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


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:
    or more precisely here:
    See also the Heine-Borel theorem.

    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.

  10. Pingback: FTOA: Let $f(z) = |a_nz^n + .. + a_1z + a_0|$. Show $f(z)$ has a minimizer $z^*$ and complex sum converges for $|z| rightarrow infty$ - MathHub

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