# Branch of logarithm which is real when z>0 I am familiar with the complex logarithm and its branches, but still this confuses me. I read this in a textbook: "For complex z!=0,log(z) denotes that branch of the logarithm which is real when z>0.What does this mean?

Branch of logarithm which is real when z>0
I am familiar with the complex logarithm and its branches, but still this confuses me. I read this in a textbook:
"For complex $z\ne 0,log\left(z\right)$ denotes that branch of the logarithm which is real when $z>0$
What does this mean?
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Carly Yang
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 $G=\mathbb{C}\setminus \left\{0\right\}$, 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 $\mathbb{C}\setminus \left\{0\right\}$, and the usual choice is to remove the nonpositive reals from $\mathbb{C}$, i.e. $G=\mathbb{C}\setminus \left\{-\mathrm{\infty },0\right]$. 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.
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Maribel Mcintyre
Let $\text{Log}z=\mathrm{ln}|z|+i\text{Arg}z$ 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 ${\mathbb{R}}^{+}=\left(0,+\mathrm{\infty }\right)$. Clearly, $x↦\phi \left(x\right)=log\left(x\right)-\text{Log}\left(x\right)$ is a continuous function on ${\mathbb{R}}^{+}$ that satisfies $\mathrm{exp}\left(\phi \left(x\right)\right)=1$ so, $\phi$ takes its values in $2\pi i\mathbb{Z}$, and since it is continuous, it must be constant. Thus, there is $k\in \mathbb{Z}$ such that and this happens on the largest connected open set that contains ${\mathbb{R}}^{+}=\left(0,+\mathrm{\infty }\right)$, 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 $log\left(z\right)$ would not be real for real $z$