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I'm looking for the sum and the product of the roots of

$$ z^7 + (3-i)z^6+\pi z^3+e = 0 $$

I think this can be done without calculating the roots (the sum should be -3+i and the product -e) but I don't understand it. How do you solve this? And what's the theory behind it?

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up vote 3 down vote accepted

If we factorise the polynomial as $$f(z)=(z-a)(z-b)(z-c)(z-d)(z-e)(z-f)(z-g)$$ and we are in any context where $pq=0 \implies p=0\text{ or }q=0$ we find that the roots of $f(z)=0$ are $a,b,c,d,e,f,g$.

Then, multiplying out we get $$f(z)=z^7-(a+b+c+d+e+f+g)z^6+\dots -abcdefg$$ and the result comes from equating coefficients.

Vieta's relations are simply a generalisation of this including the form other coefficients and different degrees. Of course the sum and product are easiest to work with, and in some contexts are closely related to ideas of trace and norm.

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If a monic polynomial $P$ has roots $\alpha_1,\cdots,\alpha_n$, then $P(x)=(x-\alpha_1)\cdots(x-\alpha_n)$. Now expand this product and get $P(x)=x^n-(\alpha_1+\cdots+\alpha_n)x^{n-1}+\cdots+(-1)^n(\alpha_1\cdots\alpha_n)$.

If $P$ is not monic, then $\dfrac1{a_n}P$ is monic and has the same roots as $P$. Then the argument above applies to give that the sum of the roots is $-\dfrac{a_{n-1}}{a_n}$ and the product of the roots is $(-1)^n\dfrac{a_0}{a_n}$.

See also Vieta's formulas.

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