Complex polynomial

  • Thread starter cdummie
  • Start date
  • #1
147
5

Homework Statement


It is known that roots of complex polynomial:

##P_n (z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0##

are the following complex numbers:

##\alpha_1, \alpha_2, \cdots, \alpha_n ##

Find the product:

##\prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1)##

Homework Equations




The Attempt at a Solution


I am pretty sure that i have to use Vieta formulas somehow, and since roots are all complex then it means that there are n/2 pairs of complex-complex conjugate roots. If i multiply all of this i would end up with product of all roots ##\alpha_1* \alpha_2* \cdots* \alpha_n ## and by Vieta formulas i can easily determine that one, but there are more elements of this product after multiplying, what can i do with them?
 

Answers and Replies

  • #2
535
72
Ok so it has ##n## roots without counting multiplicity.
So your polynomial has the form ##P_n(z) = A(z) (z-\alpha_1)...(z-\alpha_n) ##.
Explain why ##A = 1##
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966

Homework Statement


It is known that roots of complex polynomial:

##P_n (z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0##

are the following complex numbers:

##\alpha_1, \alpha_2, \cdots, \alpha_n ##

Find the product:

##\prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1)##

Homework Equations




The Attempt at a Solution


I am pretty sure that i have to use Vieta formulas somehow, and since roots are all complex then it means that there are n/2 pairs of complex-complex conjugate roots
This is only true if all coefficients are real numbers but you said "complex polynomial" which means that is not necessarily true.

If i multiply all of this i would end up with product of all roots ##\alpha_1* \alpha_2* \cdots* \alpha_n ## and by Vieta formulas i can easily determine that one, but there are more elements of this product after multiplying, what can i do with them?
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement


It is known that roots of complex polynomial:

##P_n (z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0##

are the following complex numbers:

##\alpha_1, \alpha_2, \cdots, \alpha_n ##

Find the product:

##\prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1)##

Homework Equations




The Attempt at a Solution


I am pretty sure that i have to use Vieta formulas somehow, and since roots are all complex then it means that there are n/2 pairs of complex-complex conjugate roots. If i multiply all of this i would end up with product of all roots ##\alpha_1* \alpha_2* \cdots* \alpha_n ## and by Vieta formulas i can easily determine that one, but there are more elements of this product after multiplying, what can i do with them?

The requirement of having complex-conjugate root pairs is true only if all the coefficients are real, and can fail if some of the coefficients are non-real complex numbers. For example, the equation ##x^2 + (2+i)x - 3 i = 0## has roots ##\alpha_1 \doteq 0.522 +0.788\, i## and ##\alpha_2 \doteq -2.552 - 1.788\, i ##.
 
  • #5
147
5
Ok so it has ##n## roots without counting multiplicity.
So your polynomial has the form ##P_n(z) = A(z) (z-\alpha_1)...(z-\alpha_n) ##.
Explain why ##A = 1##

I am not really sure, is it maybe because i have coefficient 1 with zn?
 
  • #6
535
72
yes, but if you want to convince yourself, you can justify that ## \text{deg}{(A)} = 0##, so that ##A## is a non-zero constant, and with your argument above, ##A = 1##.

Ok, so now you have the answer, what is ##P_n(-1)## ?
 
  • #7
147
5
yes, but if you want to convince yourself, you can justify that ## \text{deg}{(A)} = 0##, so that ##A## is a non-zero constant, and with your argument above, ##A = 1##.

Ok, so now you have the answer, what is ##P_n(-1)## ?

Well, it's ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##
But how this gets me any closer to the solution?
 
  • #8
535
72
You must have worked on this problem for too long :-)
Think about it 10 more minutes

EDIT: or don't think about it the next 10 minutes :-)

EDIT (+10 minutes) : What did we just say ? ##P_n(z) = (z-\alpha_1)...(z-\alpha_n)##. So what is ##P_n(-1)##?
 
Last edited:
  • #9
147
5
You must have worked on this problem for too long :-)
Think about it 10 more minutes

EDIT: or don't think about it the next 10 minutes :-)

EDIT (+10 minutes) : What did we just say ? ##P_n(z) = (z-\alpha_1)...(z-\alpha_n)##. So what is ##P_n(-1)##?
I am sorry, i really don't know anything than this ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0## which is obvious and it still isn't clear to me how this is going to help me solving this.
 
  • #10
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
I am sorry, i really don't know anything than this ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0## which is obvious and it still isn't clear to me how this is going to help me solving this.

Does this formula (up to a possible sign) express your desired product in terms of the coefficients ##a_0, a_1, \ldots, a_{n-1}##? Did the problem ask you to do more than that?
 
  • #11
147
5
Does this formula (up to a possible sign) express your desired product in terms of the coefficients ##a_0, a_1, \ldots, a_{n-1}##? Did the problem ask you to do more than that?

I really don't understand what are you trying to say, do you mean that ##a_0 + a_1 + \ldots + a_{n-1} = \prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1) ##?
 
  • #12
Samy_A
Science Advisor
Homework Helper
1,242
510
I am sorry, i really don't know anything than this ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0## which is obvious and it still isn't clear to me how this is going to help me solving this.
You know that ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##.

You also know another way to find ## P_n(-1)##:
What did we just say ? ##P_n(z) = (z-\alpha_1)...(z-\alpha_n)##. So what is ##P_n(-1)##?

Now put the two together.
 
  • #13
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
I really don't understand what are you trying to say, do you mean that ##a_0 + a_1 + \ldots + a_{n-1} = \prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1) ##?

No, absolutely not. YOU said that ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0 ## (not ##\sum a_j##). My question amounts to asking: why do you want to compute ##P_n(-1)##? What do you gain by doing that? How is that related to the product you started with?
 
  • #14
147
5
You know that ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##.

You also know another way to find ## P_n(-1)##:


Now put the two together.

Let's see: I have ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0## and ## P_n(-1) =(-1-\alpha_1) \cdots(-1-\alpha_n)##

Now, taking all -1's out i have ##P_n(-1) =(-1)^n(\alpha_1+1) \cdots(\alpha_n+1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##

Multiplying both sides by ##(-1)^{-n}## i have:

##(\alpha_1+1) \cdots(\alpha_n+1) = 1 + a_{n-1}(-1)^{-1} + \cdots + a_1(-1)^{1-n} + a_0(-1)^{-n}##

Now, if this is correct i have another question, my idea was to use Vieta's formulas, i mean i have a given product which i have to find:

##\prod=(\alpha_1 + 1)\cdots(\alpha_n+1)## i am sure that there would be ##\alpha_1\alpha_2\cdots\alpha_n## but what is with rest of it , i mean i would have 1 at the end, but in order to use Vieta's formulas i should get something like ##\alpha_1\alpha_2 + \alpha_1\alpha_3 + \cdots + \alpha_1\alpha_n + + \alpha_2\alpha_3 + \alpha_2\alpha_4 +\cdots + \alpha_{n-1}\alpha_n## and then sum of products of three and so on all the way to the product of all of them. Is this correct?
 
  • #15
Samy_A
Science Advisor
Homework Helper
1,242
510
Let's see: I have ## P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0## and ## P_n(-1) =(-1-\alpha_1) \cdots(-1-\alpha_n)##

Now, taking all -1's out i have ##P_n(-1) =(-1)^n(\alpha_1+1) \cdots(\alpha_n+1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0##

Multiplying both sides by ##(-1)^{-n}## i have:

##(\alpha_1+1) \cdots(\alpha_n+1) = 1 + a_{n-1}(-1)^{-1} + \cdots + a_1(-1)^{1-n} + a_0(-1)^{-n}##

Now, if this is correct i have another question, my idea was to use Vieta's formulas, i mean i have a given product which i have to find:

##\prod=(\alpha_1 + 1)\cdots(\alpha_n+1)## i am sure that there would be ##\alpha_1\alpha_2\cdots\alpha_n## but what is with rest of it , i mean i would have 1 at the end, but in order to use Vieta's formulas i should get something like ##\alpha_1\alpha_2 + \alpha_1\alpha_3 + \cdots + \alpha_1\alpha_n + + \alpha_2\alpha_3 + \alpha_2\alpha_4 +\cdots + \alpha_{n-1}\alpha_n## and then sum of products of three and so on all the way to the product of all of them. Is this correct?
Correct.

Using Vieta's formulas should also work, as you indeed get all possible combinations of products of roots when you expand ##\prod(\alpha_1 + 1)\cdots(\alpha_n+1)##.
 

Related Threads on Complex polynomial

  • Last Post
3
Replies
64
Views
3K
  • Last Post
Replies
7
Views
961
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
5
Views
915
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
736
Replies
6
Views
3K
  • Last Post
Replies
7
Views
2K
Replies
4
Views
3K
Top