0

What is the fastest way from

For every $\emptyset\neq B\subset\mathbb{C}$, if there exists $z\in\mathbb{C}$ with $\left| b\right|\leq\left| z\right|$ for all $b\in B$, then there exists $z_{min}\in\mathbb{C}$ with $\left| b\right|\leq\left| z_{min}\right|$ for all $b\in B$, and $\left| z_{min}\right|\leq\left| z\right|$ for all $z$ as above.

to

For any $a_0,\ldots, a_{n}\in\mathbb{C}$ with $a_n\neq 0$, $n\in\mathbb{N}$, there exists $z_0\in\mathbb{C}$ with $\sum\limits_{i=0}^n a_iz_0^i=0.$

assuming merely familiarity with the basic algebra of $\mathbb{C}$, but none with the concepts of limit or continuity, and in particular no knowledge of further analytic theorems?

Martin Argerami
  • 217,281
Damian Reding
  • 8,894
  • 2
  • 24
  • 40

1 Answers1

1

Your completeness axiom holds in $\mathbb R$, while the Fundamental Theorem of Algebra does not. This precludes any "elementary" proof, as some structure of $\mathbb C$ has to be used in an essential way. As far as I can tell, there is no algebraic proof of the Fundamental Theorem of Algebra.

Community
  • 1
Martin Argerami
  • 217,281