Solutions for a Complex Numbers Equation I have the equation
$z^{8}=\bar{z}$
I had to solve it, and to find the sum and product of the solutions. I did all that and found that the solutions are (in degrees): cis of: 0, 40, 80, 120, ..., 320.
The sum was 0 and the product 1 (I used arithmetic and geometric series).
Now I need to tell what would happen if the equation was:
$z^{n}=\bar{z}$ where n is natural.
I understand that it depends if n is odd or even, but I am not sure how it affect the sum and product. Does it affect the solutions themselves ?
Thank you

You have $|z|^8=|z|$, so either $z=0$ or $|z|=1$. 
In the latter case, the equation becomes $z^9=1$, because $\bar{z}=z^{-1}$.
The general case is exactly the same. The final problem is to find the sum of the $(n+1)$-th roots of $1$, which is fairly easy.
