Timeline for Does the exponential function have a (compositional) square root?
Current License: CC BY-SA 4.0
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| Apr 28, 2020 at 18:01 | comment | added | Ayman Moussa | Continuous case without covering : $h(\cdot+2i\pi)=h(\cdot)+2i\pi k$ for some $k\in \mathbf{Z}$ and actually $k=0$, if not $h(\mathbf{C})=\mathbf{C}^\times$ implies $-2i\pi k\in h(\mathbf{C})$ and thus $0\in h(\mathbf{C}+2i\pi)=h(\mathbf{C})$, contradiction. Hence $h(\mathbf{B})=\mathbf{C}^\times$ where $\mathbf{B}=\{0\leq \text{Im}(z) < 2\pi\}$. Fix $\ell\in\mathbf{Z}^\times$ s.t. $f(0)+2i\ell \pi\in \mathbf{C}^\times=h(\mathbf{B})$, so $f(z_0)=f(0)+2i\ell \pi$ for some $z_0\in\mathbf{B}$. Applying $f$ entails $e^{z_0}=e^0$, only possible if $\mathbf{B}\ni z_0=0$, contradiction. | |
| Oct 6, 2018 at 7:24 | history | edited | Denis Serre | CC BY-SA 4.0 |
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| Oct 5, 2018 at 21:01 | history | edited | Denis Serre | CC BY-SA 4.0 |
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| Oct 5, 2018 at 14:51 | history | edited | Denis Serre | CC BY-SA 4.0 |
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| Oct 5, 2018 at 13:57 | history | answered | Denis Serre | CC BY-SA 4.0 |