Area of a Hexagon in the Complex Plane given by the solutions of $x^6+x-1=0$

Question

I have this problem I friend gave me a little while ago and I don't even know where to begin.

This was the problem he gave me:

Let $z_0, z_1, z_2, z_3, z_4, z_5$ be the solutions to the equation $x^6+x-1=0$. Find the area of the hexagon given by $z_0z_1z_2z_3z_4z_5$ in the complex plane.

Also I have no clue how to even tag this problem, so feel free to edit them :)

Answer 1

As $p(x):=x^6+x-1$ is a polynomial with real coefficients, its roots are

• either real : $\approx -1.1347$ and $\approx 0.7781,$
• or pairs of complex conjugate numbers : $\approx -0.4511 \pm 1.0024i$ and $\approx 0.6294 \pm 0.7358i.$

Therefore, their representative points are either on the real axis or symmetrical with respect to it. See (red) hexagon on this figure :

As a consequence, an approximate answer can be obtained by just computing the upper part of the area (above the real axis) as a sum of trapezoidal areas, then double it.

If you look at the similar question, the opinion expressed there is that no general method exists giving an exact answer. You will find there the following formula for the numerical computation of the area :

$$A=\frac12 \Im\left(\sum_{k=1}^{n}r_k\overline{r_{k+1}}\right)$$

($\Im$ meaning "imaginary part" ; please note that indices are to be understood "modulo n" in order that $r_{n+1}=r_1$). This formula isn't especialy interesting here.

Edit 1 :

Let us consider the derivative $p'(x)=6x^5+1$ of polynomial $p(x)$. Its $5$ roots $-\frac{1}{\sqrt[5]{6}} e^{i2k\pi/5}, \ k=0,1,2,3,4$ form a pentagon included into the hexagon by the celebrated Gauss-Lucas theorem ; therefore, the area of the hexagon is larger than the area of the pentagon which can be computed exactly (it is an exception !) as $\frac{5}{2}\frac{1}{6^{2/5}}\sin \frac{2 \pi}{5}$). I wonder whether the real question of your friend is indeed to minorate the area of the hexagon..., unless he just wanted a numerical approximation of the area $A \approx 2.67261847$...

Edit 2 :

Some thoughts about the peculiarity of polynomial $p$. The fact that this polynomial has many zero coefficients invites to attempt a factorization :

$$p(x)=(x^2+ax+p)\left(x^4-ax^3+bx^2+cx-\frac{1}{p}\right)\tag{*}$$

where the quadratic factor yields the real roots.

There are 3 constraints on $b$ and $c$ :

$$b=a^2-p, \ c=a(2p-a^2), \ c+a(b-p)=0$$

from which we deduce the following condition linking $a$ and $p$

$$a^4p-3a^2p^2+(p^3+1)=0$$

which can be considered as a quadratic equation in $A:=a^2$.

Therefore, one can express $a$ as a function of $p$. As a consequence, all the coefficients in (*) can be expressed as functions of $p$.

Open question : Can this be used for reaching an exact expression of the area ?

Remark : numerical values of the coefficients of the fourth degree polynomial in (*) can be obtained at once with the very handy function "deconv" of Matlab:

$$ deconv([1,0,0,0,0,1,-1],[1,\underbrace{0.35663453}_a,\underbrace{-0.8829170}_p])$$

Matlab program for the construction of the hexagon above :

L=[1,0,0,0,0,1,-1], % list of coefficients
r=roots(L); 
[K,A]=convhull(real(r),imag(r)), % convex hull
r=r(K), % trigonometric ordering of the roots
plot(r,'-or');
Implementation of formula (1) :
s=0;
for k=1:6
   s=s+r(k)*conj(r(k+1));
end;
imag(s)/2, % area (same value as A, up to the sign)