Note that Maple uses the uppercase letter I, rather than the lowercase letter i, to denote the imaginary unit: ${\mathrm{I}}^{ 2} \= -1$.

Since $z^{ n} \= 1$ is a polynomial with complex coefficients and a degree of n, it must have exactly n complex roots according to the Fundamental Theorem of Algebra.

To solve for all the ${\mathit{n}}^{\mathbf{th}}$ roots of unity, we will use de Moivre's Theorem: ${\left(\cos \left(x\right)\+\mathrm{I}\mathbf{}\sin \left(x\right)\right)}^n\=\cos \left(n\mathbf{}x\right)\+\mathrm{I}\mathbf{}\sin \left(n\mathbf{}x\right)$, where x  is any complex number and n is any integer (in this particular case x will be any real number and n will be any positive integer).

First, convert the complex number z to its polar form: $z\=\left|z\right|\mathbf{}\left(\cos \left(\mathrm{\theta }\right)\+\mathrm{I}\mathbf{}\sin \left(\mathrm{\theta }\right)\right)$, where $\left|z\right|$ is the modulus of z and q is the angle between the positive real axis (Re) and the line segment joining the point z to the origin on the complex plane. Since $z^{ n} \= 1$, it must be true that $\left|z\right| \= 1$, and so the previous equation simply becomes  $z\=\cos \left(\mathrm{\theta }\right)\+\mathrm{I}\mathbf{}\sin \left(\mathrm{\theta }\right)$.

Also, converting the real number $1 \= 1 \+0\cdot \mathrm{I}$ to polar form, we get $1\=\cos \left(2\mathbf{}\mathrm{\pi }\mathbf{}k\right)\+\mathrm{I}\mathbf{}\sin \left(2\mathbf{}\mathrm{\pi }\mathbf{}k\right)$ for any integer k.

Now, $z^{ n} \= {\left(\cos \left(\mathrm{\theta }\right)\+\mathrm{I}\mathbf{}\sin \left(\mathrm{\theta }\right)\right)}^n \= \cos \left(2\mathbf{}\mathrm{\pi }\mathbf{}k\right)\+\mathrm{I}\mathbf{}\sin \left(2\mathbf{}\mathrm{\pi }\mathbf{}k\right) \= 1$ and so using de Moivre's Theorem, this equation becomes $z^{ n} \= \cos \left(n\mathbf{}\mathrm{\theta }\right)\+\mathrm{I}\mathbf{}\sin \left(n\mathbf{}\mathrm{\theta }\right)\= \cos \left(2\mathbf{}\mathrm{\pi }\mathbf{}k\right)\+\mathrm{I}\mathbf{}\sin \left(2\mathbf{}\mathrm{\pi }\mathbf{}k\right) \= 1$. From this form of the equation, we can see that $n\mathbf{}\mathrm{\θ} \= 2\mathbf{}\mathrm{\π}\mathbf{}k$ , or equivalently, $\mathbf{}\mathrm{\θ} \= \frac{2 \mathrm{\π} k}{n}$.

Therefore, the $n^{ \mathrm{th}}$ roots of unity can be expressed using the formula $z_k\= \cos \left(\frac{2\mathbf{}\mathrm{\pi }\mathbf{}k}{n}\right)\+\mathrm{I}\mathbf{}\sin \left(\frac{2\mathbf{}\mathrm{\pi }\mathbf{}k}{n}\right)$, for $k \= 0\, 1\, 2\, ..\. \, n-1$.   

Using Euler's formula: ${\ⅇ}^{\mathrm{I}\mathbf{}\mathrm{\θ}}\=\cos \left(\mathrm{\theta }\right)\+\mathrm{I}\mathbf{}\sin \left(\mathrm{\θ}\right)$, we can write this formula for the $n^{ \mathrm{th}}$ roots of unity in its most common form: $z_k \= {\mathrm{e}}^{ \mathrm{I} \left(\frac{2\mathbf{}\mathrm{\π}\mathbf{}k}{n}\right)}$, for $k \= 0\, 1\, 2\, ..\. \, n-1$.