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.

• the area with sign depends on the order of the points, so it cannot be expressed in terms of the symmetric functions $a_k$.
– rita
Nov 23 '11 at 9:28
• But the square of the area is clearly a symmetric function... Nov 23 '11 at 9:31
• The volume, or rather its square, is a symmetric polynomial in the three pairs $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$. So the geometric object underlying your hoped-for formula is not the symmetric cube of the complex affine line, but the symmetric cube of the affine plane. In other words, you need generators of the algebra $K[X_1,Y_1,X_2,Y_2,X_3,Y_3]^{\mathfrak S_3}$, where the symmetric groups acts via the obvious permutations.
– ACL
Nov 23 '11 at 10:58
• Wouldn't you expect $V$ to be the root of a polynomial of degree $12$ whose coefficients are polynomials in the $a_k$ and $\bar a_k$? After all, the $a_k$ and $\bar a_k$ are fixed by two independent actions of $S_3$ and $V$ is only fixed by the diagonal $A_3$ action, so, in the worst case (which might hold), you'd expect the polynomial to have degree $(6\times6)/3 = 12$. Unless that polynomial factors, I wouldn't expect that there'd be a simpler formula. Nov 23 '11 at 13:22
• Just out of curiosity, I checked (using Maple). The polynomial relating $(V,a_1,a_2,a_3,\bar a_1,\bar a_2,\bar a_3)$ is, indeed, irreducible of degree $12$ in $V$. (It is, of course, even in $V$.) It has total degree $22$ in these $7$ variables and is a sum of 646 monomials in these variables, with a typical integer coefficient having $7$ digits (when the highest common factor of the coefficients is $1$). I suppose I could paste the Maple output into an answer box if you wanted to see it, but, is there really any point to doing this? Nov 23 '11 at 16:38

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.

• There is a typo in the product of $V$ and $V^\top$. The top left entry should be 3, not 1. I wonder if it affects the outcome... Nov 25 '11 at 3:26
• @Dima: Oh, you are right; thanks for pointing this out. I have edited the answer to correct this. It doesn't really change anything, except that P now has $1175$ terms instead of merely $646$. Nov 25 '11 at 4:28
• It would be interesting to know more about $Q$. After all, if $S$ can be found as a function of $a_k$ and $\overline{a}_k$ then the determinant expression for $VV^\top$ might be more useful than the 1175-term expansion. Could you perhaps post the expansion for $Q$? Thanks in advance. Nov 25 '11 at 6:42
• @Dima: $Q$ itself has $66$ terms, but the coefficients are quite small, so it's not completely crazy to think that it might contain some useful information. However, I'm a little reluctant to put the whole thing in the answer because it's still quite long, and I'd have to struggle with the typesetting or else leave it as a 'long line'. Let me think about whether I can do something better. One thing that I noticed is that the formulae simplify quite a lot if you assume that the centroid of the triangle is at $z=0$. Maybe I'll post those formulae first, so that you can see the rough shape. Nov 25 '11 at 12:53
• @Robert, perhaps you can just post the Maple(?) code you wrote to produce $Q$. Nov 26 '11 at 7:31

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.

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 $V$.

• This is not very likely to work since if you take $a,b,c$ to be reals umbers you are likely to get $3$ real roots and the area will be zero, which will not be usefull. Of course if you take the real parts of $a,b,c$ and the imaginary paorts of $a,b,c$ to be algebraicly independant you can do it, but the number of variables is getting too big. Nov 23 '11 at 11:44
• This was already in Igor's comments. No secrets from you: My choice was $a=\log(2)$, $b=\exp(1)$ and $c=\pi\sqrt{-1}$. Believe you or not, but the zeroes were all unreal. Even more, I also tried the case $a=0$, $b=\exp(1)$ and $c=\pi\sqrt{-1}$ (which is mentioned in your post) and still no result in degree less than 7. Nov 23 '11 at 11:51
• you can simplify the computations even further by reducing to the case $a=0$ and $c=1$, but then you cna not expect the answer to be algebraic in $b$, but it should be algebraic in $Re b$ and $Im b$ Nov 23 '11 at 12:08

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.

• It seems to me that the question is fully Russian: Dima, Igor, Wadim and Kassabov. Nov 23 '11 at 11:47
• Sorry, but I am not russian :) Nov 23 '11 at 12:05
• What is the prizinal polynomial? Nov 23 '11 at 14:00
• igor: I've meant original Nov 23 '11 at 14:16
• @kassabov: I was conjecturing that, but my ignorance is deep, so there might have been some associated polynomial I was not aware of :) Nov 23 '11 at 16:24

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.