Timeline for Gauss proof of fundamental theorem of algebra
Current License: CC BY-SA 3.0
5 events
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| Jan 10, 2023 at 12:22 | comment | added | David E Speyer | @Kostya_I I think this works! Let me say it in other words. Let $\Gamma$ be a connected component of $R$. Suppose that $f$ is bounded above on $\Gamma$. Then we can choose $z_k \in \Gamma$ such that $f(z_k)$ approaches $\text{sup}_{z \in \Gamma} f(z)$. These $z_k$ lie in a bounded region of $\mathbb{C}$ (details omitted) so we can choose a convergent subsequence; let its limit be $z$. Parametrizing $\Gamma$ near $z$ gives a contradiction. Similarly, $f$ is not bounded below on $\Gamma$. So $f$ is both positive and negative on $\Gamma$, and $f$ must be zero somewhere on $\Gamma$, QED. | |
| Jan 10, 2023 at 10:38 | comment | added | Kostya_I | ... is obtained by looking at a neighborhood of $z$: the part of $\Gamma$ in that neighborhood looks like a graph in right coordinates, the abscisses of $\phi(t_k)$ must be monotone, so $\phi$ in fact parametrizes one-half of the graph, and the other half is untouched since we coundn't have escaped without passing through $z$. In particular, $b<\infty$, but then the parametrization can be continued beyond $b$, a contradiction. | |
| Jan 10, 2023 at 10:32 | comment | added | Kostya_I | Is that really difficult? Start with some point $y\in \Gamma$, and let $\phi_0:I_0\to \mathbb{C}$ be a natural parametrisation of a piece of $\Gamma$ near $y$. Consider all possible extensions of $\phi_0$ to natural parametrisations of sub-curves of $\Gamma$ by larger intervals. Any two such extensions must agree on their common domain, therefore, we can consider a maximal extension $\phi:(a,b)\to\mathbb{C}$. If there is a sequence $t_k\nearrow b$ such that $\phi(t_k)$ is bounded, then we can pick a subsequence such that $\phi(t_k)\to z\in\Gamma$, and a contradiction... | |
| Apr 13, 2017 at 12:50 | history | edited | CommunityBot |
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| Jan 20, 2017 at 18:29 | history | answered | David E Speyer | CC BY-SA 3.0 |