Triangle over complex line ``A triangle $T\subset \mathbb{C}^n$ is the closed convex hull of three points. A triangle lies on a complex line if its vertices $a,b,c\in\mathbb{C}^n$ satisfy
$$
\zeta(b-a)=c-a, \text{ for some }\zeta\in\mathbb{C}.''
$$
Now, on the real context (or rather, for $\zeta\in\mathbb{R}$) this would mean that $a,b,c$ are points over the same line. But what does this mean on the complex ambient, intuitively/geometrically?  Also, what would be such a thing as a triangle not in a complex line?
 A: As you observe, if $\zeta$ is real, then this is a degenerate triangle: all convex combinations of $a,b,c$ lie on the smallest line segment containing all three of them. The definition is also a little shaky if $b = a$, so I'll rule out that case.
For $\zeta$ not real, it's useful to do a little simplifying by changing coordinates. First, move the origin to $a$, so that we have
$$
a' = a - a = 0\\
b' = b - a\\
c' = c - a
$$
Now $$
\zeta(b') = c'.
$$
Next, "divide through by $b'$", by which I mean "change to a coordinate system in which the first coordinate vector is $b'$. That amounts to multiplying all three vectors by some matrix $M$:
$$
a'' =M 0 = 0\\
b'' = M b' = (1,0,\ldots, 0)\\
c'' = Mc'.
$$
Now we have
$$
\zeta (b'') = \zeta (M b') = M (\zeta b') = M c' = c''.
$$
Now $a''$ and $b''$ lie on the real line; the complex line containing this consists of all vectors of the form $(p + qi, 0, \ldots, 0)$.
So if $c$ lies in this complex line, then $c''$ is just $$
c'' = (p + qi, 0, \ldots, 0) = (p+qi) (1, 0, \ldots, 0) = \zeta b'' 
$$
where we choose $\zeta = p + qi$ to make things work out.
So in the right coordinate system, the claim is obvious.
