Engineering Math

1.9 The Fundamental Theorem Of Algebra

The fundamental theorem of algebra states that every non constant polynomial having coefficients in ℂ has a zero in ℂ. If ℂ is replaced by ℝ, this is not true because of the example, x² + 1 = 0. This theorem is a very remarkable result and notwithstanding its title, all the most straightforward proofs depend on either analysis or topology. It was first mostly proved by Gauss in 1797. The first complete proof was given by Argand in 1806. The proof given here follows Rudin [22]. See also Hardy [15] for a similar proof, more discussion and references. The shortest proof is found in the theory of complex analysis. First I will give an informal explanation of this theorem which shows why it is is reasonable to believe in the fundamental theorem of algebra.

To begin with, here is the informal explanation. Dividing by the leading coefficient a_n, there is no loss of generality in assuming that the polynomial is of the form

p(z) = zn +a zn− 1 + ⋅⋅⋅+ a z +a n− 1 1 0

If a₀ = 0, there is nothing to prove because p

= 0. Therefore, assume a₀≠0. From the polar form of a complex number z, it can be written as . Thus, by DeMoivre’s theorem,

zn = |z|n(cos(nθ)+ isin(nθ))

It follows that zⁿ is some point on the circle of radius

ⁿ

Denote by C_r the circle of radius r in the complex plane which is centered at 0. Then if r is sufficiently large and

= r, the term zⁿ is far larger than the rest of the polynomial. It is on the circle of radius ⁿ while the other terms are on circles of fixed multiples of ^k for k ≤ n − 1. Thus, for r large enough, A_r = describes a closed curve which misses the inside of some circle having 0 as its center. It won’t be as simple as suggested in the following picture, but it will be a closed curve thanks to De Moivre’s theorem and the observation that the cosine and sine are periodic. Now shrink r. Eventually, for r small enough, the non constant terms are negligible and so A_r is a curve which is contained in some circle centered at a₀ which has 0 on the outside.

PICT

Thus it is reasonable to believe that for some r during this shrinking process, the set A_r must hit 0. It follows that p = 0 for some z.

For example, consider the polynomial x³ + x + 1 + i. It has no real zeros. However, you could let z = r

and insert this into the polynomial. Thus you would want to find a point where

3 (r(cos t+ isint)) + r(cost +isint)+ 1+ i = 0 + 0i

Expanding this expression on the left to write it in terms of real and imaginary parts, you get on the left

3 3 3 2 ( 3 2 3 3 ) r cos t− 3r costsin t+ rcost+ 1+ i 3r cos tsint− r sin t+ rsin t+ 1

Thus you need to have both the real and imaginary parts equal to 0. In other words, you need to have

( ) r3cos3t− 3r3costsin2t+ r cost +1,3r3cos2tsin t− r3sin3t+ rsint+ 1 = (0,0)

for some value of r and t. First here is a graph of this parametric function of t for t ∈ [0,2π] on the left, when r = 4. Note how the graph misses the origin 0 + i0. In fact, the closed curve surrounds a small circle which has the point 0 + i0 on its inside.

Next is the graph when r = .5. Note how the closed curve is included in a circle which has 0 + i0 on its outside. As you shrink r you get closed curves. At first, these closed curves enclose 0 + i0 and later, they exclude 0 + i0. Thus one of them should pass through this point. In fact, consider the curve which results when r = 1.386 which is the graph on the right. Note how for this value of r the curve passes through the point 0 + i0. Thus for some t, 1.3862

is a solution of the equation p = 0.

Now here is a rigorous proof for those who have studied analysis.

Proof. Suppose the nonconstant polynomial p

= a₀ + a₁z + + a_nzⁿ,a_n≠0, has no zero in ℂ. Since lim

|z|

→∞ = ∞, there is a z₀ with

|p(z0)| = min |p (z)| > 0 z∈ℂ

Then let q

= . This is also a polynomial which has no zeros and the minimum of is 1 and occurs at z = 0. Since q = 1, it follows q = 1 + a_kz^k + r where r consists of higher order terms. Here a_k is the first coefficient which is nonzero. Choose a sequence, z_n → 0, such that a_kz_n^k < 0. For example, let −a_kz_n^k = . Then

| k | k |q(zn)| = |1+ akz + r(z)| ≤ 1− 1∕n + |r(zn)| = 1+ akzn + |r(zn)| < 1

for all n large enough because

is small compared with since the latter involves higher order terms. This is a contradiction. ■