How to find the area of a (maximum area convex) polygon, built on the roots of a given polynomial in the complex plane?

For example, consider the equation:


It has one real and four complex roots and makes a nice convex pentagon in the complex plane (thanks, Wolfram Alpha):

enter image description here

Using the formula for the area of a convex polygon:

$$A=\frac{1}{2} \left( \begin{array}| x_1 & x_2 \\ y_1 & y_2 \end{array} + \begin{array}| x_2 & x_3 \\ y_2 & y_3 \end{array} + \dots + \begin{array}| x_n & x_1 \\ y_n & y_1 \end{array} \right)$$

I obtained for this case (using numerical values of the roots):


Another simple case - roots of unity. They just make regular polygons and the general formula for the area is well known.

However, I would like to know if it's possible to find out this area without computing the roots, using only the coefficients of the polynomial? (The coefficients are meant to be rational).

I know that polynomials with only real roots will all have $A=0$, and for the polynomials with several real roots some of them will be inside our maximum area polygon.

There is a useful theorem (see Rouche's theorem ), according to which:

For a monic polynomial $$z^n+a_{n-1} z^{n-1}+\dots+a_1 z+a_0$$

All its roots will be located inside the circle $|z|=1+\max |a_k|$.

But this theorem gives relatively large area, and can't be used to approximate the area of the polygon.

  • $\begingroup$ I doubt this is directly helpful, but a somewhat related problem would be to calculate the area of the Minkowski sum of $[-r_i,r_i]$ where $r_i$ are the roots enumerated with duplicity (and such an interval is just the convex hull of the two points - i.e. a symmetrical line segment through the origin), then one can calculate it for a polynomial $P$ of degree $n$ as follows: Define $Q_{c}(x)=x^nP(c/x)$. Now, calculate $R(c)=\text{res}(P,Q_c)$ where $\text{res}$ is the resultant. Then, you sum the absolute value of the imaginary part of each root of $R$ $\endgroup$ – Milo Brandt Jul 17 '16 at 14:27
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    $\begingroup$ The area is given by $$ \frac{1}{2}\text{Re}\sum_{j=1}^{n}\zeta_j \overline{\zeta_{j+1}}$$ with $\zeta_1,\zeta_2,\ldots,\zeta_n,\zeta_{n+1}=\zeta_1$ being the the vertices of the convex envelope of the roots. However, we have to detect which roots lie inside the convex envelope and which roots do not, so I do not think there is a nice closed formula. $\endgroup$ – Jack D'Aurizio Jul 17 '16 at 14:34
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    $\begingroup$ Following up on Jack's remark: In particular, you should not expect a polynomial in the roots, for when one root is strictly inside the convex hull of the others, the area (as a function of the roots) is independent of that root. $\endgroup$ – John Hughes Jul 17 '16 at 14:55
  • $\begingroup$ What I mean to say was that the formula for the area is unlikely to be a polynomial in the roots (which doesn't seem to be stated between the highlighted areas); it's still possible, of course, that it's a polynomial in the coefficients. Sorry for wasting your time by being unclear. $\endgroup$ – John Hughes Jul 17 '16 at 16:56
  • $\begingroup$ @JohnHughes, I see what you mean now, thank you for clarification $\endgroup$ – Yuriy S Jul 17 '16 at 16:58

The polygon with vertices of roots of a polynomial must be symmetric about real axis because of the conjugate roots.

Area of the polygon equals to 2 times the region in upper half plane.

The formula of area of polygon must be abandoned if polygon area is hoped to be computed with coefficients. This is because that formula requires the vertices of polygon which is impossible to be computed by the coefficients of the polynomial if the degree of the polynomial is equals or higher than 5 due to Abel-Ruffini theorem.

It is possible for us to find out area generated by polynomial of degree 3 or 4(1 and 2 do not give a region) with the coefficients.

Here are some hints.


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