Solutions to $z^5-1=0$ using "root of unity"? I have recently stumbled upon a problem which says:
Given that $z^5-1=0$, find all solutions for $z\in\mathbb{C}$.
I have researched this problem a bit, and I have discovered that this has something to do with a Root of Unity. What is a root of unity and how do I solve one? Also, is there a formula for finding all complex solutions of $z^n=1$, where n is a constant? If so, what does this have to do with a root of unity?
EDIT
Another question: I see the $\cos(x)+i\sin(x)$ formula for all $z^n=1$, but is there a formula for all $z^n=c$ for some $c \in \mathbb{R}$, or at least positive $c \in \mathbb{R}$?
 A: Definition (Root of unity). An element $\zeta$ of a ring $R$ with unity $1$ such that $\zeta^m = 1$ for some $m \ge 1$.  The least such $m$ is the order of $\zeta$.  -Encyclopedia of Mathematics
In the field of complex numbers, roots of unity of order $m$ take the form $$\cos\frac{2\pi k}{m} + i \sin\frac{2\pi k}{m},$$
where $k=0,1,\dots,m-1$.
We may write $1$ in polar form as $e^{2\pi i k}$ and therefore $\zeta^m=e^{2\pi i k}\Rightarrow\zeta=e^{\frac{2\pi i k}{m}}$. With Euler's formula, we can decompose the exponential into real and imaginary parts to get our final result, which is stated above. If we add the condition that $k$ must be relatively prime to $m$, we only get the so-called primitive roots.
The same can be done with $\zeta^m=c,$ where $c$ is a real number. Write $c$ as $ce^{2\pi i k}$ to get the slightly generalized result $\zeta=\sqrt[m]{c}\,(\cos\frac{2\pi k}{m} + i \sin\frac{2\pi k}{m})$.

Complex roots of unity 5 $\qquad\qquad\qquad\qquad\:\:$ The mapping $z\mapsto z^5-1$


A: Given one solution of $z^n=c$, the other $n-1$ can be found by multiplying by the $n-1$ respective $n$-th roots of unity other than $1$.
In general those are the first $n-1$ powers of a primitive $n$-th root.
So take an example:  $z^5=3$.
Then take one solution, say $z=\sqrt[5]3$.  Then if $\zeta_1,\dots,\zeta_4$ are the primitive fifth roots of unity, in this case the ones other than $1$, we have that the five roots are $\sqrt[5]3\zeta_1,\dots,\sqrt[5]3\zeta_4, \sqrt[5]3$.
And note that we have $\zeta_k=\zeta_1^k$.
(All of this follows rather easily from the fact that for any $n$-th root of unity, the $n$-th power is $1$.  And using Euler's formula: $e^{i\theta}=\cos\theta+i\sin\theta$, we also get that $e^{2\pi i k/n}$ is primitive iff $(n,k)=1$.)
