Question

If $z=\frac{\sqrt{3}-1}{2}$ and ${(z^{95}+i^{67})}^{94}=z^n$, then the smallest integral value of $n$ is

• A. $2$
• B. $9$
• C. $10$
• D. $6$

Answer 1

The correct option is C $10$

$z=\frac{\sqrt{3}-i}{2}=\frac{i\sqrt{3}+1}{2i}=(-i)\left(\frac{1+i\sqrt{3}}{2}\right)=i\left(\frac{-1-i\sqrt{3}}{2}\right)=i\omega$

$z^n=(i\omega {)}^n=i^n{\omega }^n$ ---------------(1)

$z^{95}=i^{95}{\omega }^{95}=i^{94+1}{\omega }^{93+2}=(i^2{)}^{47}\times i\times ({\omega }^3{)}^{31}\times {\omega }^2=-i\times {\omega }^2$ (Using $i^2=-1$ and ${\omega }^3=1$ )

$i^{67}=(i^2{)}^{33}\times i=-i$

$(z^{95}+i^{67}{)}^{94}=(-i{\omega }^2-i{)}^{94}=i^{94}({\omega }^2+1{)}^{94}=(-1{)}^{47}(-\omega {)}^{94}$ (Using $1+\omega +{\omega }^2=0$)

$=-({\omega }^3{)}^{31}\ast \omega =-\omega$ --------(2)

Given that $(z^{95}+i^{67}{)}^{94}=z^n$

From (1) and (2),

$\Rightarrow$ $-\omega =i^n{\omega }^n$

$⟹{\omega }^{n-1}{i}^n=-1$

Let $n-1=3a$ and $n=2b$ where $a,b\in I$. This is because ${\omega }^3=1$ and $i^2=-1$.

$2b-1=3a$

$b=\frac{3a+1}{2}$

When $a=3$, $b=\frac{10}{2}=5$

Hence smallest integral value of $a$ and $b$ is $3$ and $5$ respectively.

$\therefore$ smallest integral value of $n$ is $10$