why gives me that system of equations:
$-\tfrac{1}{x} +2ux =0$
$-\tfrac{1}{y}+2uy=0$
$1+2uz=0$
a)--> $x^2 = y^2= -z $ ?
And using the equation: $x^2+y^2+z^2 = 1$ gives b)$z = 1 (+,-) \sqrt{2}$
I dont understand a) --> and from a) to b)
Would be great, if someone could explain
There's something wrong. I used the equations you gave \begin{align} \begin{cases} 2 u x-\frac{1}{x}&=0\\ 2 u y-\frac{1}{y}&=0\\ 2 u z &=-1\\ x^2+y^2+z^2 &=1 \end{cases} \end{align} whose solution are \begin{align} (x,y,z,u)= \begin{cases} \left(x , + x , -x^2 , 1+\frac{x^2}{2} \right)\\ \left(x , - x , -x^2 , 1+\frac{x^2}{2} \right) \end{cases} \end{align} where $x^4 = 3-2\sqrt{2} \implies x = \left\{ -\sqrt{\sqrt{2}-1},\sqrt{\sqrt{2}-1},-i \sqrt{1+\sqrt{2}},i \sqrt{1+\sqrt{2}}\right\}$.
The 8 points above are the solutions of your system.
I don't know what you wanted to know. Please, rephrase your question so we can better help you.