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The complex logarithm $L(z)$ is given by $$L(z)=\ln(r)+i\theta$$ where $z=re^{i\theta}$ and $\ln(x)$ is the real natural logarithm. It is well known that $L(z)$ then sends each $z$ to infinitely many values, each of which are different by an integer multiple of $2\pi i$. If we imagine going along the unit circle in the $z$-plane starting from $1$ and working anti-clockwise, the image of it under $L(z)$ will be a segment of the vertical line $i\theta$ with $\theta$ a real number. However we clearly see how $1$ has mapped to $2$ different values, and how we could continue on mapping it to infinitely many values by winding around $0$ indefinitely. We then call $0$ a Logarithmic branch point of the mapping.

My question is, do there exist mappings that have logarithmic branch points that themselves have nothing to do with logarithms i.e. the logarithm does not appear explicitly somewhere in it's definition or it does not behave like a logarithm. Does a logarithmic branch point indicate logarithmic behavior?

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    $\begingroup$ Should be:Different by an integer multiple of $2\pi i$. $\endgroup$ – Thomas Andrews Aug 26 '13 at 14:35
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    $\begingroup$ Do you feel the arctangent function behaves like a logarithm? If not, it would qualify. $\endgroup$ – Marc van Leeuwen Aug 26 '13 at 15:07
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    $\begingroup$ $arctan(z)=\frac{1}{2i}ln(\frac{1+iz}{1-iz})$, so I don't think that qualifies. See math.ethz.ch/education/bachelor/lectures/fs2012/other/ka_itet/… $\endgroup$ – Kieran Cooney Aug 26 '13 at 15:12
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    $\begingroup$ Opps... The "image of it under $L(z)$ will be a segment of the vertical line" $i\theta$, not $1+i\theta$! $\endgroup$ – peterwhy Aug 26 '13 at 16:19
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    $\begingroup$ The only function I can think off is the branch points of the incomplete elliptic integrals. Since its inverse are elliptic functions which are doubly periodic in $\mathbb{C}$, every time one loops around a branch point, one will pick up same period like what happens for "Log / Exp". $\endgroup$ – achille hui Aug 26 '13 at 17:28
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I think I may have an answer;

$f(z)=z^p$ with $p$ an irrational number. If $p$ is rational, then the order of the branch point $f(z)$ at $z=0$ is just the denominator in lowest terms. So if we take the limit as $p$ goes to an irrational number, we should get an infinite order branch point.

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  • $\begingroup$ This will be an infinite sheeted Riemann surface, but I'm not sure what you mean by non-logarithmic. Note that $z^p = \exp(p\log z)$. $\endgroup$ – mrf Sep 21 '13 at 8:22

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