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

and [tex]S_{k}[/tex] is the sum of the products of these roots taken k by k,

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

So for n = 3 this will give:

[tex]S_{1} = c_{1} + c_{2} + c_{3}[/tex]

[tex]S_{2} = c_{1}c_{2} + c_{1}c_{3} + c_{2}c_{3}[/tex]

[tex]S_{3} = c_{1}c_{2}c_{3}[/tex]

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

[tex]S_{p} = (-1)^p . \frac{a_{n-p}}{a_{n}} => S_{p+1} = (-1)^{p+1} . \frac{a_{n-(p+1)}}{a_{n}}[/tex]

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!