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.