area of triangle from coefficients of its cubic? Three points $z_1$, $z_2$, $z_3$ on the complex plane are given by the coefficients $a_k$'s of the cubic polynomial $f(z)=(z-z_1)(z-z_2)(z-z_3)=\sum_{k=0}^3 a_k z^k$.  How does one express the (signed) area $V$ of the triangle with vertices $z_1$, $z_2$, $z_3$ in terms of $a_k$'s and $\overline{a}_k$'s? One is tempted to try to expand $V^2$ in the symmetric functions in the roots of $f(z)\overline{f}(z)$, as well as these of $f(z)$ and of $\overline{f}(z)$, e.g. starting from 
$$
V=\frac{\sqrt{-1}}{4}\det
\begin{pmatrix}
1& 1&1 \\
z_1& z_2& z_3 \\
\overline{z}_1& \overline{z}_2& \overline{z}_3\end{pmatrix},
$$
(this not so well-known formula can be found in R.Deaux, Introduction to the Geometry of Complex Numbers, Ungar, New York, 1956, pp.59-60), but this rather calls for some kind of joint invariants of $f$ and $\overline{f}$ to be used. 
Any pointers etc. are much appreciated.
Added: the motivation comes from a moment problem: suppose one is given a part of the sequence $\mu_n=\int_\Delta t^n dx dy$, where $t:=x+\sqrt{-1}y$, and wants to find the triangle $\Delta$, e.g., its vertices $z_i$'s. P.Davis in his paper "Triangle formulas in the complex plane" (Math.Comp. 18(1964)) shows that the first 4 moments $V=\mu_0$,...,$\mu_3$ determine $\Delta$; this is one more parameter than needed to determine the $z_i$'s. We can do better, but were unsure how $V$ depends upon $\mu_1$,...,$\mu_3$, which boils down to this very question.
 A: Not quite an answer, but a cute fact: if the three points (I use $x, y, z$ below, for typing convenience) are all on the unit circle, then the square of the area equals (using the OP's formula, and the observation that in this case $\overline{z} = 1/z$)
$-\dfrac{(x-y)^2 (x-z)^2 (y-z)^2}{16x^2y^2z^2}.$
The numerator is a multiple of the square of the discriminant of the polynomial, and can be easily written in terms of symmetric functions as 
$-a^2 b^2 + 4 b^3 + 4 a^3 c - 18 a b c + 27 c^2.$ The denominator is obvious.
For reasons given in my comment this formula does not generalize to "general" triples in the complex plane.
A: There is a clear experimental maths strategy to attack a problem like this. A "generic" polynomial $(z-z_1)(z-z_2)(z-z_3)=z^3-az^2+bz-c$ can be replaced with a very concrete one: Choose $a$, $b$ and $c$ to be (at least presumably) $\mathbb Q$-algebraically independent complex numbers. Compute numerically the zeroes $z_1$, $z_2$, $z_3$ of the polynomial $z^3-az^2+bz-c$ and the corresponding value $V(z_1,z_2,z_3)$. Then we expect $V$ to satisfy an algebraic equation with coefficients from $\mathbb Q[a,b,c]$, and this can be guessed efficiently with either LLL or PSLQ.
I did not try hard after seeing Igor's response and comments, but it seems that there is no algebraic equation of degree $\le6$ for $V$.
A: For simplicity let assume that the $f(z) = z^3 - az +b $ and has roots $z_1, z_2, z_3$.
For each $i$ consider the polynomial 
$f_i(z) = f(z/z_1) = z^3 -a_i z + b_i$ 
which has roots $1$ and $-1/2 \pm \lambda_i$ for some $\lambda_i$
The area the triangle with the roots of $f_i$ is $3/4  |Im \lambda_i$
and its discriminat is $D_i = 4 \lambda_i^2 (9/4 - \lambda_i^2)^2$ and
$b_i = 1/4 - \lambda_i^2$.
Thus the area of this triangle is
$A_i = \frac{3}{4} Im \frac{\sqrt{\pm D_i}}{b_i +2}$
However $D_i$, $A_i$ and $b_i$ can be easily expressed using the corresponding values for the original polynomial and the root $z_i$, i.e.
$b_i = b / z_i^3$ $D_i = D/ z_i^6$ and $A_i = A/|z_i|^2$
thus we have
$$
A = \frac{3}{4} |z_i|^2 Im \frac{\sqrt{\pm D}}{b+2z_i^3} = \frac{3}{4} |z_i|^2 Im \frac{\sqrt{\pm D}}{2az_i -b} 
$$
Now we can sum over the roots and get
$$
A = \frac{1}{4} Im \left(\sqrt{\pm D} \sum \frac{|z_i|^2}{2az_i -b} \right)
$$
and we are letf with expressing the sum in the brackets as a function on $a$ and $b$.
My algebra gives 
$$
\sum \frac{|z_i|^2}{2az_i -b} = \frac{\pm \sum \frac{b}{|z_i|^2}}{b^3 + 4a^3b}
$$
but I do not see any easy way to deal with the expression
$\sum \frac{1}{z_i \bar z_i}$.
Of course one can solve the cubic equation and work it out, but there should be an easier way.
A: For $t$ real, the area determined by the roots of $x^3-x^2+tx-t=(x-1)(x^2+t)$ is either $\sqrt{t}$ or $0$ according as $t \gt 0$ or $t \lt 0.$ At first that made me think that there is no hope. But now I think that maybe the polar coordinates of the coefficients are better to look at. One can go back and forth after a fashion using $\arctan.$ Another thought is that given $x^3+ax^2+bx+c$ one should work with $a^6$,$b^3$ and $c^2$ to account for the effect of scaling.
