Fundamental theorem of algebra: a proof for undergrads? The fundamental theorem of algebra is the statement that a complex polynomial of positive degree has at least one root. I do not know complex analysis but I searched for proofs of the statement and came across proofs using complex analysis which seemed rather short and elegant. This is to say: I am aware that there exist very easy proofs using tools of complex analysis. Since I do not know complex analysis yet I started to wonder if there are any known proofs that are short and easy that use only tools that a first year undergrad knows? Like real analysis and linear algebra? 
 A: Here are three accessible proofs, via Keith Conrad:
http://www.math.uconn.edu/~kconrad/blurbs/fundthmalg/fundthmalgcalculus.pdf
http://www.math.uconn.edu/~kconrad/blurbs/fundthmalg/fundthmalglinear.pdf
http://www.math.uconn.edu/~kconrad/blurbs/fundthmalg/propermaps.pdf
(the last one requires a bit more sophistication, but isn't too bad)
A: There is a proof using linear algebra due to Derksen :
H.Derksen, The fundamental Theorem of Algebra and Linear Algebra, Amer. Math. Monthly, 110, (2003), 620-623. http://www.math.lsa.umich.edu/~hderksen/preprint.html
A somewhat expanded version of it is also available (due to S. Kumaresan) :
http://main.mtts.org.in/expository-articles (See #15 under "Analysis")
A: 

Following proof is an ad hoc proof designed for FTA so that every undergraduate can follow it without studying general path integrals in the Real or Complex plane.


If we suppose that our polynomial $P(z)$ doesn't vanish in the Complex plane, then $1/P(z)$ is a continuous function. Since the components of the polynomial $P(x,y)=u(x,y)+iv(x,y)$ satisfy the Cauchy-Riemann conditions, we can infer that the components of $1/P(x,y)=u/(u^2(x,y) + v^2(x,y))+i(-v(x,y))/(u^2(x,y)+v^2(x,y))$ also satisfy the Cauchy-Riemann conditions. Therefore, by the simple special case of Green's theorem in a disk, we have the Cauchy Integral Theorem in a disk for the function $1/P(z)$ and in turn the Gauss Meanvalue Theorem for the function $1/P(z)$.
We know that $1/P(0)$ is not zero, but due to the Gauss Meanvalue Theorem  $1/P(0) \rightarrow 0$ as the radius of the disk goes to infinity - a desired contradiction.
A: Here is a proof (probably the most well known one) mentioned in Lang's Analysis book.
Let $ f(z) = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_1 z + a_0 $ be a complex polynomial of degree $ n > 0 $. (We can write $ f(0) $ instead of $ a_0 $)
We'll show $ f $ must have a root.

*

*$ |f(z)| \to \infty $ as $ |z| \to \infty $

$ |f(z)| = |z|^n \left| a_n + \dfrac{a_{n-1}}{z} + \ldots + \dfrac{a_0}{z^n}\right| $, and terms $ |z|^n \to \infty $ and $ \left| a_n + \dfrac{a_{n-1}}{z} + \ldots + \dfrac{a_0}{z^n} \right| \to |a_n| (\neq 0) $ as $ |z| \to \infty $.


*

*So there is an $ R > 0 $ such that $ |f(z)| > |f(0)| $ whenever $ |z| > R $. Also $ z \mapsto |f(z)| $, defined on the compact disk $ \{ z : |z| \leq R \} $, attains its minimum at some point $ z_0 $ in the disk.


*Now $ | f(z_0) | \leq | f(z) | $ for all $ z \in \mathbb{C} $ (that is, $ \min_{z \in \mathbb{C}} |f(z)| $ exists and is $ | f(z_0) | $)

For points with $ |z| \leq R $ we anyways have $ |f(z_0)| \leq |f(z)| $.
For points with $ |z| > R $, we have $ |f(z_0)| \leq |f(0)| < |f(z)| $

We'll now show $ f(z_0) = 0 $.

Say to the contrary $ f(z_0) \neq 0 $.
Writing $ f(z) $ as a polynomial in $ (z-z_0) $ [by replacing each $ z^j $ with $ ((z-z_0)+z_0)^j $ and expanding], $$ f(z) = f(z_0) + b_1 (z-z_0) + \ldots + b_n (z-z_0)^n $$
As $ b_n \neq 0 $, there is a least $ j $ such that $ b_j \neq 0 $. Call it $ p $.
So $$ f(z) = f(z_0) + b_p (z-z_0)^p + (z-z_0)^{p+1} g(z-z_0) \, ; \, b_p \neq 0 $$ where $ g(z-z_0) $ is a polynomial in $ (z-z_0) $.
Putting $ z = z_0 + t \delta $ where $ \delta \in \mathbb{C} $ and $ t \in (0,1) $, and dividing by $ f(z_0) $,
$$ \dfrac{f(z_0 + t \delta)}{f(z_0)}  = 1 + c_p \delta^p t^p +\delta^{p+1} t^{p+1} G(t\delta) \, ; \, c_p \neq 0 $$
Fixing $ \delta $ such that $ c_p \delta^p = (-1) $, $$ \dfrac{f(z_0 + t \delta)}{f(z_0)} = 1 - t^p + t^{p+1} \varphi(t) $$ [where $ \varphi(t) = \delta^{p+1} G(t \delta) $ is a complex polynomial in variable $ t \in (0,1) $]
Notice $ \varphi(t)
 $ is bounded [$|\varphi(t)|$ is always $ \leq $ sum of absolute values of coefficients of $ \varphi $], say by $ K > 0 $.
Now taking absolute value,
$$\begin{align} \left| \dfrac{f(z_0 + t \delta)}{f(z_0)} \right| &\leq 1 - t^p + t^{p+1} |\varphi(t)| \\ &< 1 - t^p + t^{p+1} K \\ &= 1 - t^p (1 - tK) \end{align} $$ and RHS is $ < 1 $ for all $ t \in (0,1) $ with $ 1-tK > 0 $.
So we could find a point $ z_0 + t \delta $ at which $ |f(z_0 + t \delta) | < |f(z_0)| $, contradicting that $ |f(z_0)| = \min_{z \in \mathbb{C}} |f(z)| $.
Therefore $ f(z_0) $ must indeed be $ 0 $, as needed.
