<s>[INST] 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 $\alpha \in \{0,2\pi\}$
$$1\angle (\alpha)+ 1\angle ( \phi+\alpha)+ 1\angle ( 2\phi+\alpha)+ 1\angle ( 3\phi+\alpha)+ 1\angle ( 4\phi+\alpha)=0$$
$$ 1\angle (\phi+\alpha)+ 1\angle ( 3\phi+\alpha)+1\angle ( 5\phi+\alpha) +1\angle ( 7\phi+\alpha)+1\angle ( 4\phi+\alpha)=0$$
 [/INST] 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).
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