## Another proof of the Fundamental theorem of algebra

Motivated by Zorn Leung’s post, let me show another proof of the famous Fundamental theorem of algebra. It is also mentioned in a mathoverflow post.  The proof can be found in a paper by Terkelsen (1976), but the idea should have appeared in earlier works.

Theorem (Fundamental theorem of algebra)
Every polynomial $p(z)=a_n z^n+a_{n-1} z^{n-1}+\cdots + a_1 z+ a_0$ with leading coefficient $a_n\neq 0$ has a complex root.

Proof. Consider minimizing the function $f:\mathbb{C}\rightarrow \mathbb{R}$ given by

$f(z)\triangleq |p(z)|$              $\forall z\in \mathbb{C}$.

We have for $z\neq 0$,

$\displaystyle f(z)=|z|^n \cdot \left| a_n + \frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2}+\cdots + \frac{a_1}{z^{n-1}}+\frac{a_0}{z^n}\right|$

As $|z|\rightarrow \infty$, the sum above converges to $|a_n|>0$. Thus $f(z)\rightarrow \infty$. By this and elementary analysis, one sees that $f$ has a minimizer $z^*\in \mathbb{C}$.

First we assume $z^*=0$. Then as $0$ is a minimizer of $f$,

$\displaystyle |a_0|=f(0)\leq f(z)=\left|p(z)\right|$         $\forall z\in \mathbb{C}$.

Hence $|a_0|$ is the minimum value of $f$. If $a_0=0$, then $z=0$ is a root of $p$ and we are done. If $a_0\neq 0$, then we write

$p(z)=a_0 + a_k z^k + z^{k+1} q(z)$

where $a_k$ is the first nonzero coefficient after $a_0$ and $q$ is a polynomial. Let $w\in \mathbb{C}$ be a k-th root of $\displaystyle -\frac{a_0}{a_k}$. Then for any $t\in \mathbb{R}$,

$p(tw)=a_0 + a_k t^k w^k + t^{k+1}w^{k+1}q(tw)=(1-t^k)a_0+t^k[tw^{k+1}q(tw)]$.

If $t>0$ is small, then $|q(tw)|$ is small. Hence for a sufficiently small $t\in (0,1)$, $|tw^{k+1}q(tw)|<|a_0|$, and so

$|p(tw)|<(1-t^k)|a_0|+t^k|a_0|=|a_0|$

which is a contradiction. We have justified if $0$ is the minimizer of $f$, then $p(0)=a_0=0$.

If $z^*\neq 0$, then consider the polynomial $r(z)=p(z+z^*)$. One has that $0$ is a minimizer of $|r|$. By the above case, $r(0)=0$ and hence $p(z^*)=0$. The proof is completed.

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### 10 Responses to Another proof of the Fundamental theorem of algebra

1. KKK says:

Very nice! In my view this is the best proof of FTA in that nearly no knowledge of complex analysis is required, except that we can take the k-th root of any number.

2. Hon Leung says:

Agree. I think the proof (or the idea of the proof) can be taught to some “smart” high school students.

3. lee king leung says:

Very nice!
Very nice

4. KKK says:

Small typo: f(z) should be $|z|^n |\cdots |$ instead of $|z| | \cdots |$. And in the statement of FTA, p should be non-constant (of course).

5. Nicolas Lykke Iversen says:

Could you show formally why the sum converges to a_n and why f(z) has a minimizer z^* ? I really like your proof, but I need to understand these two steps.

6. KKK says:

As $z\to \infty$, $\frac{a_{n-i}}{z^i}\to 0$ if $i>0$, thus the finite sum $|a_n +\frac{a_{n-1}}{z}+\cdots + \frac{a_0}{z^n} |$ converges to $|a_n|$.

As $f(z)\to \infty$ as $z\to \infty$, there is a sufficiently large closed ball $B$ centered at $0$ such that $f(z)$ is large (say $>f(0)+10^9$) outside $B$. As $B$ is closed and bounded, a minimizer of $f(z)$ in $B$ exists and is actually a minimum point of $f$.

7. Nicolas Lykke Iversen says:

Could you point me to some theorem/lemma proving that f has a minimizer in B if it is closed and bounded ?

8. Nicolas Lykke Iversen says:

Also, could one show formally that the sum converges to |a_n| using epsilon without to many trouble ?

9. KKK says:

For the first question, see here:
http://en.wikipedia.org/wiki/Extreme_value_theorem
or more precisely here:
http://en.wikipedia.org/wiki/Compact_space#Theorems
For any constant $C\ne 0$ and $\varepsilon>0$, if $|z|^k> \frac{|C|}{\varepsilon}$, then $\left|\frac {C} {z^k}\right|<\varepsilon$. Using this, and the triangle inequality, I think you should have no trouble working out the estimate.