# 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.

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

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

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

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.

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