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.



area of triangle from coefficients of its cubic?
NB: Note that my $a\_k$ have different signs from those defined in the question. For me,
$$
(z - z\_1)(z-z\_2)(z-z\_3) = z^3 - a\_1\ z^2 + a\_2\ z - a\_3,
$$
so that $a\_k$ is the $k$-th elementary symmetric function of the $z\_i$. This doesn't really affect the answer in any significant way.


While I don't think that the final result relating $V$ to the $a\_k$ and $\bar a\_k$ is that interesting or useful, people might want to know a way to derive it. Here is what I did:
Start with the formula
$$
V=\frac{\sqrt{-1}}{4}\det
\begin{pmatrix}
1&1&1 \\
z\_1& z\_2& z\_3 \\
\overline{z}\_1&\overline{z}\_2&\overline{z}\_3\end{pmatrix},
$$
and note that multiplying the matrix on the right by its transpose yields
$$\begin{pmatrix}
1&1&1 &\\
z\_1&z\_2&z\_3\\
\bar z\_1& \bar z\_2 &\bar z\_3 \end{pmatrix}
\begin{pmatrix}
1&z\_1&\bar z\_1\\
1&z\_2&\bar z\_2\\
1&z\_3&\bar z\_3 \end{pmatrix}
=\begin{pmatrix}
3&a\_1&\bar a\_1 \\
a\_1&{a\_1}^2-2a\_2& S \\
\bar a\_1&S&{\bar a\_1}^2-2\bar a\_2
\end{pmatrix}
$$
where $S = z\_1\bar z\_1 + z\_2\bar z\_2 + z\_3\bar z\_3$. Thus, one has the polynomial relation
$$
R := 16V^2 - 3S^2 + 2 a\_1\bar a\_1 S + {a\_1}^2{\bar a\_1}^2 
- 4{a\_1}^2\bar a\_2 - 4{\bar a\_1}^2a\_2 + 12 a\_2 \bar a\_2 = 0.
$$
It remains to find a relation between $S$ and the $a\_k$ and $\bar a\_k$. To do this, note that $S$ will be a root of the polynomial
$$
Q := \prod\_{\pi\in S\_3} \bigl(S - z\_1{\bar z\_{\pi(1)}} - z\_2{\bar z\_{\pi(2)}} - z\_3{\bar z\_{\pi(3)}}\bigr).
$$
Note that $Q$ is a polynomial of degree $6$ in $S$ that is symmetric in the $z\_i$ and the $\bar z\_i$ separately. Hence, $Q$ can be regarded as a polynomial of degree $6$ in $S$ with coefficients that are polynomials in the $a\_k$ and $\bar a\_k$. The resulting expression for $Q$ as a polynomial in $S$, the $a\_k$, and the $\bar a\_k$ has $66$ terms. (Actually expressing $Q$ this way is not easy by hand. However, it's very easy to implement the algorithm for writing a symmetric polynomial in three variables as a polynomial in the elementary symmetric functions in those variables on a computer, which is what I did.)


Finally, $R$ and $Q$ are polynomials in $S$ with coefficients that are polynomials in the variables $V, a\_1,a\_2,a\_3,\bar a\_1,\bar a\_2,\bar a\_3$. Set
$$
P(V,a\_1,a\_2,a\_3,\bar a\_1,\bar a\_2,\bar a\_3) := \text{Resultant}\_S(R,Q).
$$
Then $P=0$ is the desired relation. Computation (using Maple) shows that it has the following properties: $P$ is irreducible, is even and of degree $12$ in $V$, is of total degree $20$, and contains $598$ monomial terms, with typical integer coefficient in the millions.


Added remark: The formulae simplify dramatically if one assumes that the centroid of the triangle is at $z=0$, i.e., that $a\_1 = \bar a\_1 = 0$. (This is analogous to the way that the cubic formula itself simplifies when one removes the quadratic term. Moreover, this can always be easily arranged in the usual way by translation.) When $a\_1 = \bar a\_1 = 0$, we have
$$
R = 16V^2 - 3S^2 + 12 a\_2 \bar a\_2
$$
and 
$$
\begin{array}
\\
Q &= S^6-6{a\_2}{\bar a\_2}S^4-27{a\_3}{\bar a\_3}S^3+9{a\_2}^2{\bar a\_2}^2S^2\\
 &\qquad {} +81{a\_2}{a\_3}{\bar a\_2}{\bar a\_3}S
 -27{a\_3}^2{\bar a\_2}^3 - 27{a\_2}^3{\bar a\_3}^2 - 4{a\_2}^3{\bar a\_2}^3.
\end{array}
$$
The formula for $P$ is still not that nice, though; it has $17$ terms, with coefficients in the millions. Rather than input it here, I'll just recommend that those interested compute the resultant of this reduced $R$ and $Q$ with respect to $S$ to get it.


Another added remark (the real case): If, in addition, one assumes that the $a\_k$ are real (and that $a\_1=0$), then the polynomial $P$ factors as
$$
16V^2 (16V^2{+}9{a\_2}^2)^2
(4096V^6{+}4608{a\_2}^2 V^4{+}1296{a\_2}^4V^2{-}19683{a\_3}^4{-}2916{a\_2}^3{a\_3}^2).
$$
In particular, note that $V=0$ is always a root and that $P=0$ has a positive real root $V$ if and only if $27{a\_3}^2+4{a\_2}^3$ is positive (and $a\_3\not=0$), i.e., if and only if the original cubic has only one real (nonzero) root, just as one would expect.


