Q.Find the least positive integer $n$ such that $\left(\dfrac{1+i}{1-i}\right)^n = 1$.

$n$th roots of a complex number. If $\omega=\rho(\cos\phi+i\sin\phi)$ satisfies $\omega^n=z$ where $z=r(\cos\theta+i\sin\theta)$, write $z$ with its general argument $\theta+2k\pi$ (since a single value of $\theta$ would miss roots) and apply de Moivre's theorem to $\omega^n$: comparing modulus and argument gives $\rho^n=r$ and $n\phi=\theta+2k\pi$, so

Hence the $n$th roots of $z=r(\cos\theta+i\sin\theta)$ are

Although $k$ can be any integer, only $k=0,1,\dots,n-1$ give distinct values (larger $k$ repeats the same $n$ roots cyclically). Geometrically, all $n$ roots share the modulus $r^{1/n}$, so they lie on a circle of radius $r^{1/n}$ centred at the origin, equally spaced at angular intervals of $\dfrac{2\pi}n$ — i.e. at the vertices of a regular $n$-gon.

$n$th roots of unity. Setting $z=1=\cos0+i\sin0$ specialises the formula to

Writing $\omega=e^{2\pi i/n}$ (the primitive $n$th root), the $n$ roots are exactly $1,\omega,\omega^2,\dots,\omega^{n-1}$ — a geometric progression with common ratio $\omega$ — and they are the vertices of a regular $n$-gon inscribed in the unit circle.

Standing facts about the $n$th roots of unity (all provable from the GP sum/product formulas):

• Sum $1+\omega+\omega^2+\cdots+\omega^{n-1}=0$ (a finite GP with ratio $\omega\ne1$, sum $\dfrac{\omega^n-1}{\omega-1}=\dfrac{1-1}{\omega-1}=0$).
• Product $1\cdot\omega\cdot\omega^2\cdots\omega^{n-1}=(-1)^{n-1}$.
• They all satisfy $|z|=1$ and $z^n=1$. …