# 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

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 $\prod(\alpha_1 + 1)\cdots(\alpha_n+1)$.