For $n\in N$ let $\sum_n(1)$ be the standard-simplex. Let there be a point $b\in R^n$ and a basis {$a_1,...,a_n$} of $R^n$. The $n-simplex$ set up in this point b by the basis is the set
\begin{equation*}
S_n(b;a_1,...,a_n):=\{b+\sum_{k=1}^nx_ka_k|(x_1,...,x_n)\in \sum_n(1)\}.
\end{equation*}
1.1.: Show that
\begin{equation*}
v_n(S_n(b;a_1,...,a_n))=\frac{|det(a_1,...,a_n)|}{n!}.
\end{equation*}
1.2.: $z_0,z_1,z_2\in C$ are the vertices of a triangle $\Delta$ in the complex plane. Show that the area is
\begin{equation*}
F=\frac{1}{2}|Im(\overline{(z_1-z_0)}(z_2-z_0))|.
\end{equation*}
My ideas so far:
1.1: I know that the general formula for $v_n$ of an $n-simplex$ is $\frac{a^n}{n!}$, but I can't see how the determinant plays a role in here.
1.2.: 
I tried to put in general coordinates for $z_0,z_1,z_1$ and trying to see if I can find something out of it. So, for $z_i=a_i+ib_i$ I get $\frac{1}{2}(-b_1+b_0)(b_2-b_0)$ and can't seem to see a connection to the genreal way of trying to find the area of a triangle since this only has values in the y-coordinate (in this case the $C$-axis).
Would like if someone had some hints about these.
You can think to determinant as to an operator that gives you the hyper-volume of ${\{\overrightarrow a_1, \overrightarrow a_2, \dots, \overrightarrow a_n}\}$ basis compared to hyper-volume of standard basis ${\{\overrightarrow e_1, \overrightarrow e_2, \dots, \overrightarrow e_n}\} $, which is 1 because product of othogonal vectors of length 1. In fact you are applying a transformation, call it $\phi$, to pass from ${\{\overrightarrow e_1, \overrightarrow e_2, \dots, \overrightarrow e_n}\}$ basis to ${\{\overrightarrow a_1, \overrightarrow a_2, \dots, \overrightarrow a_n}\}$ basis and the determinant says the scaling factor the hyper-volume generated of vectors done by $\phi$.
In the same way $S_n{(\overrightarrow b; \overrightarrow a_1, \overrightarrow a_2, \dots, \overrightarrow a_n)}$ is the final risult of $\phi$ applied on $S_n({\overrightarrow e_1, \overrightarrow e_2, \dots, \overrightarrow e_n})$.
About the second question: call $a = z_1-z_0$ and $b = z_2-z_0$. Expanding them in real and imaginary parts, calculate $F$ $$ \begin{align} F &= \frac{1}{2} |Im( (\overline{z_1-z_0}) (z_2-z_0) )| \\ &= \frac{1}{2} |Im( \bar a b )| \\ &= \frac{1}{2} |Im( (a_r - i a_i) (b_r + i b_i) )| \\ &= \frac{1}{2} |Im( a_rb_r + ia_rb_i - ia_ib_r + a_ib_i )| \end{align} $$ The imaginary part here is the determinant of $a$ and $b$ thought as vectors of components $a=(a_r, a_i)$, $b=(b_r, b_i)$.