Hello everyone I have to solve the following equations in complex numbers:
$z_1 + z_2 + z_3 + z_4 = 0$
$z_1^2 + z_2^2 +z_3^2 +z_4^2 = 0$
$z_1^3 + z_2^3 +z_3^3 +z_4^3 = 0$
$z_1^4 + z_2^4 +z_3^4 +z_4^4 = 1$
I need to find all the answer for those equations.
I tried to use the fact that $z_1 = -(z_2 + z_3 + z_4)$ and place it in all the other equations and it become a more complex.
Someone have an idea how to solve this problem?
On
Roots of unity are handy here. In terms of the Iverson bracket, the choice $z_k=c\exp\frac{2\pi ik}{4}=ci^k$ gives $\sum_{k=1}^4z_i^p=c^p\sum_ki^{kp}=4c^p[4|p]$. This matches your problem with $c=4^{-1/4}=\tfrac{1}{\sqrt{2}}$.
Let $P(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$ be a polynomial whose zeros are $z_1, z_2, z_3$ and $z_4.$ Then by Newton's Sum Identities, $a_3=a_2=a_1=0,$ and $$a_4+4a_0=0.$$ Note that, both $a_0, a_4$ anr nonzero and arbitrary up to this last relationship. Hence by choosing $a_0=-1,$ we get $$P_*(x)=4x^4-1=(2x^2-1)(2x^2+1).$$