A question on Complex numbers.

Question

If $$z=\dfrac{\sqrt{3}-i}{2}$$ then $$(z^{95}+i^{67})^{94}=z^n$$ then, $\text{find the smallest positive integral value of}$ $n$ $\text{where}$ $i=\sqrt{-1}$

$\text{My Attempt:}$ First of all I tried to convert $z$ into $\text{Euler's Form}$ so, $z=e^{-i(\frac{π}{6})}$ Then, I raised $z$ to the $\text{95th}$ power. Then I'm getting stuck. And, not being able to proceed. Help.

Answer 1

$$z=e^{-i\pi/6}\implies z^{95}=e^{-95i\pi/6}=e^{-16i\pi+i\pi/6}=e^{i\pi/6}\\i^{67}=i^{64}(-i)=-i$$So we have $$z^{95}+i^{67}=e^{i\pi/6}+e^{-i\pi/2}=\frac{\sqrt3+i}{2}-i=\frac{\sqrt{3}-i}2=z$$

$$z^{94}=z^{-2}=z^{10}$$So the smallest positive integer for which $(z^{95}+i^{67})^{94}=z^n$ is $n=10$.

Answer 2

$z^6=-1$

$z^{12m-1}=z^{-1}=e^{i\pi/6}$

$i^{4n+3}=i^3=-i$

$z^{12m-1}+i^{4n+3}=$cis$(\pi/6)-i=$cis$(-\pi/6)=z$

where cis $=\cos +i\sin$

So, we have $z^{n-94}=1$

$\implies12$ divides $n-94$