I am trying to find the branch points and choose a branch cut for the function $f(z) = \log(z^2)$.

I know that both $z = 0$ and $z = \infty$ are branch points, so it seems reasonable to just choose the negative real axis as a branch cut as with $\log{z}$. However, as $\theta$ goes from $0$ to $2\pi$, $\arg{z^2}$ goes from $0$ to $4\pi$ so I am not sure whether the branch cut stated above is sufficient.

My understanding of all this is still not very strong, so any tips or intuition are appreciated. Are the two branch points and the branch cut stated above correct?

  • $\begingroup$ Hint: What is $e^{2\log z}$? $\endgroup$ Feb 16, 2014 at 2:26

1 Answer 1


Note that $z$ must satisfy that $z^2 \not \in (-\infty,0)$ taking the usual cut. What values of $z$ make that happen?

  • $\begingroup$ Ah so then $\{z|\Re{z} = 0\}$ is the cut I'm looking for? Is the fact that it partitions the complex plane into two pieces acceptable? That strikes me as a little bit weird because there is no way to get from $z = 1 + i$ to $z = -1 + i$ for example. $\endgroup$ Feb 16, 2014 at 1:42
  • $\begingroup$ Yes, a disconnected domain has an odd feel about it. You can extend the definition to the branch, but you'll lose continuity across the imaginary axis. We typically want to define it in a way to be holomorphic. $\endgroup$
    – Darrin
    Feb 16, 2014 at 2:02
  • $\begingroup$ Great. Thanks for the help. $\endgroup$ Feb 16, 2014 at 2:38

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