2 edited body

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 prizinal 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.

1

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 prizinal 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.