FTA proof not by contradiction I have a simple (maybe not easy) question.
Is there a direct proof of Fundamental Theorem of Algebra? 
If there is not, can we conclude if it is possible to create one?
 A: The Fundamental Theorem of Algebra is usually stated as follows:

Every non-constant complex polynomial has a zero.

State it using the contrapositive as follows:

If a complex polynomial has no zero, then it is constant.

Then this proof using Liouville's theorem is not by contradiction:


*

*If a complex polynomial $p$ has no zero, then $1/p$ is a bounded entire function.

*By Liouville's theorem, $1/p$ is constant and so $p$ is constant.
The canonical proof of Liouville's theorem is not by contradiction.
A: How about using planar Brownian motion.
Let $p:\Bbb C\to\Bbb C$ be a non-constant polynomial. Then $\lim_{|z|\to\infty}|p(z)|=\infty$, so there exists $R>0$ such that if $|z|\ge R$ then $|p(z)|>1 $.
Two properties of a planar Brownian motion (call it $(B_t)_{t\ge 0}$) are crucial: (i) it is recurrent, and in particular  for each $\epsilon>0$ the set of times $\{t: |B_t|<\epsilon\}$ is unbounded above, with probability 1; (ii) there is a (random) reparameterization of time $\tau(t)$ such that $p(B_{\tau(t)})$ is again a Brownian motion. 
Combining (i) and (ii), it is true that with probability 1 there are times $t_1<t_2<\cdots$ with $|f(B_{\tau(t_n)})|\le 1/n$. Fix one sample point $\omega$ with this property. Then $z_n:=B_{\tau(t_n)}(\omega)$ satisfies  $|p(z_n)|\le 1/n$. In particular,  $|p(z_n)|\le 1$ so that  $|z_n|\le R$. The sequence $\{z_n\}$, being bounded, has a convergent subsequence$\{z_{n_k}\}$ with limit $z_\infty$, and then $p(z_\infty)=\lim_kp(z_{n_k})=0$.
