Understanding the fundamental theorem of algebra

In summary, the fundamental theorem of algebra states that for a nth order complex polynomial like ## p(z) = a_{0} + a_{1}z + ... + a_{n}z^n ## there are n distinct complex solutions. To prove this theorem, we use topological arguments.
  • #1
Dear all,
I am trying to understand the fundamental theorem of algebra from the text (Alan F. Beardon, Algebra and Geometry) attached in this post. I have understood till the first two attachments and my question is from the 3rd attachment onwards.

I will briefly describe what is said in the first two attachments. In the first two attachments, it is stated that for a nth order complex polynomial like
## p(z) = a_{0} + a_{1}z + ... + a_{n}z^n ##
there are n distinct complex solutions.

The fundamental theorem of Algebra proves this fact. It also says that this same equation can be written in another format like

## p(z) = a_{n}(z - z_{1}) ... (z - z_{n}) ##

so that each z1, z2, ..., zn are complex solutions.

We start off by saying that a complex polynomial like ## p(z) ## has at least a single root.
We will use topological arguments to prove this.

If ## p(z) ## as in the first equation above were written in complex form with z substituted by ## z = re^{iθ} ## it would be

## p(z) = a_{0} + (a_{1}r^{iθ}) + (a_{2}r^2e^{i2θ}) + ... + (a_{n}r^ne^{inθ}) ##

In the above equation ## a_{0} ## is a non-zero term so that p(0) is not 0 but why z=0 be a root?
If r value in ## re^{iθ} ## is very big then the ##(a_{n}r^ne^{inθ}) ## term will dominate and it will be a big circle that has rotated n times. if the r value is small then the constant ## a_{0} ## term will dominate. and the small r will form a small circle around it as shown in the second figure in the second attachment.

Now, If the radius, r increases gradually every 0 to 2pi rotation. For some r and θ, p(z) will intersect the origin, 0 + 0i.

This will be the first root of p(z). I see that the magnitude of the origin from ## a_{0} ## will be fixed and the only changes are going to be in the angle θ. so the next solutions will be the angle at which p(z) intersects the 2nd time, 3rd time and so on. So r is fixed and only θ changes for each root? Am I right about this??

Next, in the 3rd paragraph of the third attachment. We use proof by induction to show that if there is one root for n = 1. then there are roots from 1 to n for n > 1. Right??

n = 1 is the anchor step. It is correct when we substitute in p(z) = 0 below.

## p(z) = a_{0} + (a_{1}r^{iθ}) = a_{n}(z - z_{1}) ##

so the single solution for n =1 must be ## r^{iθ} = - a_{0} / a_{1} ## similarly ## z = z_{1} ## from above equation.

by putting n = n + 1, we must show that the R.H.S = L.H.S in Induction step of the proof but I don't know how to do it. :( can anyone show me?? thanks..

How did they arrive at the formula ## p(z) - p(z_{1}) = \sum\limits_{k=1}^n a_{k}(z^k - z_{1}^k) ##

In the above equation, z_{1} is the first root of p(z) when n = 1 so why is it in the above equation??

Danke..
 

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  • #2
PcumP_Ravenclaw said:
How did they arrive at the formula ## p(z) - p(z_{1}) = \sum\limits_{k=1}^n a_{k}(z^k - z_{1}^k) ##
..
You can rewrite the right hand side as ## \sum\limits_{k=1}^n a_{k}(z^k) - \sum\limits_{k=1}^n a_{k}(z_{1}^k)##
 
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  • #3
so This is equivalent to ## p(z) = p(z_{1}) + \sum\limits_{k=1}^n a_{k}(z^k) - \sum\limits_{k=1}^n a_{k}(z_{1}^k) ##. When this equation is compared with ## p(z) = a_{0} + a_{1}z + ... + a_{n}z^n ## The term ## a_{0} ## equals ##p(z_{1}) - \sum\limits_{k=1}^n a_{k}(z_{1}^k)## and the terms ## a_{1}z + ... + a_{n}z^n ## equals ## \sum\limits_{k=1}^n a_{k}(z^k) ## ?
 
  • #4
PcumP_Ravenclaw said:
so This is equivalent to ## p(z) = p(z_{1}) + \sum\limits_{k=1}^n a_{k}(z^k) - \sum\limits_{k=1}^n a_{k}(z_{1}^k) ##. When this equation is compared with ## p(z) = a_{0} + a_{1}z + ... + a_{n}z^n ## The term ## a_{0} ## equals ##p(z_{1}) - \sum\limits_{k=1}^n a_{k}(z_{1}^k)## and the terms ## a_{1}z + ... + a_{n}z^n ## equals ## \sum\limits_{k=1}^n a_{k}(z^k) ## ?

Yes you are correct . But more simply, the ## a_{0}'s## cancel and you are just left with the difference of terms of degree 1 and higher.

Notice that the right hand side has a factor of ##z-z_{1}##
 
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  • #5
PcumP_Ravenclaw said:
it is stated that for a nth order complex polynomial like
## p(z) = a_{0} + a_{1}z + ... + a_{n}z^n ##
there are n distinct complex solutions.

It doesn't say "distinct".
 
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  • #6
I scribbled some numbers and tried to understand the connection between
## p(z) = a_{0} + a_{1}z + ... + a_{n}z^n ## and ## p(z) = p(z_{1}) + \sum\limits_{k=1}^n a_{k}(z^k) - \sum\limits_{k=1}^n a_{k}(z_{1}^k) ## . I have attached my work in this post. So it seems to me that the only solution for ## p(z) - p(z_{1}) = \sum\limits_{k=1}^n a_{k}(z^k - z_{1}^k) ## is ## z_{1} ##. what about the remaining n - 1 solutions??
 

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  • #7
Once you factor out ## z - z_{1}## , ## a_{1} ## will be a constant
 
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  • #8
PcumP_Ravenclaw said:
## p(z) = a_{0} + a_{1}r^{iθ} + a_{2}r^2e^{i2θ} + ... + a_{n}r^ne^{inθ}##

In the above equation ##a_{0}## is a non-zero term so that p(0) is not 0 but why z=0 be a root?
If r value in re^{iθ} is very big then the (anrneinθ)(a_{n}r^ne^{inθ}) term will dominate and it will be a big circle that has rotated n times. if the r value is small then the constant a0 a_{0} term will dominate. and the small r will form a small circle around it as shown in the second figure in the second attachment.

Now, If the radius, r increases gradually every 0 to 2pi rotation. For some r and θ, p(z) will intersect the origin, 0 + 0i.

This will be the first root of p(z). I see that the magnitude of the origin from a0 a_{0} will be fixed and the only changes are going to be in the angle θ. so the next solutions will be the angle at which p(z) intersects the 2nd time, 3rd time and so on. So r is fixed and only θ changes for each root? Am I right about this??
 
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  • #9
PcumP_Ravenclaw said:
## p(z) = a_{0} + a_{1}r^{iθ} + a_{2}r^2e^{i2θ} + ... + a_{n}r^ne^{inθ}##
In the above equation ##a_{0}## is a non-zero term so that p(0) is not 0 but why z=0 be a root?
Why do you think that z = 0 should be a root?
 
  • #10
PcumP_Ravenclaw said:
This will be the first root of p(z). I see that the magnitude of the origin from ## a_{0} ## will be fixed and the only changes are going to be in the angle θ. so the next solutions will be the angle at which p(z) intersects the 2nd time, 3rd time and so on. So r is fixed and only θ changes for each root? Am I right about this??
No. Remember that ##r## is the modulus of ##z##. If ##r## were fixed, that means all of the roots would be the same distance from the origin. Clearly, that can't be true in general, e.g., p(z) = (z-2)(z-1).

Suppose you let ##p(z) = a_0 + q(z)##. For every root of p(z), it's must be that ##q(z)=-a_0##. So, for instance, ##q(z_1) = q(r_1e^{i\theta_1})=-a_0## and ##q(z_2) = q(r_2 e^{i\theta_2}) = -a_0##. It doesn't follow, however, that ##r_1 = r_2##. You've confused ##\|q(z)\|##, which is fixed for all the roots of p(z), with ##r=\|z\|##.

How did they arrive at the formula ## p(z) - p(z_{1}) = \sum\limits_{k=1}^n a_{k}(z^k - z_{1}^k) ##
I'm not sure you understood what lavinia said in post 4. You have
\begin{align*}
p(z) &= a_0 + a_1 z + \cdots + a_n z^n \\
p(z_1) &= a_0 + a_1 z_1 + \cdots + a_n z_1^n
\end{align*} What do you get when you subtract the second equation from the first?
 
  • #11
If you start out with the fact that every complex polynomial has at least a single root (say z = a0), then you can rewrite your polynomial p(z) as (z - a0)*p1(z), where the degree of p1 is one less than the degree of p. Now p1 is also a polynomial and will therefore have at least a single root (say z = a1). Continue this argument until the resulting pn has degree 1, and the theorem is proved.
 
  • #12
PcumP_Ravenclaw said:
Dear all,
I am trying to understand the fundamental theorem of algebra from the text (Alan F. Beardon, Algebra and Geometry) attached in this post. I have understood till the first two attachments and my question is from the 3rd attachment onwards.

I will briefly describe what is said in the first two attachments. In the first two attachments, it is stated that for a nth order complex polynomial like
## p(z) = a_{0} + a_{1}z + ... + a_{n}z^n ##
there are n distinct complex solutions.
Was this typo? While any nth degree polynomial equation has n complex solutions, in general they are NOT all distinct.
 

What is the fundamental theorem of algebra?

The fundamental theorem of algebra is a mathematical theorem that states that every non-constant polynomial equation with complex coefficients has at least one complex root. In other words, it guarantees the existence of solutions to polynomial equations.

Why is the fundamental theorem of algebra important?

The fundamental theorem of algebra is important because it is a fundamental result in mathematics that has many applications in algebra, geometry, and other fields. It also serves as the basis for many other theorems and concepts in mathematics.

How is the fundamental theorem of algebra proven?

The fundamental theorem of algebra is typically proven using techniques from complex analysis, which is a branch of mathematics that studies functions of complex numbers. The proof involves showing that a polynomial equation with complex coefficients can be rewritten as a product of linear factors, and then using the properties of complex numbers to show that at least one of these factors must have a complex root.

Can the fundamental theorem of algebra be generalized to other fields?

Yes, the fundamental theorem of algebra has been generalized to other fields, such as the real numbers, rational numbers, and finite fields. These generalizations state that every non-constant polynomial equation in these fields has at least one solution in that field.

Are there any exceptions to the fundamental theorem of algebra?

No, there are no exceptions to the fundamental theorem of algebra. This means that every non-constant polynomial equation with complex coefficients will have at least one complex root. However, it is important to note that the theorem only guarantees the existence of solutions, not the method for finding them.

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