Why does $z^n-1=0$ have at max n solutions? $z\in\mathbb{C}$ I know that there is a Theorem which says that a Polynom of Degree n has at most n Solutions, however we have not proved it yet in our class. Is there Maybe another explaination for this Special case?
 A: Because it is a non-constant, single-variable, polynomial with complex coefficients of degree $n$ and the fundamental theorem of algebra says it has $n$ roots in $\mathbb{C}$. Another way to say that is complex numbers is algebraically closed. By a successive factorization argument you can then show that the polynomial has exactly $n$ roots. 
A: By the factor theorem ,  $f(z)$ is divisible by $(z-r)$ for each root $r$.
If there were more than $n$ roots, $r_1,\dots,r_k$, then $f(z)=p_k(z)(z-r_1)\dots(z-r_k)\implies \operatorname{deg}f\ge k\gt n$.
A: Additions in $\color{blue}{\textrm{blue}}$ 
$\color{blue}{\textrm{Suppose that}~ z^n-1 = 0~ \textrm{has}~m > n~\textrm{solutions}~ a_1, a_2, \ldots, a_m}$.

Suppose that $a_1$ is a solution of $\require{enclose}\enclose{horizontalstrike}{z^n - 1 =0}$. Then:
$\color{blue}{\textrm{Consider the solution}~a_1~\textrm{. Then:}}$
$$z^n-1 = (z-a_1)p_1(z) = zp_1(z)-a_1p_1(z),$$
where $p_1(z)$ is a polynomial. Since the degree of $zp_1(z)-a_1p_1(z)$ should be equal to $n$, then $p_1(z)$ has degree $n-1$.
Now, suppose that $\require{enclose}\enclose{horizontalstrike}{a_2}$ is another solution. Then:
$\color{blue}{\textrm{Now, consider the solution}~a_2.~\textrm{Then:}}$
$$z^n-1 = (z-a_1)(z-a_2)p_2(z)=(z^2 \ldots)p_2(z).$$
This time, $p_2(z)$ should have degree $n-2$.
In general, given $k$ solutions $a_1, a_2, \ldots, a_k$, we ca write:
$$z^n-1 = p_k(z)\prod_{i=1}^{k}(z-a_k),$$
where the degree of $p_k(z)$ is $n-k$. Of course, this can be reiterated up to $p_n(z)$, which has degree $n-n = 0$, i.e. it is $p_n(z) = p,$ a constant. Formally:
$$z^n - 1 = p\prod_{i=1}^{n}(z-a_n).$$
Then, you have $n$ solutions $a_1, a_2, \ldots, a_n$. Of course, if some $a_i$ coincide, then you have at most $n$ solutions.
$\color{blue}{\textrm{In conclusion, the number of solutions cannot be}~m>n, \textrm{but}~ m\leq n}$.
A: There's an explanation if you represent them in polar co-ordinates and consider that multiplying two complex numbers involves adding their arguments (ie angles) and multiplying their magnitudes (distances from the origin). It turns out they need to have $1$ as their magnitude and be multiples of$\frac{360°}{n}$ apart on the resulting circle, so only $n$ of them will fit.
(This is effectively the same as gimusi's answer.)
Edit: In fact, there are exactly $n$ of them, and they're equally spaced round the circle. The reason should be obvious if you pick one of the candidate angles and multiply it by $n$.
A: First fact: for complex (nonzero) polynomials $f(z)$ and $g(z)$, the degree formula holds:
$$
\deg(f(z)g(z))=\deg f(z)+\deg g(z)
$$
where $\deg$ denotes the standard polynomial degree. 
Proof. If we write
$$
f(z)=az^m+f_0(z),\qquad g(z)=bz^n+g_0(z)
$$
where $f_0$ and $g_0$ group together the lower degree terms, $a\ne0$ and $b\ne0$, then
$$
f(z)g(z)=abz^{m+n}+h(z)
$$
where again $h(z)$ has degree less than $m+n$. Thus $f(z)g(z)$ has degree $m+n$. QED
Second fact (basic and well known: if $a$ is a root of the polynomial $f(z)$, then $f(z)$ is divisible by $z-a$.
Now we prove by induction on the degree of $f(z)$ the following statement.

Let $f(z)$ be a nonzero polynomial with coefficients in $\mathbb{C}$. Then the number of distinct roots of $f$ cannot exceed the degree of $f$.

The statement is obvious for polynomials of degree $1$. Assume we know it for polynomials of degree $n-1$. Let $f(z)$ have degree $n$ and let $a_1,a_2,\dots,a_m$ be its pairwise distinct roots. By the second fact, we have $f(z)=(z-a_m)g(z)$ and $g(z)$ has degree $n-1$. Now, for $k=1,\dots,m-1$,
$$
f(a_k)=(a_k-a_m)g(a_k)=0
$$
and, since $a_k-a_m\ne0$, we conclude $g(a_k)=0$. Therefore $a_1,\dots,a_{m-1}$ are pairwise distinct roots of $g(z)$. By the induction hypothesis, we have
$$
m-1\le n-1
$$
and therefore $m\le n$. QED
Note. This proof applies with no change to polynomials having coefficients in an arbitrary domain.
A: We have that
$$z^n=1\iff z^n=e^{i2k\pi}$$
and then
$$z_k=e^{i\frac{2k\pi}n} \quad k=0,\ldots,n-1$$
and those are the $n$ roots of unity. 
It is trivial to see that the solution is periodic since $e^{i2kπ/n}$ for $k=k_0$ and $k=k_0+n$ represents the same complex number. 
Refer also to roots of unity for details.
