Undergrad Principal branch of the log function

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The discussion centers on the principal branch of the logarithm in complex analysis, specifically the conditions under which the logarithm function aligns with the ordinary natural logarithm. It highlights that the interval I must contain 0 for the properties of the real logarithm to hold true. If 0 is excluded from the interval, the logarithm behaves differently in the complex plane, leading to unique values that do not correspond to real logarithmic behavior. The principal value is typically defined within the range of -π to π for the imaginary part. Understanding these nuances is crucial for grasping the differences between real and complex logarithmic functions.
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I'm learning complex analysis right now, and I'm reading from Joseph Taylor's Complex Variables.

On Theorem 1.4.8, it says "If a log is the branch of the log function determined by an interval I, then log agrees with the ordinary natural log function on the positive real numbers if and only if the interval I contains 0."

Can anyone help me understand what this means? Is it just saying that the properties of the real log function only hold if the interval I contains 0? I don't understand why.
 
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220px-Riemann_surface_log.svg.png

(from: https://de.wikipedia.org/wiki/Logarithmus#Komplexer_Logarithmus; engl.: https://en.wikipedia.org/wiki/Logarithm#Complex_logarithm)

You see, that the logarithm ##w## for a value ##z##, i.e. ##e^w=z## has all solutions ##w+2k\pi i\; , \;k\in\mathbb{Z}##. To make it unique, we have to choose one of these branches, i.e. one value for ##k##. This prinicpal value is usually determined by ##- \pi < \operatorname{Im}(w) \leq \pi\,.## Of course we could also say ##-\pi/2 < \operatorname{Im}(w) \leq 3/2 \pi## or any other interval with ##0## in it. But if we climb up, such that zero isn't part of the interval anymore, then we are in the complex world, i.e. we ran around the circle once before regarding the logarithm, but this "run around once" has no real counterpart, because in the reals we would arrive at where we started, whereas in the complexes we climbed up a branch, which is not identical to where we started.
 

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We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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