Complex Roots

**Coda202** · March 12th 2009, 05:00 PM

Let f be in R[x] and suppose a+bi is a complex root of f with a,b in R and b not equal to 0. Prove that a-bi is also a root of f.

I let h = (x-(a+bi))(x-(a-bi)). I know I need to show h is in R[x] in order to apply the division algorithm to f and h in R[x], but I'm stumped on how to do that.

**ThePerfectHacker** · March 12th 2009, 08:38 PM

Originally Posted by Coda202

Let f be in R[x] and suppose a+bi is a complex root of f with a,b in R and b not equal to 0. Prove that a-bi is also a root of f.

I let h = (x-(a+bi))(x-(a-bi)). I know I need to show h is in R[x] in order to apply the division algorithm to f and h in R[x], but I'm stumped on how to do that.

Let $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_1x+a_0$ we are told that $f(\alpha) = 0$ therefore $a_n\alpha^n + ... + a_1\alpha + a_0 = 0$ where $\alpha = a+bi$ .

This means, $\overline{a_n\alpha^n + ... + a_1\alpha +a_0} = 0 \implies a_n \beta^n + ... + a_1\beta + a_0 = 0$ where $\beta = \overline{\alpha} = a-bi$ .

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