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Consider the product of complex linear monic polynomials times polynomials of degree less than $n$, that is $\big( (z-\lambda), p(z)\big)\mapsto (z-\lambda)p(z)$. If we represent a polynomial by its coefficients, this produces a map $$h:\mathbb{C}\times\mathbb{C}^n\to \mathbb{C}^{n+1},$$ \begin{align} & \begin{bmatrix} \lambda \\ p_{0} \\ p_{1} \\ \vdots \\ p_{n-2} \\ p_{n-1} \end{bmatrix} \mapsto \begin{bmatrix} \phantom{p_0}-\lambda p_0, \\ p_0-\lambda p_1 \\ p_1-\lambda p_2\\ \vdots \\ p_{n-2}-\lambda p_{n-1}\\ p_{n-1} \phantom{-\lambda p_{n-1}} \end{bmatrix} \end{align} which is, by the Fundamental Theorem of Algebra, surjective onto the space of polynomials of positive degree at most $n$; in fact, a.e. with fiber of cardinality $n$. So the Gaussian measure of the image of $h$ with multiplicity (i.e., the integral of the multiplicity function $\# h^{-1}$) w.r.to Gaussian probability measure) is exactly $$n=\frac{1}{\pi^{n+1}} \int_{\mathbb{C}^{n+1}} \# h^{-1}(q)e^{-\|q\|^2} dq.$$ On the other hand, if we apply the Change of Variable formula, we find the following multiple integral with respect to the $2(n+1)$-dimensional Lebesgue measure on $\mathbb{C}\times\mathbb{C}^n\sim \mathbb{R}^{2(n+1)}$: $$\frac{1}{\pi^{n+1}}\int_{\mathbb{C}\times\mathbb{C}^n} \Big|\sum_{k=0}^{n-1}c_k\lambda^k\Big|^2e^{{-\sum_{k=0}^{n}| c_{k-1}-\lambda c_k|^2}} \mathrm d(\lambda,c_0,\dots,c_{n-1}),$$ where for the sake of notation we put $c_{-1}=c_n=0$ in the sum. If we compute directly this integral and find the value $n$, and use the obvious bound $\# h^{-1}\le n$ (meaning that a polynomial can't have more than $n$ roots), we have that the image of $h$ has full measure, so it is dense. Due to the elementary a priori bounds on the roots of a polynomial, we conclude that $h$ is indeed surjective onto the polynomials of positive degree, thus proving the Fundamental theorem of Algebra by means of multiple integrals.

The computation of the above integral turns out to be an elementary, yet not completely trivial exercise of iterated integrals and linear algebra (I will link the complete computation later). I wonder if there is a more clever way to perform the computation (of course, without using the FTA, nor any proof of it). Also, I am pretty sure -quia nihil sub sole novi, that somebody should already have done this computation, and I'd be glad to learn who.

edit Here is the elementary computation, for this question, and to whom is interested.

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    $\begingroup$ @FedorPetrov I uploaded it few hours ago, it is still on hold. Does this work? dropbox.com/s/v0ra4duxh85iloq/… $\endgroup$ Jan 28, 2021 at 20:18
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    $\begingroup$ Doesn't this just prove that almost every polynomial has zeroes? (I.e. the set of those degree $n$ polynomials without zeroes has measure 0) $\endgroup$ Jan 30, 2021 at 20:19
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    $\begingroup$ @AchimKrause yes, but a polynomial close enough to a polynomial without roots also does not have roots (it is bounded from below). $\endgroup$ Jan 30, 2021 at 20:49
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    $\begingroup$ @LSpice it=a close polynomial. If $p(x)$ of degree $n$ does not have roots, then we have $|p(x)|>c>0$ for all $x\in \mathbb{C}$. Then $|q(x)|>c/2$ whenever $q$ of degree $n$ has coefficients close enough to those of $p$ (consider separately a large disc and its complement). $\endgroup$ Jan 30, 2021 at 22:13
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    $\begingroup$ This reminds me of a proof of FTA by P. Pushkar formulated entirely in terms of real numbers: each real polynomial of even degree at least $2$ can be factored into a product of real quadratic polynomials. Your use of the fiber of $h$ is analogous to Pushkar's use of the degree of a smooth proper map. His paper is at mathnet.ru/php/… and I wrote up a translation of it into English in the last section of kconrad.math.uconn.edu/blurbs/fundthmalg/propermaps.pdf. Properness of the mapping is proved by "compactification". $\endgroup$
    – KConrad
    Jan 31, 2021 at 14:31

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