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) = x^{n} + a_{1}x^{n-1} + ... + a_{n-1}x + 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) = x^{n} + a_{1}x^{n-1} + ... + a_{n-1}x + a_{n}(2) Assume that for all

x, f(x) ≠ 0(3) We know that there exists a value

x_{0} such that

w_{0}=f(x_{0}) and

w_{0} is the smallest absolute number. [See Lemma 2

here for proof]

(4) From step #2, we can assume that

absolute(w_{0}) is greater than

0.

(5) We can plot the minimal point,

w_{0}, 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 = x_{0} + ζIn 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(x_{0} + ζ) = (x_{0} + ζ)^{n} + a_{1}(x_{0} + ζ)^{n-1} + ... + a_{n}(11) From the equation in #10, we can rearrange the values to get the following:

w = f(x_{0}) + c_{1}ζ + c_{2}ζ^{2} + ... + c_{n}ζ^{n} = w_{0} + c_{1}ζ + c_{2}ζ^{2} + ... + c_{n}ζ^{n}(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 = w_{0} + cζ^{v} + c'ζ^{v'} + c''ζ^{v''} + ...(13) Since we assume that

w_{0} is nonzero, we can divide both sides by

w_{0} to get:

w/w_{0} = 1 + (cζ^{v})/w_{0} + (c'ζ^{v'})/w_{0} + (cζ^{v''})/w_{0} + ...(14) Now let us define some values to make this equation more manageable:

Let

q = c/w_{0}Let

ξ = (c'ζv' + c''ζv'' + ...)/(cζ^{v})So that:

w/w_{0} = 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 λAnd use:

1_{θ} = cos θ + i sin θSo 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/w_{0} = 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/w_{0} = 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/w_{0} is greater than

0 and less than

1.

(28) But this is a contradiction because it means that

w is less than

w_{0} which is impossible from step #3.

The reasoning here is that if

w/w_{0} = fraction, this implies that

w = w_{0} * fraction which implies that

w is less than

w_{0}(25) So we are forced to reject our assumption in step #2.

QED

References