My question is really simple. I didn't understand why do we say the principal branch of the logarithm has the negative axis removed (the branch cut). Usually, the argument of the principal branch of the logarithm without the branch cut is defined on $\{z:-\pi\lt \arg (z)\le\pi\}$. So the negative axis is still there since we are allowed to use $\pi$ as argument.
Could someone clarifies me this?
Best Answer
Quite a funny point! My suggestion: When defining the principal branch you probably want to say that you get a holomorphic function in an open domain and if you include the negative real axis, then you get a discontinuous function. On the other hand, as you say it is natural to assign a value to, say $log(-1)$ by choosing one of the two branches. The assignment is simply not a continuous function of $z$.
A more technical issue: A 'natural' definition of $\log z$ is that the complex derivative should be $\frac{1}{z}$. So wherever $\log$ is defined we should have: $$ \frac{d}{dz} \log (z) = \frac{1}{z} $$ We obviously have to omit zero. Now, if $\gamma:[0,1]\rightarrow {\Bbb C}\setminus\{0\}$ is a smooth path then integrating the above we must have: $$ \log(\gamma(1))-\log(\gamma(0))= \int_\gamma \frac{dz}{z} $$ Suppose that we could define our 'log' in all of ${\Bbb C}\setminus\{0\}$ then choosing e.g. $\gamma{t}=ae^{2\pi it}$, $a>0$ (which winds around the origin once) we get the contradiction: $$ 0 = \log(a) - \log(a) = \int_0^1 \frac{2\pi i \;e^{2\pi i t}}{e^{2\pi i t}} dt= \int_0^1 2\pi i \; dt = 2\pi i$$ This contradiction happens for any closed path encircling the origin non-trivially. Thus we can not define $\log$ on such a path and we have to resort to making a cut on some path $c$ (the cut) from 0 to $\infty$ and not allow $\log$ to be defined on that cut.