why is it that the conjugate of a+bi is a-bi? 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)?
 A: For a complex number $a+bi$, the complex number $a-bi$ is more properly called the complex conjugate.  This is a distinct use of the word "conjugate" to that in a group, where $x^{-1}gx$ is also called a conjugate of $g$.
Having said that, you could take the multiplicative group $\mathbb{C}^{\ast}$ of units of the field $\mathbb{C}$ and form the semi-direct product with the group $\langle\alpha\rangle$ of order $2$ acting by complex conjugation, and then complex conjugation in that semi-direct product would be realised by group conjugation by $\alpha$: $z^{\alpha} = \bar{z}$, for $z\in\mathbb{C}^{\ast}$.
EDIT: Oops, from your comment, I guess one should instead take the group $U$ of units in the ring $\mathbb{Z}[i]$ of Gaussian integers, rather than $\mathbb{C}^{\ast}$.  But the same idea applies.
A: 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.
