If I understand correctly the domain of the principal branch of exp(Log(Z)) is restricted from -/+ pi because arg(z) is periodic, therefore without the restriction the function would not be one to one. Also the negative real axis is removed because the Limit, as one approaches the negative real axis is not the same if approached from above or from below. This makes some sense to me

What I am really confused about is the removal of the negative real axis and what effects this has, does this mean you cannot take the Log of any negative number such as (-3+0i) because it is undefined with the principal branch cut.

  • $\begingroup$ What does it mean limit is not the same when we approach from above or below? $\endgroup$
    – moli
    May 26, 2021 at 15:39

2 Answers 2


The situation is a little bit more complex than you are describing. Also there are two different concepts you are talking about: just calculating $\operatorname{Log}(-3)$ using the principal branch of the complex logarithm and studying $\operatorname{Log}$ as a function.

  1. There ist not THE principal branch of the complex Logarithm. Depending on how you define the argument, your Logarithm can change. Defining the argument of a complex number $z\in\mathbb C$ as the unique $\varphi\in(-\pi,\pi]$ such that $z=r\cdot e^{i\varphi}$ is just a convention. One might define the argument using $\varphi\in [-\pi,\pi)$ or $\varphi\in [0,2\pi)$ or $\varphi \in [x,x+2\pi)$ for any $x\in\mathbb R$.
  2. The Logarithm of a complex number $z\in\mathbb C^*$ can now be defined as $$\operatorname{Log(z)}=\ln(|z|)+i\arg(z).$$ This yields a function $$\operatorname{Log}:\mathbb C^*\rightarrow\mathbb C,~z\mapsto\operatorname{Log}(z).$$ which is usually called the principal branch Keep in mind that our definition of $\operatorname{Log}$ depends on our definition of $\arg$, so if you decide to define $\arg$ using $\varphi\in (-\pi,\pi]$ and a friend of yours decides to define $\arg$ using $\varphi\in [0,2\pi)$, you will get different principal branches. If you are just interested in the value of $\operatorname{Log}(z)$ for any $z\in\mathbb C^*$, you can use the principal branch without any problems, you don't have to exclude the negative real axis for just calculating the value.
  3. If you want to study the principal branch as a function, you will soon start ask about things like continuity. Only now it is neccesary, to exclude some part of the complex plane to get a continuous, holomorphic function. Again, the way you defined $\arg$ might change what you have to cut out. If you chose to have $\varphi\in (-\pi,\pi]$, then $\operatorname{Log}$ is continuous on $$\mathbb C\setminus\{x\in\mathbb R~|~x\leq 0\}.$$ If you instead chose to have $\varphi \in[0,2\pi)$, then $\operatorname{Log}$ is continuous on $$\mathbb C\setminus\{x\in\mathbb R~|~x\geq 0\}.$$

Yes, one removes that in order to keep continuity, and differentiability. As pointed out you'll have different limits if you come from above or below and therefore you have no unrestricted limit and without limit you can't define a function continuously there.

This is because one of the central points in complex analysis is the differentiability of functions, so that one very much like to keep.

The principal branch is just something that has been chosen to be the "normal" way of creating a branch. If the situation calls for it one could always create another branch. You do this by first selecting a cut path (other than the negative real axis) and also which value that should be taken at a particular point.

  • $\begingroup$ Thank you, If you wanted to take the log(-3+0i), can you just take a different branch cut ? $\endgroup$
    – user299113
    Jan 19, 2016 at 10:27
  • $\begingroup$ @user299113 Yes, thats what one normally would do if you need it to be defined for negative real numbers. You could then for example cut along the negative imaginary axis. $\endgroup$
    – skyking
    Jan 19, 2016 at 10:35
  • $\begingroup$ Thank you for your answer, I have spent far to long thinking about this one question and I am running out of study time for an upcoming exam. Now I can move on. $\endgroup$
    – user299113
    Jan 19, 2016 at 10:38

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