# How to find area of a polygon built on the roots of a given polynomial?

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:

$$2x^5+3x^3-x+1=0$$

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

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):

$$A=1.460144\dots$$

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.

• 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 $R$ – Milo Brandt Jul 17 '16 at 14:27
• 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. – Jack D'Aurizio Jul 17 '16 at 14:34
• 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. – John Hughes Jul 17 '16 at 14:55
• 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. – John Hughes Jul 17 '16 at 16:56
• @JohnHughes, I see what you mean now, thank you for clarification – Yuriy S Jul 17 '16 at 16:58