Find a branch of f(z)=\log(z^3−2) that is analytic at z=0.

garnentas3m 2022-01-03 Answered
Find a branch of \(\displaystyle{f{{\left({z}\right)}}}={\log{{\left({z}^{{3}}−{2}\right)}}}\) that is analytic at \(\displaystyle{z}={0}\).

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Expert Answer

redhotdevil13l3
Answered 2022-01-04 Author has 5471 answers
Note that, \(\displaystyle{z}={0}\) is not a branch point of \(\displaystyle{f{{\left({z}\right)}}}\). To find the branch points of \(\displaystyle{f{{\left({z}\right)}}}\), solve the equation
\(\displaystyle{z}^{{3}}-{2}={0}\Rightarrow{z}^{{3}}={2}{e}^{{{2}{k}\pi{i}}}\Rightarrow{z}={2}^{{\frac{{{1}}}{{{3}}}}}{e}^{{{\frac{{{2}{k}\pi{i}}}{{{3}}}}}},{k}={0},{1},{2}\)
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Anzante2m
Answered 2022-01-05 Author has 256 answers
Or without integration, just take \(\displaystyle{\log{}}\) to be the ''natural branch'', i.e. the one with a branch cut along the positive real axis. Or any branch cut that avoids \(\displaystyle−{2}\) for that matter.
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Vasquez
Answered 2022-01-11 Author has 8850 answers

This branch can be defined (at least, in the open unit disk centered at 0) as follows.
\(f(z):=\int^z_0\frac{3t^2}{t^3-2}dt+\log(-2)\)
where the integration is taken over the interval [0,z] and \(\log(−2)=\log2+\pi i.\)

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