I suspect you haven't thought carefully enough about this. By definition, a point $x$ in some convex set $C \subseteq X$ is extreme if it cannot be written as $x = ty + (1-t)z$ for $y,z \in C$ and $t \in (0,1)$. In other words, the property depends only on the structure of $X$ as a real vector space, no complex numbers will ever enter the picture. So, at least for restriction of scalars it makes no difference. If, instead, you start with a real vector space $X$ and take its complexification $X^{\mathbb{C}} = X \otimes_{\mathbb{R}} \mathbb{C}$, we have $X^{\mathbb{C}} \cong X \oplus iX$ (as real vector spaces) and an embedding $i: X \hookrightarrow X^{\mathbb{C}}$ by $x \mapsto (x,0)$. In particular, scalar multiplication by real numbers on $X$ is unchanged (for that reason, it is called an extension of scalars), so again a point $x$ in $C$ is extreme if and only if $i(x)$ is extreme in $i(C)$. If you had something else in mind, please edit your question appropriately.