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
 A: 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.
A: So, as noted, $z = 0$ is always a solution. For $z \ne 0$, if $z^n = \bar{z}$,
$$
z^{n+1} = |z|^2 = 1.
$$
So the solutions are the $n+1$th roots of unity. The sum of these is always zero, which can be seen from
$$
e^{\frac{2\pi i}{n+1}}\sum_{m=0}^n e^{\frac{2\pi i m}{n+1}} = \sum_{m=1}^n e^{\frac{2\pi i m}{n+1}} + e^{2\pi i \frac{n+1}{n+1}}  = \sum_{m=1}^n e^{\frac{2\pi i m}{n+1}} + 1 = \sum_{m=0}^n e^{\frac{2\pi i m}{n+1}}
$$
and $e^{\frac{2\pi i}{n+1}} \ne 1$. Their product, on the other hand, is 
$$
\prod_{m=0}^ne^{\frac{2\pi i m}{n+1}} = \exp\left(\frac{2\pi i}{n+1}\sum_{m=0}^nm\right) = \exp\left(\frac{2\pi i}{n+1}\sum_{m=0}^nm\right) = e^{i\pi n} = (-1)^n
$$ So we have


*

*Sum of solutions: 0

*Product of nonzero solutions: $(-1)^n$

*Product of all solutions: 0

