Fundamental Theorem of Algebra: At least one solution

In today's blog, I continue the proof for the Fundamental Theorem of Algebra. Today, I will show the proof that all polynomials in the complex domain have at least one root that leads to 0.

Today's proof is taken from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma 1: If f(x) = xⁿ + a₁x^n-1 + ... + a_n-1x + a_n where a_i and x are complex numbers, then there exists at least one solution r such that f(r)=0.

Proof:

(1) Let f(x) = xⁿ + a₁x^n-1 + ... + a_n-1x + a_n

(2) Assume that for all x, f(x) ≠ 0

(3) We know that there exists a value x₀ such that w₀=f(x₀) and w₀ is the smallest absolute number. [See Lemma 2 here for proof]

(4) From step #2, we can assume that absolute(w₀) is greater than 0.

(5) We can plot the minimal point, w₀, on the plane of complex numbers (see here for more details if needed)

(6) From this point, we can define a small circle K with radius R.

(7) Now, for any point x in K, x = x₀ + ζ

In other words, complex numbers form a one-to-one mapping between the possible values for f(x) and the Cartesian coordinate system. If we use the form r(cos θ + i sin θ), then we see that r is the radius of K [See here if more information needed]

(8) Using the plane of complex numbers, we know that there exists ρ, θ such that ζ = ρ(cos θ + isin θ) (see here for review of how cos θ + isin θ can be used in this situation)

(9) In the above case, ρ = the absolute magnitude of ζ

(10) So, for any value x, there exists a value w and a value ζ such that:

w = f(x) = f(x₀ + ζ) = (x₀ + ζ)ⁿ + a₁(x₀ + ζ)^n-1 + ... + a_n

(11) From the equation in #10, we can rearrange the values to get the following:

w = f(x₀) + c₁ζ + c₂ζ² + ... + c_nζⁿ = w₀ + c₁ζ + c₂ζ² + ... + c_nζⁿ

(12) Now, it is quite possible that some of the c_i values are 0 so we can rearrange the values so the first nonzero coefficient is c and the power is v, the next is c' and the power is v', and so on where each c,v are nonzero and v is less than v' is less than v'', etc.:

w = w₀ + cζ^v + c'ζ^v' + c''ζ^v'' + ...

(13) Since we assume that w₀ is nonzero, we can divide both sides by w₀ to get:

w/w₀ = 1 + (cζ^v)/w₀ + (c'ζ^v')/w₀ + (cζ^v'')/w₀ + ...

(14) Now let us define some values to make this equation more manageable:

Let q = c/w₀

Let ξ = (c'ζv' + c''ζv'' + ...)/(cζ^v)

So that:

w/w₀ = 1 + qζ^v(1 + ζξ)

(15) Now, since q, ζ are complex numbers, we can represent them both using r(cos + isin) form (see here if more information needed)

So that there exists h, λ such that:

q = h(cos λ + i sin λ)

From step #8, there exists ρ, θ such that:

ζ = p(cos θ + i sin θ)

(16) To shorten the equation we can use:

1_λ = cos λ + i sin λ

And use:

1_θ = cos θ + i sin θ

So that we have:

q = h * 1_λ

ζ = ρ * 1_θ

(17) Using step #16, we get:

qζ^v = h*1_λ*(ρ*1_θ)^v = h*p^v*1_λ*(1_θ)^v

(18) Now, using Euler's Formula (see Lemma #1, Lemma #2 here if needed), we know that:

(1_θ)^v = 1_vθ

1_vθ*1_λ = 1_{λ + vθ}

(19) Within the circle K, we can now consider only the values of x that are associated with θ = (π - λ)/v. [Since a circle includes all values of θ between 0 and 2π, see here if needed]

(20) In this case:

1_{λ + vθ} = 1_{λ + v(π - λ)/v} = 1_{λ - λ + π} = 1_π

(21) Now 1_π = cos (π) + isin(π) = -1 + i*0 = -1. [See here if needed]

(22) So, in this case, combining step #21 with step #17:

qζ^v = h*ρ^v*(-1) = -h*ρ^v

(23) Combining step #22 with step #14:

w/w₀ = 1 - hρ^v(1 + ζξ)

(24) Now, we can set the radius of K to any value so we can constrain ζξ to be as close to 0 as we wish so that for all purposes, we have:

w/w₀ = 1 - hρ^v

[The idea here is that the radius of K was selected arbitarily in step #6, any nonzero radius r will do]


(25) Now, we can choose any value for ρ so we choose a value such that ρ is greater than 0 and less than (1/h)^(1/v).

We can do this since ρ is the magnitude of ζ (see step #9). In step #19, we constrained θ and in step #24, we constrained the maximum magnitude of the circle K.

Even with all of the above constraints, we are still left with a set of values and we can select a value of x such that the magnitude is less than the radius of K and less than (1/h)^(1/v) but still greater than 0.

(26) But then hρ^v is greater than 0 and less than h*(1/h) = 1 since:

h*[(1/h)^(1/v)]^v = 1

(27) But this means there is a value of x such that w/w₀ is greater than 0 and less than 1.

(28) But this is a contradiction because it means that w is less than w₀ which is impossible from step #3.

The reasoning here is that if w/w₀ = fraction, this implies that w = w₀ * fraction which implies that w is less than w₀

(25) So we are forced to reject our assumption in step #2.

QED

References

• Heinrich Dorrie, 100 Great Problems of Elementary Mathematics