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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?

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  • $\begingroup$ The principal branch of the logarithm is an analytic function on its domain of definition, which has to be open. Period. $\endgroup$ – Christian Blatter Aug 28 '16 at 13:44
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Let me propose an explanation I gave to myself (while studying this excellent book [ref.]), which can be summarized as follows.
Take the exponential, defined as $$ w = \exp \left( z \right) = \exp \left( {x + i\,y} \right) = e^{\,x} \left( {\cos y + i\sin y} \right) = e^{\,x} \left( {\cos y^ * + i\sin y^ * } \right) $$ where $y^*$ indicates the value of $y$ reduced $mod(2\pi)$.
Whether you take $0 \leqslant y^ * < 2\pi $ or $-\pi < y^ * \leqslant \pi $ is a matter of convention. In the latter case $$ y^{\, * } = \pi - \bmod \left( {\pi - y,\;2\,\,\pi } \right) = \bmod (y,\;2\,\,\pi ) - 2\,\,\pi \left\lfloor {\frac{1} {2} - \left\{ {\frac{y} {{2\,\,\pi }}} \right\}} \right\rfloor $$ The $exp(z)$ so defined is analytic, surjective, periodic in $y$, defined in the whole plane.

logz

If you now consider the map from $z$ to $w$, it is clear that going backwards you will have infinite values for $z$, stacked on the vertical axis at $x$ and separated by $2\pi$. The referred book calls this the multivalued logarithm Ln(w).
If you want to limit it to a single value (there indicated by $ln(w)$), you shall place a radial frontier necessarily of the type "( ]" (or v.v.), somewhere in the $w$ plane. That will translate into a horizontal band in the $z$ plane.
The standard is considered to be placed at $+/- \pi$, and that is.
It is also clear that, crossing the frontier, you have a jump in $ln(w)$.


[ref]:" Theory of analytic functions" - A.I. Markuševič

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  • $\begingroup$ Univoque? What does it mean? $\endgroup$ – Santiago Aug 28 '16 at 5:54
  • $\begingroup$ @Santiago, you are right, sorry, that's not the right adjective. Changed to surjective. $\endgroup$ – G Cab Aug 28 '16 at 11:27
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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.

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  • $\begingroup$ So do you mean the principal branch of the logarithm is not continuous? $\endgroup$ – user42912 Aug 27 '16 at 19:18
  • $\begingroup$ yes, $\log(-1+0^+i) = +i\pi$ but $\log(-1+0^- i) = -i\pi$ (if my notation is clear?) $\endgroup$ – H. H. Rugh Aug 27 '16 at 19:20
  • $\begingroup$ So what is the special with these arguments: $-\pi\lt \arg(z)\le \pi$? In another words why do we choose these values to be the argument of the principal logarithm? $\endgroup$ – user42912 Aug 27 '16 at 19:24
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    $\begingroup$ Nothing special but if $z=re^{i\phi}$ then you have to say $\log(z)=\log(r) + i \phi + 2\pi i k$ for some $k\in Z$ and there is simply NO WAY to avoid whatever global definition you give to be discontinuous. After all going around once $\phi$ increase (or decrease) by $2\pi$ $\endgroup$ – H. H. Rugh Aug 27 '16 at 19:26
  • $\begingroup$ Yes I know that. I was talking specifically about the principal branch $Log(z)=\log(r)+i\theta$. What I hadn't understood is why we pick specifically these arguments since they fail even to be continuous. $\endgroup$ – user42912 Aug 27 '16 at 21:53
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It is useful for many applications that the argument take values in an open subset of the real line, but $-\pi < \theta \le \pi$ is not an open subset.

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It's generally quite a good idea to think about exponential function in terms of geometry. If $z = x + iy$ with $x, y \in \mathbb R$, $w = e^z = e^x e^{iy}$, where $(e^x, y)$ are polar coordinates of $w$, which means that the line $\mathbb R + i y_0$ is being transformed onto $\mathbb R_+ \cdot e^{iy_0}$ whereas the line $x_0 + i \mathbb R$ is being mapped onto the circle $e^{x_0} \cdot S^1$.

enter image description here

The exponential function $\exp \colon \mathbb C \to \mathbb C^\times$ is a surjection, because every point $w \neq 0$ may be represented as $w = |w| e^{i \varphi} = e^{\ln |w| + i \varphi}$. We know (from other properties of exponential map) that $\exp z_1 = \exp z_2$ iff $y_2 - y_1$ is an integer multiple of $2 \pi i$. It means that $\exp$ maps the stripes $S = \{z : |\Im z| <\pi\}$ in a bijective manner onto the complex plane cut along the negative real axis, $\mathbb C^- = \mathbb C \setminus (-\infty, 0]$.

enter image description here

The restriction $\exp \colon S \to \mathbb C^-$ has therefore the inverse function, $\ln \colon \mathbb C^- \to S$, called the principal branch of logarithm. To compute its value we have to return to the representation $$w = |w| e^{i\varphi} = e^{\ln |w| + i \varphi},$$ where $-\pi < \varphi < \pi$ and deduce the formula $\ln w = \ln |w| + i \varphi$.

The images are taken from the book "Analysis 1" by Konrad Königsberger.

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    $\begingroup$ thanks for putting my sketch into more formal terms. However you should correct $z_2-z_1$ into $y_2-y_1$ $\endgroup$ – G Cab Aug 28 '16 at 13:41
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As the other answeres have explained well, it’s not possible to define the logarithm as an analytic function on any open set that contains the whole unit circle. Rather, the domain of definition of the analytic logarithm must be a set $D$ with the following three properties: (1) $D$ is open in the complex plane; (2) $D$ is simply connected; and (3) $0\notin D$.

But there’s nothing sacrosanct about the negative real axis. If you wanted to talk about properties of the logarithm when you’re close to negative real numbers, you could omit the negative imaginary axis, for instance, or indeed any arc that goes from $0$ out to infinity.

But remember this: no matter what you do, you will never be able to get your logarithm to be a homomorphism, that is, you will never be able to get $\log(zw)=\log z+\log w$ for all $z$, $w$, and $zw\in D$.

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