# Complex numbers system of equations problem with 5 variables

Let $z_0$,$z_1$,$z_2$,$z_3$ and $z_4$ such that $z_i\in C$ that hold:

$$(1)|z_0|=|z_1|=|z_2|=|z_3|=|z_4|=1$$

$$(2)z_0+z_1+z_2+z_3+z_4=0$$ $$(3) z_0z_1+ z_1z_2+z_2z_3+z_3z_4+z_4z_0=0$$

Prove that the solutions $z_i$ of this equation lay on the corners of regular pentagon.

I have tried with insertion of complex numbers with property that $z_i=1\angle \phi_i$ with $\phi_i=i 360^°/5$ and $i\in\{0,1,2,3,4\}$

I am interested if I should use $z_i=1\angle ( \phi_i+\alpha)$ with $\alpha \in \{0,2\pi\}$

$$1\angle (\alpha)+ 1\angle ( \phi+\alpha)+ 1\angle ( 2\phi+\alpha)+ 1\angle ( 3\phi+\alpha)+ 1\angle ( 4\phi+\alpha)=0$$

$$1\angle (\phi+\alpha)+ 1\angle ( 3\phi+\alpha)+1\angle ( 5\phi+\alpha) +1\angle ( 7\phi+\alpha)+1\angle ( 4\phi+\alpha)=0$$

The equations $(1)$, $(2)$, and $(3)$ are invariant under rotations of the plane of complex numbers around the origin, and so is the conclusion, thus we can assume that $z_0=1$ (this is equivalent to the introduction of the new unknowns $a$, $b$, $c$, $d$ in the answer by Omran Kouba). Writing $z_1=x_1+iy_1$ etc., we obtain eight equations in eight real unknowns $x_1$, $y_1$, $\ldots$, $x_4$, $y_4$, two of them linear and the remaining six quadratic. Mathematica returned four solutions; two of them are
\qquad\qquad$$\qquad\qquad and the complex conjugates (mirror images across the real axis) of these two are the other two solutions. The same answer was given by Omran Kouba. The interesting thing to note here is that not only are the z_k's the vertices of the regular pentagon, they can be that in only four ways out of the possible twenty-four. A few words about the failed attempt. The z_k's (before the normalization z_0=1) are the five roots of the equation z^5+s_2z^3-s_3z^2+s_4z-s_5=0, where s_1=0, s_2, s_3, s_4, s_5 are the elementary symmetric polynomials in z_k's. Because of |z_0|=\cdots=|z_4|=1 we know that |s_5|=1, and we also know that s_2'=z_0z_1+z_1z_2+\cdots+z_4z_0, which is 'half\mspace{1mu}' the expression for s_2, is 0. I tried to derive from z_0\overline{z}_0=\cdots=z_4\overline{z}_4=1, s_1=0 and s_2'=0 that s_2''=z_0z_2+z_1z_3+\cdots+z_4z_1=0, and that then also s_3=s_4=0. Writing s_5=v^5, for a suitable v with |v|=1, this would give us$$ \{z_0,z_1,z_2,z_3,z_4\}\:=\:\{v,v\omega,v\omega^2,v\omega^3,v\omega^4\}~,\tag{A} $$where \omega=\exp(2\pi i/5). After producing quite a few pages of messy formulas I desisted from this futile attempt. I shoved the idea into a dark underground chamber of my mathematical mind, where it will have chance to mutate, given enough time, into something that will actually work. Continued. ~The condition (1) means that \overline{z}_k=z_k^{-1} for 0\leq k\leq4. The other two conditions are (2) s_1=0, and (3) s_2'=0. Conjugating s_1=0 and multiplying by s_5 we get s_4=0. Suppose, for a moment, that instead of (3) we have the condition s_2=0. Conjugating this, then multiplying by s_5, we get s_3=0, and we are through, since we have \text{(A)}. In this case z_0, z_1, \ldots, z_4 are vertices of a regular pentagon in any of the 24 possible cyclic arrangements. The condition s_2'=0 is stronger than s_2=0, since it implies that z_k^2=z_{k-1}z_{k+1}=z_{k-2}z_{k+2} and z_k^2+z_{k+1}z_{k+2}+z_{k-1}z_{k-2}+z_{k+1}z_{k-2}+z_{k-1}z_{k+2}=0 for all k\, (indices are integers modulo 5). Let us define a,b,c,d as follows:$$ a=\frac{z_0}{z_4},~b=\frac{z_1}{z_4},~c=\frac{z_2}{z_4},~d=\frac{z_3}{z_4}. $$The proposed equations are equivalent to the equations$$ |a|=|b|=|c|=|d|=1\tag{1}  a+b+c+d=-1\tag{2}  a b+b c+ c d+ d+a=0\tag{3} $$The equations (2) and (3) can be written in the form$$ \left\{\eqalign{\phantom{(1+b)}a + \phantom{(1+c)}d&\,=\,-c-b-1 \cr (1+b)a + (1+c)d&\,=\,-bc }\right.\tag{4} $$Let us consider two cases: • b=c, in this case the substitution of the first equation in (4) in the second yields (1+b)(1+2b)= b^2 or equivalently b^2+3b+1=0, which is absurd since this equation has only real roots of absolute value different from 1 in contradiction with (1). • b\ne c, here the system (4) can be solved with respect to a and d and we get$$ a=\frac{b+(c+1)^2}{b-c},\qquad d=\frac{(b+1)^2+c}{c-b}.\tag{5} $$Noting that \overline{b}=1/b and \overline{c}=1/c, we conclude from (5) that we have also$$ \overline{a}=\frac{b (c+1)^2+c^2}{c (c-b)},\qquad \overline{d}=\frac{ b^2 + (1 + b)^2 c}{b (b - c)}.\tag{6} $$Now the equation a\overline{a}=|a|^2=1 becomes$$ \left(\frac{b+(c+1)^2}{b-c}\right)\left(\frac{b (c+1)^2+c^2}{c (c-b)}\right)=1\,, $$which is equivalent to (c^2+3c+1)(b^2+b(c^2+c+1)+c^2)=0 , but c^2+3c+1\ne0 since |c|=1, hence$$b^2+b(c^2+c+1)+c^2=0\,.\tag{7}$$In similar way, the equation d\overline{d}=|d|^2=1 becomes$$ \left(\frac{(b+1)^2+c}{c-b}\right)\left(\frac{ b^2 + (1 + b)^2 c}{b (b - c)}\right)=1\,, $$which is equivalent to (b^2+3b+1)(c^2+c(b^2+b+1)+b^2)=0 . So,$$c^2+c(b^2+b+1)+b^2=0\,.\tag{8}$$Subtracting (8) and (7) yields (bc-1)(c-b)=0, but we have seen that c\ne b, so we must have cb=1 or equivalently$$c =\frac{1}{b}=\overline{b}\,.\tag{9}$$Replacing back in (7) we get b^2+c+1+b+c^2=0 or equivalently$$b^4+b^3+b^2+b+1=0\,.$$Thus, b is a fifth root of 1 different from 1, that is$$b\in\{\omega,\omega^2,\omega^3,\omega^4\}\quad\hbox{ where\omega=\exp \left(\frac{2\pi i}{5}\right)\,.}$$It follows that c=b^{-1}=b^4 and c^2=b^8=b^3. From (5) we conclude that$$\eqalign{ a&\,=\,\frac{b+b^3+2b^4+1}{b-b^4}\,=\,\frac{b^4\!-b^2}{b-b^4}\,=\,b^3\,,\cr d&\,=\,b^2\,,}$$where we used the identities 1+b+b^2+ b^3+b^4=0 and b^5=1. We get the solution (z_0,z_1,z_2,z_3,z_4) with$$ z_0=b^3z_4\,,~z_1=b z_4\,,~z_2=b^4 z_4\,,~z_3=b^2 z_4\,,$which are the vertices of a regular pentagon in some order, (A regular pentagon, or a regular pentagonal star). • Nice solution, I wonder if it can be simplified. The "which is equivalent to" between (7) and (8) seems to be pure luck. – Ewan Delanoy May 22 '14 at 16:35 • @Ewan Delanoy. It is not pure luck: the "which is equivalent to" between$(7)$and$(8)$is$(b\leftrightarrow c)$-symmetric to the "which is equivalent to" between$(6)$and$(7)\$. – chizhek Aug 27 '14 at 16:59