possible to pick the branch that is real (i.e. is the well-known real logarithm) for positive reals.
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Maribel Mcintyre
Answered 2022-10-15 Author has 3 answers
Let Log z = ln | z | + i Arg z be the principal branch of the logarithm, that corresponds to a cut along the negative real numbers. Now, consider any branch of l o g that is defined on a connected neighborhood of R + = ( 0 , + ). Clearly, x φ ( x ) = l o g ( x ) Log ( x ) is a continuous function on R + that satisfies exp ( φ ( x ) ) = 1 so, φ takes its values in 2 π i Z , and since it is continuous, it must be constant. Thus, there is k Z such that l o g ( z ) = Log  z + 2 i π k and this happens on the largest connected open set that contains R + = ( 0 , + ), and on which both functions are analytic (by analytic continuation).
Thus, the statement says that we choose the logarithmic function the corresponds to k = 0 because otherwise l o g ( z ) would not be real for real z
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