<s>[INST] Let $z$ be a complex number, s.t. $z^2-z+1=0$. Find $z^5 + z$. 
Let $z$ be a complex number, s.t. $z^2-z+1=0$.
Find $z^5 + z$.

As $z= z^2+1$, so $z\neq 0$, also $z^2\neq 0$, as $z\neq 1$.
Anyway, we can divide $z^2-z+1=0$ by $z$ to get:
$z+\frac1z-1=0$
We need to derive $z^5 + z$ in terms of $z+\frac1z$, so on division get the quotient.
$$
\require{enclose}
\begin{array}{rll}
    z^4 -z^2 +2 && \hbox{} \\[-3pt]
   z+\frac1z \enclose{longdiv}{z^5 + z}\kern-.2ex \\[-3pt]
      \underline{z^5+ z^3} && \hbox{} \\[-3pt]
      -z^3 +z && \hbox{} \\[-3pt]
      \underline{-z^3-z} && \hbox{} \\[-3pt]
      2z && \hbox{} \\[-3pt]
      \underline{2z+\frac2z} && \hbox{} \\[-3pt]
      -\frac2z
  \end{array}
$$
So, $z^5 + z= (z+\frac1z)(z^4 -z^2 +2)-\frac2z$
Seems it needs to further find division of $(z^4 -z^2 +2)$ too by $z+\frac1z$. Still remainders at each step (like, $\frac2z$) will be a further issue.
Alternatively (second approach), multiply by $(z+1)$ to get: $(z+1)(z^2-z+1)= z^3 +1=0\implies z^3=1$
$z(z^4+1)= z((z^2+1)(z^2 -1)+2)$
This approach fails too as remainder $2$ is obtained.
Next (third approach), $z+\frac1z=1\implies z^2 +\frac1{z^2}= -1$
This too fails to be useful in solving $z^5+z$.
Next (fourth approach)
, $z^2=z-1\implies z^4 = z^2-2z +1\implies z^4+1= z^2-2z +2\implies z(z^4+1)= z^3-2z^2 +2z$.
Can use value of $z^3=1$, to get:
$z^3-2z^2 +2z\implies 1-2z^2+2z\implies 1-2(z^2 -z) = 3$.
Seems the least intuitive approach works only.
Alternatively (fifth approach) to have a working intuitive approach, use the trigonometric /polar form approach:
the roots of $z^2-z+1$ are: $z=\frac12 \pm \frac{\sqrt3}{2}i$, with two roots $z_1, z_2$.
$e^{i x} = \cos(x) + i \sin(x)$
$z_1 = \frac12 + \frac{\sqrt3}{2}i = e^{ i\pi/3}$
$z_2 = \frac12 - \frac{\sqrt3}{2}i = e^{- i\pi/3}$
\begin{align}
z_1^5 + z_1 &= e^{5 i\pi/3} + e^{i\pi/3} \\
&= (sin(-\pi/3)+\cos (\pi/3))+(sin(\pi/3)+\cos (\pi/3))
 = 2\cos (\pi/3)
 = 1
\end{align}
\begin{align}
z_2^5 + z_2 &= e^{-5 i\pi/3} + e^{-i\pi/3} \\
&= (sin(\pi/3)+\cos (\pi/3))+(sin(-\pi/3)+\cos (\pi/3))
 = 2\cos (\pi/3)
 = 1
\end{align}
Now, the answer differs from earlier of $3$.
 [/INST] $z^2 - z + 1 = 0$ implies $$z^2 = z-1,$$ hence
$$\begin{align}
z^5 + z &= z((z^2)^2 + 1) \\
&= z((z-1)^2 + 1) \\
&= z(z^2 - 2z + 2) \\
&= z(z-1 - 2z + 2) \\
&= z(-z+1) \\
&= -z^2 + z \\
&= 1-z + z \\
&= 1.
\end{align}$$

All your other calculations either have errors or are essentially failing to simplify the expression.  The only one that is correct is your fifth approach.  The aim is to reduce the power of $z^5$, so the above method is how you would do it.
Multiplying $z^2 - z + 1$ by $z+1$ also works, which gives you $z^3 = -1$, thus
$$z + z^{-1} = 1$$ implies $$z^5 + z = z^3 (z^2 + z^{-2}) = -(z^2 + z^{-2}) = -(z + z^{-1})^2 + 2 = -1 + 2 = 1.$$
</s>