Question

Let $z=\frac{\sqrt{3}}{2}-\frac{i}{2}.$ Then the smallest positive integer $n$ such that $(z^{95}+i^{67}{)}^{94}=z^n$ is

Answer 1

From the hypothesis, we have
$z=\frac{\sqrt{3}}{2}-\frac{i}{2}=i(-\frac{1}{2}-\frac{i\sqrt{3}}{2})=i\omega$
where $\omega =-\frac{1}{2}-\frac{i\sqrt{3}}{2},$ which is a cube root of unity.

Now, $z^{95}=(i\omega {)}^{95}=-i{\omega }^2$
and $i^{67}=i^3=-i$
Therefore, $z^{95}+i^{67}=-i(1+{\omega }^2)=(-i)(-\omega )=i\omega$
$\Rightarrow (z^{95}+i^{67}{)}^{94}=(i\omega {)}^{94}=i^2\omega =-\omega$

Now, $-\omega =z^n=(i\omega {)}^n$
$\Rightarrow i^n\cdot {\omega }^{n-1}=-1$
$\Rightarrow n=2,6,10,14,\dots \dots$
and $n-1=3,6,9,\dots \dots$
Hence, $n=10$ is the required least positive integer.