Branch of logarithm which is real when z>0

Question

I am familiar with the complex logarithm and its branches, but still this confuses me. I read this in a textbook:

"For complex $z\neq 0, log(z)$ denotes that branch of the logarithm which is real when $z > 0$.

What does this mean?

Answer 1

It doesn't mean anything unless you specify the region $G$ you want to define the complex logarithm on. The statement seems to try to do this with $G=\mathbb C\setminus \{0\}$, but that is impossible: In general, the complex logarithm is only defined up to $2k\pi$, $k\in\mathbb Z$, and if your $G$ winds around $0$, you cannot make the function continuous. Instead, $G$ could be any simply connected subset of $\mathbb C\setminus\{0\}$, and the usual choice is to remove the nonpositive reals from $\mathbb C$, i.e. $G=\mathbb C\setminus\{-\infty,0]$. With this (or any other simply connected $G$ that contains the positive reals), it is possible to pick the branch that is real (i.e. is the well-known real logarithm) for positive reals.

Answer 2

Let $\hbox{Log} z=\ln|z|+i\hbox{Arg}z$ be the principal branch of the logarithm, that corresponds to a cut along the negative real numbers. Now, consider any branch of $log$ that is defined on a connected neighborhood of $\Bbb{R}^+=(0,+\infty)$. Clearly, $x\mapsto \varphi(x)=log(x)-\hbox{Log}(x)$ is a continuous function on $\Bbb{R}^+$ that satisfies $\exp(\varphi(x))=1$ so, $\varphi$ takes its values in $2\pi i\Bbb{Z}$, and since it is continuous, it must be constant. Thus, there is $k\in\Bbb{Z}$ such that $log(z)=\hbox{Log }{z}+2i\pi k$ and this happens on the largest connected open set that contains $\Bbb{R}^+=(0,+\infty)$, and on which both functions are analytic (by analytic continuation).

Thus, the statement says that we choose the logarithmic function the corresponds to $k=0$ because otherwise $log(z)$ would not be real for real $z$.