Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is related to a question I asked a few days ago that wasn't really answered.

Let $f(z) = \ln(-z)$

If we choose the branch where $0 \le \arg(-z) < 2 \pi $, the cut is on $(-\infty,0]$.

And if we choose the branch where $-\pi \le \arg(-z) < \pi$, the cut is on $[0,\infty)$.

I've been trying to understand the location of the cuts.

Is it because $\arg_{0}(-z)= \pi + \arg_{-\pi}(z)$ and $\arg_{-\pi}(-z) = -\pi + \arg_{0} (z)$?

EDIT: I don't know if the above equations are true. But they would explain the location of the cuts and the value of the arguments just above and just below the cuts.

share|improve this question

2 Answers 2

You would like to describe $\log$ as the inverse function of $e$. So you would like to say that $\log(re^{i\theta}) = \log r + i \theta$.

But then you have a problem because $\theta$ is only defined up to $2\pi$. Conclusion : every logarithm on $\mathbb C$ has a cut on the argument. But not on the radius, so the cut is always a half-line. This half-line correspond to where you start counting the argument.

For example, in your first example, you start counting at $\rm{arg}$ $(z)= 0$ i.e the half-line $[0, +\infty[$.

In your second example, you start at $\rm{arg}$ $(z) = \pi$, which is the half line $]-\infty, 0]$.

share|improve this answer
    
My question is about why the cuts for $\ln(-z)$ for those branch choices go in the opposite direction. –  Random Variable Mar 13 '13 at 21:56

To add to Damien's good answer: in fact, there are further options, which are sometimes useful. Namely, as long as you "cut" along some (not self-intersecting) path that connects the two "bad" points for log, 0 and $\infty$, you can define an unambiguous log on what's left (by $log(z)=\int_\gamma dz/z$, where $\gamma$ is a path from $1$ to $z$ that does not cross your cut).

And looking at $\log(-z)$ doesn't really change anything: you could use the same old cut, if you wanted, since the goal is to disambiguate (in some way or another, in the usual simple fashion as described by Damien, or, possibly, in a trickier way to suit circumstances).

Edit: added in response to comment... It is not precisely true that $\arg(-z)=\arg(-1)+\arg(z)$ as numbers, because of the ambiguity (by $2\pi\mathbb Z$) in both of these "arg"s. For that matter, do we want $\arg(-z)=\arg(z)+\pi$, or do we want $\arg(z)-\pi$? Cases can be made for both. There's simply no compulsion to choose one or the other.

And, back to the more general point, there is no compulsion to have the "cut" be a straight line from 0 to $\infty$.

An example to illustrate that our "cuts" really have no impact on the mathematical objects themselves: we know that a power series for a holomorphic functions converges up to the first "singularity". Ok. Consider the power series of $\log(z)$ at $-4+3i$. If we believe that somehow the plane is "cut" along the negative real axis, we'd think that the radius of convergence would be $3$. However, it is $5$. The function ignores our "cutting" entirely.

share|improve this answer
    
I want to understand the location of the cuts if you purposely choose those branches for $\ln(-z)$. And I think it has something to do with the fact that $\arg(-z) = \arg\Big((-1)z\Big)= \arg(-1)+\arg(z)$. –  Random Variable Mar 13 '13 at 22:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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