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Show that complex conjugate is also a root of polynomial with real coefficients 
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#1
Oct508, 12:51 PM

P: 6

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_(n1) x^(n1)+ …+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)*(abi) = 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
Oct508, 02:53 PM

Sci Advisor
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Thanks
P: 25,247

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



#3
Oct508, 02:54 PM

P: 6

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_(n1) x^(n1)+ …+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_(n1) x^(n1)+ …+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_(n1) x^(n1)(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
Oct508, 02:55 PM

P: 6

Show that complex conjugate is also a root of polynomial with real coefficients
Hahah, thanks Dick! I caught on a little late, but thanks a bunch for your reply!



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