TITLE: why is it that the conjugate of a+bi is a-bi?
QUESTION [2 upvotes]: if a+bi is an element of a group, then its conjugate is a-bi, 
how can we prove this by using the fact that the conjugate of an element g of a group is h if there is an x in the group such that h=xgx^(-1)?

REPLY [1 votes]: Given a mathematical structure $S$ and a group of automorphisms $G \subseteq \mathrm{Aut}(S)$, we often say that the elements $\sigma(x)$ for $\sigma \in G$ are the conjugates of $x$. And  if $\sigma$ is any particular automorphism, we might call $\sigma(x)$ the conjugate of $x$ by $\sigma$.
For the complex numbers, $\sigma(a+bi) = a-bi$ is an automorphism.
For your group theory, there is a standard homomorphism $G \to \mathrm{Aut}(G)$ that sends any element $g$ to the inner automorphism $\sigma_g(x) := gxg^{-1}$. This mapping from elements to inner automorphisms is so natural, that rather than talk about conjugation by $\sigma_g$, we simply call it conjugation by $g$.
More verbosely, the condition you are thinking of "$h$ is a conjugate of $g$ by an inner isomorphism". But the complex conjugation example is not an inner isomorphism of any (reasonable) group structure on the complexes, so you can't think of it that way.
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