Find $n \in \mathbb{N}$ such that $(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$ I was requested to find all $n \in \mathbb{N}$ such that $(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$. My approach was
$$(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$$
$$\implies 2^n(-\sqrt{3}+i)^n= \frac{2^n(1-i\sqrt{3})}{2}$$
$$\implies (-\sqrt{3}+i)^n = \frac{1-i\sqrt{3}}{2}$$
However, I was unable to proceed from here. Am I missing something? Thanks in advance.

Suppose $(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$, then_
$$(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$$
$$\implies 2(-\sqrt{3}+i)^n= 2^n(1-i\sqrt{3})$$
$$\implies (\frac{-\sqrt{3}+i}{2})^n = \frac{1-i\sqrt{3}}{2}$$
Take $z=\frac{-\sqrt{3}+i}{2}$. Then $\frac{1-i\sqrt{3}}{2}=i\bar{z}$
So we have the complex equation
$$z^n=i\bar{z}$$
Since $\vert z\vert =1$, we have $z^{n+1}=i$. Can you conclude from here?
Edit: Suppose there exist some $n\in\mathbb{N}$ such the inequality holds. Applying the modulus both sides we Will get that $\sqrt{13}^n=2^n$, which is clearly impossible
