# Is the principal logarithm defined for negative numbers?

I have seen the definition of a principal logarithm as the logarithm whose imaginary part is in $(-\pi, \pi]$. However, I have also seen that the principal logarithm defined as the logarithm obtained from the branch cut that removes the negative real axis. The first allows one to take logarithms of negative numbers, while the second does not. Is one definition more standard than the other, or am I misinterpreting something?

• Well, due to the cut, you can take either of $\log|p|\pm i\pi$ as the "logarithm" of $p$ for $p$ negative... (i.e., taking the exponential of either value does give $p$). Thus, one chooses a particular value for the "principal branch", and in this case, it's the one with positive imaginary part... – J. M. is a poor mathematician Sep 1 '11 at 5:58
• One thing that makes the principal branch more standard is that it extends the real log, i.e., numbers on the real axis have an argument $Argz=0$ , so that, for x in the positive real axis, Logx=ln|x|+iArgx=lnx+0=lnx. – gary Sep 1 '11 at 6:00
• @gary: The principal branch is not the only one that agrees with the usual $\ln$ on the positive real axis. Lots of others do too. The question is about the value on the negative real axis (where the arg is not 0: it is $\pi$ (for the principal branch), by the same convention that says the imaginary part of the principal branch of the logarithm is $\pi$ there. – Robert Israel Sep 1 '11 at 6:26
• @Robert: in the layout I am familiar with, we remove a half-line $theta=theta_0$ from the plane, and then we measure the argument against that line, increasing when we go clockwise and otherwise decreasing. Under this layout, only the standard branch is the only one that agrees with lnx. What layout do you use? – gary Sep 1 '11 at 6:59
• @Robert: Never mind; my comment may be too off-topic, and I don't want to distract from the OP. – gary Sep 1 '11 at 7:06

$$\lim_{\epsilon \to 0^+} \log (-z + i \epsilon) = \log(-z) \qquad \lim_{\epsilon \to 0^+} \log (-z - i \epsilon) = \log(-z) - 2 \pi i$$ for $z>0$.