If equation ${\left(Z-1\right)}^n=Z^n=1 \left(\forall n\in N\right)$ has solutions, then $n$ can be

${\left(Z-1\right)}^n=Z^n\Rightarrow \left|Z-1\right|=\left|Z\right|\Rightarrow x=\frac{1}{2}$ …(i)
$Z^n=1\Rightarrow \left|Z\right|=1\Rightarrow x^2+y^2=1$ …(ii)
From (i) & (ii),

 $\frac{1}{4}+y^2=1\Rightarrow y=\pm \frac{\sqrt{3}}{2}$
$\Rightarrow \frac{1}{2}+\frac{\sqrt{3}}{2}i$  and  $\frac{1}{2}-\frac{\sqrt{3}}{2}i$ can be the solutions
Which would exist when $n$ is a multiple of $6$ $\left\{\because arg\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\sqrt{3}=\frac{\pi }{3}\Rightarrow \mathrm{l}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{t\; value\; of} n is \mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l} \mathrm{t}\mathrm{o} \frac{2\pi }{\frac{\pi }{3}}=6\right\}$