Complex polynomial

1. Dec 4, 2015

cdummie

1. The problem statement, all variables and given/known data
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:

$\prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1)$
2. Relevant equations

3. 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?

2. Dec 4, 2015

geoffrey159

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. Dec 4, 2015

HallsofIvy

Staff Emeritus
This is only true if all coefficients are real numbers but you said "complex polynomial" which means that is not necessarily true.

4. Dec 4, 2015

Ray Vickson

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. Dec 4, 2015

cdummie

I am not really sure, is it maybe because i have coefficient 1 with zn?

6. Dec 4, 2015

geoffrey159

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. Dec 4, 2015

cdummie

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. Dec 4, 2015

geoffrey159

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: Dec 4, 2015
9. Dec 4, 2015

cdummie

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. Dec 4, 2015

Ray Vickson

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. Dec 4, 2015

cdummie

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. Dec 4, 2015

Samy_A

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.

13. Dec 4, 2015

Ray Vickson

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. Dec 5, 2015

cdummie

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. Dec 5, 2015

Samy_A

Correct.

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