A bijection $\log^+:\Bbb R_{>0}\times\Bbb R\to \Bbb C:(r,\theta)\mapsto\ln r+\mathrm i\theta$ can be defined, along with a family of smooth bijections $\mathrm c_{\alpha}:\Bbb R_{>0}\times (\alpha-\pi\,,\alpha+\pi]\to\Bbb C_{\neq0}:(r,\theta)\mapsto r\mathrm e^{\mathrm i\theta}$ ($\alpha\in\Bbb R$). Then the function $$\log^{(\alpha)}:=\log^+\circ\;\mathrm c_\alpha^{-1}$$maps $\Bbb C_{\neq0}$ injectively to $\Bbb C$, and is continuous on $\Bbb C\setminus\{r\mathrm e^{-\mathrm i\alpha}:r\geqslant0\}$. We could thus define the complex logarithm to be the family $$\log:=\{\log^{(\alpha)}:\alpha\in\Bbb R\}.$$ Any convenient member of this family can then be used as a logarithmic function on $\Bbb C_{\neq0}$ (conventionally with the dependence on $\alpha$ implicit and suppressed in notation).

The above is only my attempt to get a clear picture of the complex logarithm. The definitions I have come across hitherto are simpler than this, but not as clear as I might hope for. What is usually taken to be the simplest and clearest definition of complex logarithm as a set-theoretic object?


For fixed \alpha your function \log^{(\alpha)} is not a bijection onto C. Its range is a horizontal strip of height 2\pi.

  • $\begingroup$ Thank you. I have corrected it now (I hope!). Your correction should be posted as a comment since it is not an answer. $\endgroup$ – John Bentin Nov 25 '15 at 20:45
  • $\begingroup$ OK I understand the problem better now. My personal preference for the definition of the logarithm would be to map it not to C or a horizontal strip of C, but to an infinite cylinder obtained by identifying the top and bottom boundary of the strip. $\endgroup$ – Justpassingby Nov 25 '15 at 20:49

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