area of triangle from coefficients of its cubic?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $V$.

Answer 4

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.

Answer 5

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.