What I am trying to say is, that for a multi-valued function like $\log{z}$, it seems that generally, one chooses to make a branch cut along the negative real axis.

It seems that then people always choose to use the possible values of $\log{z}$ that have imaginary part within $(-\pi, \pi]$. But is this the case?

If I took a branch cut somewhere else, for example along the positive imaginary axis, would that force me to use certain values for the complex logarithm? Would I even be justified in taking a branch cut along the positive imaginary axis? The visualizations of $\log{z}$ showing several "layers" of values are hard for me to manipulate in my head, so I am unsure if a discontinuity even appears anywhere except the negative real axis.

I apologize if this is silly. My knowledge on this topic is weak.


1 Answer 1


You can choose a branch cut on any line going out from the origin. The reason is that $\log 0$ is not defined. Now because we want to define the logarithm as an integral we need the domain of $\log z$ to be simply connected (this has to do with equality of some integrals along paths, and a theorem called Cauchy's integral theorem). What happens when you remove the negative imaginary axis is that the new domain is simply connected (or without holes). If we do not delete anything and want $\log z$ defined on all of $\mathbb{C}\setminus\{0\}$, then $0$ is a "hole" (which gives us some problems).

This means that as you say, you can delete any line going out from the origin to remove the hole. You can even delete any curve going out from the origin, the new domain just has to be without holes.

Now you will get different values for your logarithm depending on which branch you choose. But notice that even if you delete the negative imaginary axis, you can still choose many different values for the logarithm, or more specifically values in $(-\pi+2\pi n,\pi+2\pi n]$ for some $n$.

  • $\begingroup$ Can you give an example of what values you might get if you chose a cut along a line apart from the negative real axis? $\endgroup$ Aug 19, 2021 at 21:43
  • $\begingroup$ If you choose the positive axis, you can choose values in $[0,2\pi)$. If you choose the positive imaginary axis, for example $[\pi/2,5\pi/2)$ works. Notice that it's always $[\theta,\theta+2\pi)$ or $(\theta,\theta+2\pi]$ where $\theta$ depends on which branch you choose. $\endgroup$ Aug 19, 2021 at 21:52

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