Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $ Find all real solutions of  the system of equation
\begin{cases}
{x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0.
\end{cases}
 A: From second equation we get $y=-z^2-\frac{x}{z}-\frac{1}{z^2}$ (it's obvious that $z\neq0$).
Thus, $$x^2+\left(z^2+\frac{1}{z^2}+1+\frac{x}{z}\right)^2=4$$ or
$$\left(1+\frac{1}{z^2}\right)x^2+\frac{2}{z}\left(z^2+\frac{1}{z^2}+1\right)x+\left(z^2+\frac{1}{z^2}+1\right)^2-4=0,$$ 
which gives 
$$\frac{1}{z^2}\left(z^2+\frac{1}{z^2}+1\right)^2-\left(1+\frac{1}{z^2}\right)\left(\left(z^2+\frac{1}{z^2}+1\right)^2-4\right)\geq0$$ or
$$(z^4+z^2-1)^2\leq0,$$
which gives $z^2=\frac{\sqrt5-1}{2}$,  $x=-\frac{\frac{1}{z}\left(z^2+\frac{1}{z^2}+1\right)}{1+\frac{1}{z^2}}$ and the rest is smooth.
A: This system of 2 equations has infinite solutions.
Let $x=2\cos\theta$ and $y=2\sin\theta+1$ which satisfy the first equation. Then 
$$z^4+yz^2+xz+1=0$$
$$z^4+(2\sin\theta+1)z^2+(2\cos\theta)z+1=0$$
$$(z^2+\sin\theta)^2+(z+\cos\theta)^2=0$$
$z=-\cos\theta$ and $z^2=-\sin\theta$, this leads us to find solutions of $\theta$.
$z^2=-\sin\theta$ shows $\sin\theta<0$.
$\sin^2\theta-\sin\theta-1=0$ shows $\theta=\arcsin\dfrac{1-\sqrt{5}}{2}\sim-38.17^\circ$. Only two solution $(x,y,z)$ exist. one for $\theta=(180+38.17)^\circ$ and other for $\theta=(360-38.17)^\circ$ and then
$$(x,y,z)=(2\cos\theta,2\sin\theta+1,-\cos\theta)$$
A: let $t = y-1$
$x^2+t^2 = 4$
$tz^2 + xz = -z^4 - z^2 - 1$
Solutions (x,t) exist if and only if $\frac{z^4 +z^2 +1}{\sqrt{z^4 +z^2}} \le 2$ (geometrically)
Let $k = z^4 + z^2$
$\frac{k +1}{sqrt(k)} \le 2$ equals $k^2 +2k +1 \le 4k$  equals $k^2 -2k +1 \le 0$ equals k = 1;
So, $z^4 + z^2 = 1$
