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Textbook by: Lawrence Baggett. E-mail the author

# The Fundamental Theorem of Algebra

Module by: Lawrence Baggett. E-mail the author

Summary: We can now prove the Fundamental Theorem of Algebra, the last of our primary goals. One final trumpet fanfare, please!

We can now prove the Fundamental Theorem of Algebra, the last of our primary goals. One final trumpet fanfare, please!

## Theorem 1: Fundamental Theorem of Algebra

Let p(z)p(z) be a nonconstant polynomial of a complex variable. Then there exists a complex number z0z0 such that p(z0)=0.p(z0)=0. That is, every nonconstant polynomial of a complex variable has a root in the complex numbers.

### Proof

We prove this theorem by contradiction. Thus, suppose that pp is a nonconstant polynomial of degree n1,n1, and that p(z)p(z) is never 0. Set f(z)=1/p(z),f(z)=1/p(z), and observe that ff is defined and differentiable at every point zC.zC. We will show that ff is a constant function, implying that p=1/fp=1/f is a constant, and that will give the contradiction. We prove that ff is constant by showing that its derivative is identically 0, and we compute its derivative by using the Cauchy Integral Formula for the derivative.

From part (4) of (Reference), we recall that there exists a B>0B>0 such that |cn|2|z|n|p(z)|,|cn|2|z|n|p(z)|, for all zz for which |z|B,|z|B, and where cncn is the (nonzero) leading coefficient of the polynomial p.p. Hence, |f(z)|M|z|n|f(z)|M|z|n for all |z|B,|z|B, where we write MM for 2/|cn|.2/|cn|. Now, fix a point cC.cC. Because ff is differentiable on the open set U=C,U=C, we can use the corollary to (Reference) to compute the derivative of ff at cc by using any of the curves CrCr that bound the disks Br(c),Br(c), and we choose an rr large enough so that |c+reit|B|c+reit|B for all 0t2π.0t2π. Then,

| f ' ( c ) | = | 1 2 π i C r f ( ζ ) ( ζ - c ) 2 d ζ | = 1 2 π | 0 2 π f ( c + r e i t ) ( c + r e i t - c ) 2 i r e i t d t | 1 2 π r 0 2 π | f ( c + r e i t ) | d t 1 2 π r 0 2 π M | c + r e i t | n d t M r B n . | f ' ( c ) | = | 1 2 π i C r f ( ζ ) ( ζ - c ) 2 d ζ | = 1 2 π | 0 2 π f ( c + r e i t ) ( c + r e i t - c ) 2 i r e i t d t | 1 2 π r 0 2 π | f ( c + r e i t ) | d t 1 2 π r 0 2 π M | c + r e i t | n d t M r B n .
(1)

Hence, by letting rr tend to infinity, we get that

| f ' ( c ) | lim r M r B n = 0 , | f ' ( c ) | lim r M r B n = 0 ,
(2)

and the proof is complete.

## REMARK 1

The Fundamental Theorem of Algebra settles a question first raised back in (Reference). There, we introduced a number II that was a root of the polynomial x2+1.x2+1. We did this in order to build a number system in which negative numbers would have square roots. We adjoined the “number” ii to the set of real numbers to form the set of complex numbers, and we then saw that in fact every complex number zz has a square root. However, a fear was that, in order to build a system in which every number has an nnth root for every n,n, we would continually need to be adjoining new elements to our number system. However, the Fundamental Theorem of Algebra shows that this is not necessary. The set of complex numbers is already rich enough to contain all nnth roots and even more.

Practically the same argument as in the preceding proof establishes another striking result.

## Theorem 2: Liouville

Suppose ff is a bounded, everywhere differentiable function of a complex variable. Then ff must be a constant function.

## Exercise 1

Prove Liouville's Theorem.

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