Suppose A is a complex number and $\mathrm{n}\in \mathrm{N}$, such that ${\mathrm{A}}^{\mathrm{n}}=(\mathrm{A}+1{)}^{\mathrm{n}}=1$, then the least value of n is

Let A = x + iy. Given,
$\left|\mathrm{A}\right|=1\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=1$ and $|\mathrm{A}+1|=1\Rightarrow (\mathrm{x}+1{)}^2+{\mathrm{y}}^2=1$
$\begin{array}{l}\Rightarrow \mathrm{x}=-\frac{1}{2}\text{ and }\mathrm{y}=\pm \frac{\sqrt{3}}{2}\\ \Rightarrow \mathrm{A}=\mathrm{\omega }\text{ or }{\mathrm{\omega }}^2\\ \Rightarrow (\mathrm{\omega }{)}^{\mathrm{n}}=(1+\mathrm{\omega }{)}^{\mathrm{n}}={\left(-{\mathrm{\omega }}^2\right)}^{\mathrm{n}}\end{array}$
Therefore, n must be even and is divisible by 3.