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}

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
$$(x,y,z)=(2\cos\theta,2\sin\theta+1,-\cos\theta)$$
