Finding Product of Complex Polynomial Roots

AI Thread Summary
The discussion revolves around finding the product of complex polynomial roots using Vieta's formulas. It highlights that while complex roots can form conjugate pairs when coefficients are real, this is not guaranteed for polynomials with complex coefficients. The participants explore the relationship between the polynomial evaluated at -1 and the desired product of roots, ultimately establishing that the product can be expressed in terms of the polynomial's coefficients. They conclude that the product can be derived from Vieta's formulas, confirming that all combinations of root products are accounted for in the expansion. The conversation emphasizes the importance of understanding polynomial behavior and root relationships in complex analysis.
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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?
 
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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##
 
cdummie said:

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?
 
cdummie said:

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 ##.
 
geoffrey159 said:
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?
 
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)## ?
 
geoffrey159 said:
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?
 
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)##?
 
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geoffrey159 said:
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
cdummie said:
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
Ray Vickson said:
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
cdummie said:
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)##:
geoffrey159 said:
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.
 
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  • #13
cdummie said:
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?
 
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  • #14
Samy_A said:
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
cdummie said:
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)##.
 
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