Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The fundamental theorem of algebra asserts:

Theorem Let $P$ be a polynomial of degree $\geq 1$ in $\Bbb C$. Then there exists a $z_1\in\Bbb C$ such that $P(z_1)=0$.

The proof sketch goes as follows:

Lemma 1 For each $A>0$ there is an $R>0$ such that $|z|\geq R$ implies $|P(z)|\geq A$

Proof For $z\neq 0$, write

$$P(z)=a_0z^n Q(z)$$

where $$Q(z)=\sum_{k=0}^n \frac{a_k}{a_0}z^{-k}$$

Then $Q\to 1$ for $z\to \infty$ so there is an $R_1$ for which $|z|\geq R_1$ implies $|Q(z)|\geq 1/2$. Thus for $|z|\geq R_1$ we have

$$|P(z)|=|a_0||z|^n| Q(z)|\geq \frac 1 2 |a_0||z|^n$$

From this it is clear we can make $|P(z)|>A$ by taking $|z|>R$ where

$$R=\max\left\{R_1,\left(\frac{2A}{|a_0|}\right)^{1/n}\right\}$$

Lemma 2 If $P$ is a polynomial of degree $\geq 1$ and $P(z_1)\neq 0$, given $\delta_0 >0$ there is a $z_2$ such that $|z_1-z_2|<\delta$ and $|P(z_2)|<|P(z_1|$

Proof Consider the particular case $P(0)=1$. We may then write

$$P(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+1$$

Now let $a_k$ by the first coefficient $a_0,\dots,a_{n-1}$ that is nonvanishing. Then

$$P(z)=1+a_kz^k+\cdots a_nz^n$$ so we may write $$P(z)=1+a_k z^k(1+H(z))$$ where $$H(z)=\frac{a_{k-1}}{a_k}z+\cdots+^\frac{a_0}{a_k}z^{n-k}$$

Since $H(0)=0$; by continuity $|H(z)|<\frac 1 2 $ for each $z$ with $|z|\leq \delta <\delta_0$, where $\delta$ is such that $\delta^k|a_k|<1$. We choose $z_2$ as a solution of

$$z^k=-\delta^k\frac{|a_k|}{a_k}$$

Then clearly $|z_2|=\delta$ and $$|P(z_2)|=|1-\delta^k|a_k|-\delta^k|a_k||H(z_2)||\\ \leq 1-\delta^k|a_k|+\delta^k|a_k||H(z_2)|\\ \leq 1-\delta^k|a_k|+\frac 1 2\delta^k|a_k|<1$$ which is what we wanted.

In the general case where $P(z_0)\neq 0$, we write $P(z)=\sum a_k z^k$ as $P(z)=\sum b_k(z-z_0)^k$ and note that $P(z)=b_n P_1(z)$ with $P_1(0)=1$, so we profit from what we did before.

Finally, the proof

PROOF Since $P$ is continuous so is $|P|$, and $|P|\geq 0$. Set thus $\alpha =\inf_{z\in \Bbb C}|P(z)|$. Since we can take, from the first lemma, an $R$ such that $|P(z)|\geq \alpha +1$ we may discard the region $|z|> R$ and write $$\alpha =\inf_{|z|\leq R}|P(z)|$$ By continuity of $|P|$ and compactness of the disk $K=\{z:|z|\leq R\}$, Weierstrass' theorem asserts that for some $z_1$ we have $|P(z_1)|=\alpha$. Since $|P(z_1)|=\alpha<\alpha+1$, $z_1$ is an interior point of the disk $K$, whence there is some $\delta_0$-ball around it. If $\alpha \neq 0$, the second lemma implies we can take some $z_2$ with $|z_1-z_2|<\delta_0$ and $|P(z_2)|<\alpha$ contrary to $\alpha$ being the infimum, whence $|P(z_1)|=\alpha=0$, which implies $P(z_1)=0$ and the theorem is proven.

Does anyone know where this proof first appeared? Who produced it? I've seen it already in Spivak's "Calculus" and E.G. Shilov's "Elementary Real and Complex Analysis".

share|improve this question
1  
When I read "topological proof of FTA" I thought you meant the proof that the fundamental group of $\Bbb C \setminus\{\text{Zeros of }P\}$ is non-trivial if $P$ is non-constant. –  Arthur Dec 22 '12 at 22:06
    
@Arthur Oh. I was merely stressing the compactness considerations. –  Pedro Tamaroff Dec 22 '12 at 23:37

1 Answer 1

up vote 3 down vote accepted

When I was taught this proof, it was attributed to d'Alembert by my professor (and probably textbook). However, this paper I just read -- D'Alembert's proof of the fundamental theorem of algebra makes me doubt it. I don't see the key idea ("if $|p|$ is not zero, we can make it smaller") there, in fact there is nothing about inequalities with $|p|$. According to the paper, the idea belongs to Argand, published in 1806. (And it seems that Cauchy republished it without crediting Argand.)

I now recall that complex plane was called Argand's plane in some books, perhaps justifiably so.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.