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Find a branch of $f(z)=\log(z^3-2)$ that is analytic at $z=0$. Can anyone help me on this question? I have no idea how to find a branch. The definition of branch given in lecture is

$F$ is a branch of $f$ on a domain $D$ if $F$ is a (single valued) continuous function on $D$ and if for all $z \in D$, $F(z)$ is one of the values of $f(z)$. $f$ is a multiple valued function.

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Note that, $z=0$ is not a branch point of $f(z)$. To find the branch points of $f(z)$, solve the equation

$$ z^3-2=0 \implies z^3= {2} \rm e^{ 2k\pi i } \implies z=2^{1/3} \rm e^{ \frac{2k\pi i}{3} }, \quad k=0,1,2\,. $$

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    $\begingroup$ The question is to find a single-valued branch of $\log f(z)$, not to find its branch points. $\endgroup$ – user64494 Sep 3 '13 at 18:20
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    $\begingroup$ And this answer is a nice hint how to do that, @user64494 . Not all love to give the whole answer ready. +1 $\endgroup$ – DonAntonio Sep 3 '13 at 20:25
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    $\begingroup$ @DonAntonio: Good comment. Thank you for answering him. $\endgroup$ – Mhenni Benghorbal Sep 3 '13 at 23:07
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Or without integration, just take $\log$ to be the ''natural branch'', i.e. the one with a branch cut along the positive real axis. Or any branch cut that avoids $-2$ for that matter.

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  • $\begingroup$ On top of this solution, to compute explicitly, given some $z$, you express $z^3 - 2 = re^{i\theta}$ for $0 < \theta < 2\pi$ (find argument and magnitude), and then define $\log(z^2 - 2) := \log r + i\theta$ $\endgroup$ – Evan Sep 3 '13 at 18:30
  • $\begingroup$ @ Evan: Could you ground that calculation? $\endgroup$ – user64494 Sep 3 '13 at 18:36
  • $\begingroup$ @user64494 it's a bit ugly generally but good for finding specific values (for instance to do residue computations) $\endgroup$ – Evan Sep 5 '13 at 2:52
  • $\begingroup$ s it possilbe if you (or someone) can elaborate on this answer? I dont see the reason immediately $\endgroup$ – NazimJ May 10 at 19:50
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This branch can be defined (at least, in the open unit disk centered at $0$) as follows. $$ f(z):=\int_0^z \frac {3t^2} {t^3-2}\,dt+\log(-2), $$ where the integration is taken over the interval $[0,z]$ and $\log(-2)=\log 2+\pi i.$

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