# Complex roots problem (a proof by induction)

1. Jun 14, 2005

### TomMe

Problem:

If $$c_{1}, ..., c_{n}$$ are the complex roots of $$a_{n}z^n + a_{n-1}z^{n-1} + ... + a_{1}z + a_{0} = 0$$ with $$a_{n}$$ not 0

and $$S_{k}$$ is the sum of the products of these roots taken k by k,

then $$S_{k} = (-1)^k . \frac{a_{n-k}}{a_{n}}$$. Prove this by using induction.

So for n = 3 this will give:

$$S_{1} = c_{1} + c_{2} + c_{3}$$
$$S_{2} = c_{1}c_{2} + c_{1}c_{3} + c_{2}c_{3}$$
$$S_{3} = c_{1}c_{2}c_{3}$$

Now I am stuck just at the point where I need to prove that

$$S_{p} = (-1)^p . \frac{a_{n-p}}{a_{n}} => S_{p+1} = (-1)^{p+1} . \frac{a_{n-(p+1)}}{a_{n}}$$

I thought that by looking at the n = 3 case I would see this connection between S1 and S2, but I can't see it.

Hopefully someone can help me gain some insight. Thanks!

2. Jun 14, 2005

### NateTG

Can you rewrite the $a_n$ in terms of the $c_i$?

Last edited: Jun 14, 2005
3. Jun 15, 2005

### TomMe

The only way I can see is $$a_{n} = (-1)^p . \frac{a_{n-p}}{S_{p}}$$ with $$S_{p}$$ in terms of the $$c_{i}$$.
But I can't do that since $$S_{p}$$ could be zero.

No, still don't see it.

Last edited: Jun 15, 2005
4. Jun 15, 2005

### NateTG

You know that:
$$(z-c_1)(z-c_2)(z-c_3)...(z-c_n)=a_nz^n+a_{n-1}z^{n-1}+...a_0z^0$$

right?

5. Jun 15, 2005

### snoble

close

$$a_n\cdot (z-c_1)(z-c_2)(z-c_3)...(z-c_n)=a_nz^n+a_{n-1}z^{n-1}+...a_0z^0$$

6. Jun 15, 2005

### NateTG

Oh, yeah, oops. I suppose that explains where the $\frac{}{a_n}$ comes from...

7. Jun 17, 2005

### TomMe

Yes I know this. That's what I used with the n = 3 case.
Are you saying I just need to compare both sides and that's my proof? The point of the excercise is to prove it using induction, but I don't see how this helps me.

If k = 1 then yes, I can see that $$S_{1} = c_{1} + c_{2} + ... + c_{n} = -1 . \frac{a_{n-1}}{a_{n}}$$
But why does this mean that $$S_{2} = \frac{a_{n-2}}{a_{n}}$$?
That's where I'm stuck.

8. Jun 19, 2005

### TomMe

Anyone??

9. Jun 19, 2005

### AKG

Expand $a_n(z - c_1)\dots(z - c_n)$. Let $a_n = 1$ for now, for simplicity.

n=1:

$$z - c_1 = z + a_0$$

$$c_1 = -a_0 = (-1)^1a_{1-1}$$

n = 2:

$$(z - c_1)(z - c_2) = z^2 + (-c_1 - c_2)z + (c_1c_2) = z^2 + a_1z + a_0$$

$$-c_1 - c_2 = a_1$$

$$c_1 + c_2 = -a_1 = (-1)^1a_{2-1}$$

$$c_1c_2 = a_0 = (-1)^2a_{2-2}$$

Does this help?

10. Jun 19, 2005

### shmoe

It looks like you're trying to do induction on k, which I think is doomed to fail. Try induction on n, the degree of the polynomial. It wouldn't hurt to add another index to your S, $$S_{k,n}$$.

The idea here is that your degree n polynomial is the product of $$(z-c_n)$$ and a polynomial of degree n-1. You then know the coefficients of this degree n-1 polynomial in terms of it's roots.

You could also save some writing by assuming that $$a_n=1$$. If you know the result with 1, it shouldn't be a problem to jump to the general case.

11. Jun 19, 2005

### TomMe

Well, it has crossed my mind that I was doing induction on the wrong parameter here but was convinced it had to be done this way.

Thanks for your advice. I'll work this out first thing tomorrow morning and let you know how it turns out.

12. Jun 20, 2005

### TomMe

Okay, I think I have it now by doing induction on n. (with $$a_{n} = 1$$)

One last question though: is it really necessary to prove this using induction? In my induction step I assumed that $$S_{k} = (-1)^k . a_{p-k}$$ and so I can rewrite my p-degree polynomial with coefficients in terms of the roots. But I KNOW this is true because if I work out $$(z-c_{1})...(z-c_{p})$$ I get the same answer.
Does this mean I can prove this theorem just by looking at coefficients of the n-degree polynomial?

13. Jun 27, 2005

### NateTG

Well, that really depends on what you start with. I've seen proofs that addition is commutative, but, most of the time, people don't bother with stuff like that.