polynomial with complex roots....

Right. This is really getting on my nerves.

The question is:

Let be a polynomial where the coefficientsn are real and

i). Show that if is a complex root of f then it's conjugate is also a root of f.

Would this just be that the product of a complex number and it's conjugate produce a real number. Since this polynomial has real coefficients then there can't be a complex number on it's own (or some of the coefficients would be complex)?

ii) Suppose that the degree n is odd. Show that f must have at least one real root.

Would the solution to this be that a polynomial with a degree n has n roots. If the polynomial is odd then there would be pairs of complex numbers and their conjugates and one root left over. Since this root is on it's own, it must be real to avoid complex coefficients.

I do not know how to prove this though (Crying).

Please help!

P.S: Sorry for two threads in one night, there's no gentle introduction on a maths course!