Solving equations with complex numbers I have to solve:
$Z^3+\bar{\omega^7} = 0$ and $Z^5\omega^{11}=1$
From the second equation, I got $Z^5=\omega$ and from the first I got $Z^3=-\omega^2$. I plugged in omega from the first result into $Z^3=-\omega^2$, giving me $Z=0$ or $Z^7=-1$, finally giving me 8 solutions:$0,-1,-\alpha...-\alpha^6$. 
This is incorrect. Why? 
Also, what is the correct way to solve complex equations?
 A: Let's assume that $\omega$ is a root of unity, but not necessarily a cube root of unity, and see what happens.
It's clear that $Z\not=0$.  The equation $Z^3+\overline\omega^7=0$ can be rewritten as $Z^3\omega^7=-1$, which, by squaring both sides, implies $Z^6\omega^{14}=1$.  Combining with the other equation, $Z^5\omega^{11}=1$, gives $Z\omega^3=1$, or $Z=\overline\omega^3$.
Plugging $Z=\overline\omega^3$ into $Z^3\omega^7=-1$, we find $\overline\omega^2=-1$, which means $\omega=\pm i$, and this implies $Z=\pm(-i)^3=\pm i$.
In other words, what we've shown is that in order for the simultaneous equations $Z^3+\overline\omega^7=0$ and $Z^5\omega^{11}=1$ to have a solution with $\omega$ a root of unity, we must have $Z=\omega=\pm i$.
A: Hint:$$z^3=-\overline{\omega}^7\\z^5=\frac{1}{\omega^{11}}$$ note that $$(z^5)^3=(z^3)^5$$so 
$$(z^5)^3=(z^3)^5\\
(\frac{1}{\omega^{11}})^3=(-\overline{\omega}^7)^5\\
\frac{1}{\omega^33}=-\overline{\omega}^{35}\\-\overline{\omega}^{35}\times \omega^33=1\\\to 
-(\omega \overline{\omega})^{33}.\overline{\omega}^{2}=1\\-(|\omega|^2)^{33}.\overline{\omega}^{2}=1$$ in the other hand 
$$z^5\omega^{11}=1 \to z^6\omega^{11}=z \\ (-\overline{\omega}^7)^2\omega^{11}=z\\\overline{\omega}^{11}\overline{\omega}^3\omega^{11}=z \\\overline{\omega}^3=z$$
