Prove that a complex equation has a solution of module 1 
Prove that the equation 
  $$z^n + z + 1=0 \ z \in \mathbb{C}, n \in \mathbb{N} \tag1$$ 
  has a solution $z$ with $|z|=1$ iff $n=3k +2, k \in \mathbb{N} $.


One implication is simple: if there is $z \in \mathbb{C}, |z|=1$ solution for (1) then $z=cos \alpha + i \cdot sin\alpha$ and $|z + 1|=1$. It follows $cos\alpha=-\frac 1 2$ etc.
The other implication is the one I failed to prove.
 A: If $z^n + z + 1=0 $ then $z^n = -(1+z)$ and $$|z|^n = |1+z|.$$
If $|z| = 1$, then $z = \cos a + i \sin a$, with $a \in [0, 2\pi]$. Moreover:
$$1^n = |1 + \cos a + i \sin a| \Rightarrow \sqrt{(1+\cos a)^2 + \sin^2 a} = 1 \Rightarrow \\
1 + \cos^2 a + 2 \cos a + \sin^2 a = 1\Rightarrow 
\cos a = -\frac{1}{2}\\ \Rightarrow a = \frac{2\pi}{3}\vee a = \frac{4\pi}{3}.$$
Let's plug $z = \cos a + i \sin a$ in the starting equation. We get:
$$\cos (na) + i \sin(na) + \cos (a) + i \sin(a) + 1 = 0 \Rightarrow\\
\begin{cases}
\cos(na) + \cos(a)+ 1 &= 0\\
\sin(na) + \sin(a)  &= 0 
\end{cases}$$
Let's work on the second equation, which becomes $\sin(a) =-\sin(na)$.
If $a=\frac{2\pi}{3} $, then $na = \frac{4\pi}{3}+2\pi k \vee na = \frac{5\pi}{3}+2\pi k$.
First case:
$$\frac{2\pi}{3}n = \frac{4\pi}{3}+2\pi k \Rightarrow 
\frac{2\pi}{3}n = \frac{2\pi}{3}(2 + 3k) \Rightarrow
n = 2+3k$$
Second case:
$$\frac{2\pi}{3}n = \frac{5\pi}{3}+2\pi k \Rightarrow 
\frac{2\pi}{3}n = \frac{2\pi}{3}\left(\frac{5}{2} + 3k\right) \Rightarrow
n = \frac{5}{2} + 3k.$$
The last one can't be satisfied since both $n$ and $k$ are integer.
If $a=\frac{4\pi}{3} $, then $na = \frac{\pi}{3}+2\pi k \vee na = \frac{2\pi}{3}+2\pi k$.
First case:
$$\frac{4\pi}{3}n = \frac{\pi}{3}+2\pi k \Rightarrow 
\frac{4\pi}{3}n = \frac{4\pi}{3}\left(\frac{1}{4} + \frac{3}{2}k\right) \Rightarrow
n = \frac{1}{4} + \frac{3}{2}k.$$
The last one can't be satisfied since both $n$ and $k$ are integer.
Second case:
$$\frac{4\pi}{3}n = \frac{2\pi}{3}+2\pi k \Rightarrow 
\frac{4\pi}{3}n = \frac{4\pi}{3}\left(\frac{1}{2} + \frac{3}{2}k\right) \Rightarrow
n = \frac{1}{2} + \frac{3}{2}k.$$
The last one can't be satisfied since both $n$ and $k$ are integer.
Finally, the only feasible case is:
$$a = \frac{2\pi}{3}$$
and
$$n = 2 + 3k.$$
Notice that in this case also the equation $\cos(na) + \cos(a)+ 1 = 0$ is satisfied. Indeed:
$$\cos\left((2+3k)\frac{2\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) + 1 = 0 \Rightarrow \\
\cos\left(\frac{4\pi}{3} + 2\pi k\right) + \cos\left(\frac{2\pi}{3}\right) + 1 = 0 \Rightarrow \\
\cos\left(\frac{4\pi}{3} \right) + \cos\left(\frac{2\pi}{3}\right) + 1 = 0 \Rightarrow \\
-\frac{1}{2}-\frac{1}{2} + 1 = 0.$$
A: You have to prove that if $n=3k+2$, then the equation has a solution $z$ with $|z|=1$.
So, suppose $n=3k+2$ and take $z=\frac12(-1\pm i\sqrt3)$, which is the solution you found for the other direction.  Certainly it is true that $|z|=1$, and by substituting and simplifying you can check that $z^n+z+1=0$.
A: HINT:
$$z^n+1=-z$$
As $z\ne0,$  $$z^{n/2}+z^{-n/2}=-z^{1-n/2}$$
Let $z=r(\cos t+i\sin t)$
$$2r^{n/2}\cos\dfrac{nt}2=-r^{(2-n)/2}\left(\cos\dfrac{(2-n)t}2+i\sin\dfrac{(2-n)t}2\right)$$
We need $\sin\dfrac{(2-n)t}2=0\iff\dfrac{(2-n)t}2=m\pi$ where $m$ is any integer
So, $\cos\dfrac{(2-n)t}2=\pm1$
Can you take it from here?
