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

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.

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

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

• The question is to find a single-valued branch of $\log f(z)$, not to find its branch points. – user64494 Sep 3 '13 at 18:20
• And this answer is a nice hint how to do that, @user64494 . Not all love to give the whole answer ready. +1 – DonAntonio Sep 3 '13 at 20:25
• @DonAntonio: Good comment. Thank you for answering him. – Mhenni Benghorbal Sep 3 '13 at 23:07

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.

• 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 $\log(z^2 - 2) := \log r + i\theta$ – Evan Sep 3 '13 at 18:30
• @ Evan: Could you ground that calculation? – user64494 Sep 3 '13 at 18:36
• @user64494 it's a bit ugly generally but good for finding specific values (for instance to do residue computations) – Evan Sep 5 '13 at 2:52
• s it possilbe if you (or someone) can elaborate on this answer? I dont see the reason immediately – NazimJ May 10 at 19:50

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 $\log(-2)=\log 2+\pi i.$