Question

Prove that equation as rings. 

09 Jan 2026
Answer :
Word Count : 95
Define a map (\phi:\mathbb{R}[x]\to\mathbb{C}) by (\phi(f(x))=f(i)), where (i^2=-1). For any (f(x),g(x)\in\mathbb{R}[x]), [ \phi(f+g)=(f+g)(i)=f(i)+g(i)=\phi(f)+\phi(g), ] and [ \phi(fg)=(fg)(i)=f(i)g(i)=\phi(f)\phi(g), ] so (\phi) is a ring homomorphism. Every complex number
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