Proof One
Proof that the dumber of real roots are at least one if the degree of the polynomial is odd?
Show that the number of non-real roots to a polynomial with real coefficients are, counted by multiplicity, even. Show especially that all real polynomials with an odd degree has at least one real root.
I know that this is due to the fact that complex roots comes in complex conjugate pairs. But I do not have a clue about how to show it. All hints, or name of this theorem, if there is any, would be greatly appreciated.
One idea is to use the intermediate value theorem:
Let p(x) = a0 + a1x + … anx^n be a polynomial of odd degree. We can assume that a_n is greater than zero, since -p(x) and p(x) have the same zeroes.
Then p(x) = (x^(n-1))(a0/(x^n-1) + a1/(x^n-2) + … + a_n-1 + xa_n).
As x tends to negative infinity, p(x) tends to x^(n-1)[a_n-1 + xa_n]. As x becomes small, the term inside the bracket tends to negative infinity, and the term outside becomes large (since it’s an even power), so p(x) is surely < 0 for some small a.
Similarly, as x becomes large, the term inside the bracket tends to infinity, as does the term outside, so p(x) is surely > 0 for some b > a.
Then by the intermediate value theorem, p(x) has a root on [a,b], which is the real root you want.
Your idea is also valid: by the fundamental theorem of algebra, p(x) has n roots over the complex numbers. If it has a complex root z, it has another root z-conjugate, so if all roots are complex it has an even number of roots, which is a contradiction since n is odd. This is a shorter proof, but I prefer the above because it gives a direct proof.
Edit: I realize you wanted a proof that if z = a + bi is a root of a real polynomial, so is zc = a – bi. It’s actually surprisingly simple to show.
Let p(x) = a0 + a1x + … anx^n
Since z is a root of p(x), a0 + a1z + … + anz^n = 0
If we conjugate both sides, 0 conjugate is still 0, and using the facts that (r*i)c is r*(ic) for real r, and imaginary i, and that the conjugate of a sum is the sum of the conjugates of each summand, we get that
a0 +a1(zc) + … +an(zc)^n = 0, so z^c is a root of p(x) as well.
Green’s Theorem Proof Part 1