# Show that complex conjugate is also a root of polynomial with real coefficients

1. Oct 5, 2008

### Muffins

1. The problem statement, all variables and given/known data
Suppose that f(x) is a polynomial of degree n with real coefficients; that is,

f(x)=a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n,… ,a_0∈ R(real)

Suppose that c ∈ C(complex) is a root of f(x). Prove that c conjugate is also a root of f(x)

2. Relevant equations

(a+bi)*(a-bi) = a^2 + b^2 where a, and b are always reals?
Not really sure if this helps or not.

3. The attempt at a solution

I'm really clueless on how to start approaching this. I was thinking perhaps the fundamental theorem of algebra might be of some use, or perhaps the fact that a number of complex form multiplied by it's conjugate is a real number, but I'm really not sure.

Could anyone give me a nudge in the right direction?
Any help would be greatly appreciated!! Thanks!

2. Oct 5, 2008

### Dick

The root is where f(c)=0. Take the complex conjugate of that equation.

3. Oct 5, 2008

### Muffins

Never mind! I found the actual theorem on the web, and I think this is pretty much what I was looking for anyway

Consider the polynomial
f(x)=a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n,… ,a_0∈ R(real)
where all a*x are real. The equation f(x) = 0 is thus
a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n = 0
Given that all of the coefficients are real, we have
a_n x^n(conjugate) = a_n x^n(x is conjugate)
Thus it follows that
a_n x^n(conj)+ a_(n-1) x^(n-1)(conj)+ …+a_1 x(conj)+ a_0, a_n = 0(conj) = 0
and thus that for any root ζ its complex conjugate is also a root.

4. Oct 5, 2008

### Muffins

Hahah, thanks Dick! I caught on a little late, but thanks a bunch for your reply!