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


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

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 $\alpha \in \{0,2\pi\}$

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

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


I can offer one successful solution, obtained by cheating (I fed the question to Mathematica, which promptly regurgitated the answer), and an attempt at a respectable solution that failed (so far).

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

$\qquad\qquad$first solution$\qquad\qquad$second solution

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).

  • $\begingroup$ Nice solution, I wonder if it can be simplified. The "which is equivalent to" between (7) and (8) seems to be pure luck. $\endgroup$ – Ewan Delanoy May 22 '14 at 16:35
  • $\begingroup$ @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 $(7)$. $\endgroup$ – chizhek Aug 27 '14 at 16:59
  • $\begingroup$ @chizhek Why is it not pure luck ? The factorization seems quite unexpected. Algebraic equations rarely factorize. $\endgroup$ – Ewan Delanoy Aug 27 '14 at 17:02
  • $\begingroup$ @Ewan Delanoy - Yes, there is that. Though you will be surprised to know how often I see quite complex polynomials factorize (Mathematica does the factorization, not me -- I am extremely prone to errors: wrong signs, missing terms, and such), especially when I have a premonition that they must factorize. This is usually before I really understand what is going on; once I do, the factorizations become obvious, inevitable. In the little problem here we know that the answer is correct, but do not understand how it came about. There's something, some regularity, hiding in the shadows. $\endgroup$ – chizhek Aug 27 '14 at 17:16

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