4
$\begingroup$

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.

$\endgroup$

3 Answers 3

1
$\begingroup$

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

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

$\endgroup$
3
  • 1
    $\begingroup$ The question is to find a single-valued branch of $\log f(z)$, not to find its branch points. $\endgroup$
    – user64494
    Sep 3, 2013 at 18:20
  • 1
    $\begingroup$ 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, 2013 at 20:25
  • 1
    $\begingroup$ @DonAntonio: Good comment. Thank you for answering him. $\endgroup$ Sep 3, 2013 at 23:07
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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, 2013 at 18:30
  • $\begingroup$ @ Evan: Could you ground that calculation? $\endgroup$
    – user64494
    Sep 3, 2013 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, 2013 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, 2019 at 19:50
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.