 #1
 106
 3
Main Question or Discussion Point
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 z_{1}, z_{2}, ......, z_{n} 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 nonzero 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..
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 z_{1}, z_{2}, ......, z_{n} 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 nonzero 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..
Attachments

97.9 KB Views: 364

57.9 KB Views: 375

107.7 KB Views: 335

76.9 KB Views: 361

97.5 KB Views: 363

24.3 KB Views: 384