What is the proof of the ''Fundamental Theorem of Algebra''?

Answer 1

The Fundamental Theorem of Algebra:

If P(z) = Σ­­nk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0, then P(z0) = 0 for some z0 Є C. Descriptively, this says that any nonconstant polynomial over the complex number space, C, can be written as a product of linear factors.

Proof:

First off, we need to apply the Heine-Borel theorem to C. The Heine-Borel theorem states that if S is a closed and bounded set in an m-dimensional Euclidean space (written as Rm), then S is compact.

From above, P(z) = Σ­­nk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0. Let m = inf{|P(z)| : z Є C} where inf is the infinum, or the greatest lower bound of the set.

From the triangle inequality, |P(reit)| ≥ rn(|an| - r-1|an-1| - … - r-n|a0|),

so limr --> ∞ |P(reit)| = ∞. Therefore there is a real number R that |P(reit)| > m + 1 whenever r > R.

If S = {reit : r ≤ R}, then S is compact in C, by the Heine-Borel Theorem; and let m = inf{|P(z)| : z Є S}. |P| is a continuous and real-valued function in S, so, using the result from another proof not done here, it has a minimum value on S; i.e., there is a value for z0 Є S that makes |P(z0)| = m. So, if m = 0 then the theorem is proved.

We're going to show that m = 0 by proving that m can't equal anything else, and since we know m exists, it has no choice but to be zero. So, suppose m ≠ 0 and let Q(z) = P(z + z0)/P(z0), z Є C.

Q is therefore a polynomial with degree n and |Q(z)| ≥ 1 for all z Є C.

Q(0) = 1 so Q(z) can be expressed via P's series as:

Q(z) = 1 + bkzk + … + bnzn where k is the smallest positive integer ≤ n such that bk ≠ 0.

Since |-|bk|/bk| = 1, there exists a t0 Є [0, 2π/k] such that eikt0 = -|bk|/bk.

Then Q(reit0) = 1 + bkrkeikt0 + bk+1rk+1ei(k+1)t0 + … + bnrneint0

= 1 - rk|bk| + bk+1rk+1ei(k+1)t0 + … + bnrneint0.

So, if rk|bk| < 1 then |Q(reit0)| ≤ 1 - rk(|bk| - r|bk+1| - … - rn-k|bn|).

That means that if we pick a small enough r, we can make |Q(reit0)| ≤ 1 which contradicts the statement above that |Q(z)| ≥ 1 for all z Є C. Therefore m ≠ 0 doesn't hold and P(z0) = 0

Q.E.D.

Another proofSuppose P has no zeroes. Then we can define the function f(z) = 1 / P(z), and f is analytic. By the proof above, P(z) tends to infinity as z tends to infinity; hence f(z) tends to 0 as z tends to infinity. So there is a disc S such that f, restricted to the outside of S, is bounded. Also by the proof above, f is bounded inside the disc as well; therefore f is bounded. Now we apply a theorem called Liouville's Theorem, which says that any analytic function which is defined on all of C and is bounded must be a constant. So f is a constant; therefore P is constant. But we were assuming that P is not constant, so this is a contradiction.

(To prove Liouville's Theorem: Suppose M is a bound for the function f, i.e. |f(z)| < M for all z. Suppose a and b are complex numbers, and we want to show f(a) = f(b). Use the theorem that f(a) = integral of f(z)/(z-a) / (2 * pi * i) around the circle of radius R and centre 0. Then, if R is sufficiently large:

|f(b) - f(a)|

= | integral, around circle, of (f(z) * (1/(z-b) - 1/(z-a))) | / (2*pi)

= | integral around circle of (f(z) * (b-a) / ((z-a)(z-b)) ) | / (2*pi)

<= (M * |b-a| / ((R-|a|)(R-|b|)) ) * (2*pi*R) / (2*pi)

The last line uses the formula |integral| <= |pathlength| * |maximum value|. Then we get |f(b) - f(a)| <= M * |b-a| * R / ((R-|a|)(R-|b|)). Letting R tend to infinity, we can prove that |f(b)-f(a)| is as small as we like; therefore f(a) = f(b).

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