I need to introduce a pair of invertible complex functions $\exp$ and $\log$ with the following properties:

$A$ being a branch (or strip?) of $\mathbb{C} \backslash \{0\}$:

$\forall a \in A \quad \exp(\log(a)) = \log(\exp(a)) = a$

$\forall a \in \mathbb{R}^* \quad \log(-|a|) = \log(|a|) + i\pi$

What is the proper way of introducing these two functions? I am especially concerned about the proper enunciation of the functions’ domains and codomains.

Furthermore, is there a proper way of introducing the “continuation” of the first property to a subset containing 0?

I am somewhat familiar with complex logarithms, but I still struggle with the proper definition of domains and codomains using branches (strips?). I would like the definitions to be as precise and unambiguous as can be.

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    $\begingroup$ I am afraid that your functions cannot coincide with the usual exponential and logarithm. Because exponential is not injective on $\mathbb C\setminus\{0\}$, and so it cannot have an inverse and the first property cannot hold. What you can do is considering the restriction of the exponential to a strip $\{z=x+iy\ :\ y\in [a, a+2\pi)\}$. This function is a bijection onto $\mathbb C\setminus\{0\}$. $\endgroup$ – Giuseppe Negro Jan 9 '19 at 16:16
  • $\begingroup$ @GiuseppeNegro I believe this is what I am looking for. Would you mind writing the full definitions in the form $f : A \longrightarrow B$ for both $\exp$ and $\log$ as an answer to the question? Thanks. $\endgroup$ – ismael Jan 9 '19 at 16:19
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    $\begingroup$ I disagree with the downvote, which, moreover, is not accompanied by a motivating comment. I'll see if I can convert my comment into an answer. $\endgroup$ – Giuseppe Negro Jan 9 '19 at 16:50
  • $\begingroup$ @GiuseppeNegro Thank you! I edited the original question in order to clarify things. $\endgroup$ – ismael Jan 9 '19 at 16:51
  • $\begingroup$ This is a great case of definition question. $\endgroup$ – ismael Jan 10 '19 at 15:32

For the first property to hold, the exponential must be restricted to a subset where it is injective. Now, the exponential is injective only on "horizontal strips" $$ A_\lambda:=\{x+iy\ :\ y\in [\lambda, \lambda+2\pi)\}, $$ where $\lambda\in \mathbb R$. This is a consequence of Euler's formula $$e^{x+iy}=e^x(\cos y + i \sin y).$$ For each fixed $\lambda$, since $\exp$ is a bijection of $A_\lambda $ onto $\mathbb C\setminus\{0\}$, there is an inverse function $$\log_\lambda \colon \mathbb C\setminus\{0\} \to A_\lambda.$$ WARNING: This function is discontinuous on the half-line $$\{re^{i\lambda}\ :\ r\ge 0\}.$$

For all $\lambda\in\mathbb R$, it holds that $$\log_\lambda(-z)=\log_\lambda(z)\pm i\pi,$$ where the sign is chosen in such a way that $\log_\lambda(z)\pm i\pi$ stays in $A_\lambda$.

  • $\begingroup$ Thank you! This is exactly what I was looking for. $\endgroup$ – ismael Jan 10 '19 at 14:49

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