Thank you for reporting, we will resolve it shortly
Q.
If equation $(Z-1)^{n}=Z^{n}=1 \quad(\forall n \in N)$ has solutions, then $n$ can be
NTA AbhyasNTA Abhyas 2022
Solution:
$
\begin{array}{l}
(Z-1)^{n}=Z^{n} \\
\Rightarrow|Z-1|=|Z| \Rightarrow x=\frac{1}{2} \ldots \text { (i) } \\
Z^{n}=1 \\
\Rightarrow|Z|=1 \\
\Rightarrow x^{2}+y^{2}=1 \ldots \text { (ii) }
\end{array}
$
From (i) \& (ii), we get
$
\begin{array}{l}
\frac{1}{4}+y^{2}=1 \\
\Rightarrow y=\pm \frac{\sqrt{3}}{2} \\
\Rightarrow \frac{1}{2}+\frac{\sqrt{3}}{2} i \text { and } \frac{1}{2}-\frac{\sqrt{3}}{2} i \text { can be the solutions }
\end{array}
$
which would exist, when $n$ is a multiple of 6
$\left(\because \arg \left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=\tan ^{-1} \sqrt{3}=\frac{\pi}{3} \Rightarrow\right.$ Least value of $n$ is equal to
$\left.\frac{2 \pi}{\frac{\pi}{3}}=6\right)$
