Fundamental theorem of algebra: a proof for undergrads?

Question

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?

Answer 1

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)

Answer 2

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")

Answer 3

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 $.

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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.