We have $f(x) \in R[x]$ and say that $(a+bi)$ is a zero. show that $(a-bi)$ is a zero.

Question

So in my course, we are currently doing homomorphisms so I am somehow supposed to use the properties of homomorphisms to prove this and I am honestly at loss. I was thinking that perhaps the homomorphism $R[x]->C$ where $\phi(f(x))=f(a+bi)$ would help. What i would need to show is that if $f(x) \in Ker{\phi}$ then $f(a-bi)=0$ but I have no idea how to do that and it seems like a dead-end approach. Any explanation would be appreciated. This is a duplicate question but the other did not get an answer I am looking for. As a Hint I have a+bi->a-bi is an isomorphism. I am not sure how that helps.

Answer 1

Hint: use the fact that since the coefficients are real $\overline {f(a+ib)}=f(a-ib)=0$.

To use homomorphisms, consider $c:\mathbb{C}\rightarrow \mathbb{C}$ defined by $c(z)=\overline{z}$ the complex conjugation, it is an homomorphism, and if $f\in ker(\phi)$, $(c\circ \phi)(f)=c(0)=0=\overline{f(a+b)}=f(a-ib)=0$.