Question

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

• A. $4$
• B. $12$
• C. $15$
• D. $21$

Answer 1

Answer: (B)

$\begin{array}{l}(Z-1{)}^n=Z^n\\ \Rightarrow |Z-1|=|Z|\Rightarrow x=\frac{1}{2}\dots \text{ (i) }\\ Z^n=1\\ \Rightarrow |Z|=1\\ \Rightarrow x^2+y^2=1\dots \text{ (ii) }\end{array}$
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
$(\because \arg \left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)={\tan }^{-1}\sqrt{3}=\frac{\pi }{3}\Rightarrow$ Least value of $n$ is equal to
$\frac{2\pi }{\frac{\pi }{3}}=6)$