Timeline for Complex logarithm base 1
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 40 mins ago | answer | added | Anixx | timeline score: 0 | |
| 53 mins ago | history | became hot network question | |||
| 1 hour ago | answer | added | Dan | timeline score: 0 | |
| 2 hours ago | comment | added | Jianing Song | I was wondering that, if $z^w$ is intended for multivalued function, then probably it is better to write $\exp(z)$ instead of ${\rm e}^z$. In this context I understand ${\rm e}^z=\exp(z\log e)=\{\exp(z(1+2k\pi{\rm i}))\}$. In other words, ${\rm e}^z=\exp(z)\cdot 1^z$. | |
| 3 hours ago | answer | added | Oscar Lanzi | timeline score: 0 | |
| 7 hours ago | comment | added | M.G. | Finally, I would check if perhaps these "doubly" parametrized branches don't partially overlap to get a bigger continuous or holomorphic section (off top of my head doesn't seem to be the case here, but I am not being careful). | |
| 7 hours ago | comment | added | M.G. | @AvelBulatov: right, so you get all the branches $1^w = e^{2 \pi k i w}$, $k \in \mathbb{Z}$, and you need to solve $e^{2 \pi k i w} = \xi$ functionally, i.e. $e^\zeta = \xi$, which brings you again to the logarithm, so you've basically already answered your own question, but you need to pick a branch for $\ln(z)$ again. Thus you get a family of $\log_1$-s, parametrized by $\mathbb{Z} \setminus{0} \times \mathbb{Z}$. There is no unique $\log_1(z)$. | |
| 7 hours ago | comment | added | Anixx | If you want it multi-valued, you can take it as $\frac{\ln z+2 \pi i k} {2\pi i}=\frac{\ln z} {2\pi i}+k, k\in\mathbb{Z}$. | |
| 7 hours ago | comment | added | Avel Bulatov | @M.G., yes, I know. $ z^w=e^{w*log(z)} $ | |
| 8 hours ago | comment | added | M.G. | @AvelBulatov: How much complex analysis have you covered, e.g. if you forget about $1$, do you know how to make sense of $z^w$ for complex $z$ and $w$? There is a clean ansatz to think about this type of questions. | |
| 12 hours ago | history | edited | Anixx | CC BY-SA 4.0 |
added 7 characters in body
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| 12 hours ago | comment | added | Avel Bulatov | No,1 to any INTEGER power is 1. But this is not true in general. In the general case, $1$ belongs to the set $1^z$ for any complex number $z$. For example: $1^{1/4} = \{1,\ -1,\ i,\ -i\}$, and $1^{1/(2\pi i)} = e^{\log(1)/(2\pi i)} = e^{2\pi i k/(2\pi i)} = e^{k} = \{\dotsc,\ e^{-1},\ 1,\ e,\ e^{2},\ \dotsc\}$ | |
| 13 hours ago | comment | added | Anixx | 1 to any power is 1, complex or not. | |
| 13 hours ago | history | asked | Avel Bulatov | CC BY-SA 4.0 |