I have this question which I'm stuck on, here's the question and what I did.
Find the smallest positive integer m such that $\left(\sqrt{3}+i\right)^m=\left(\sqrt{3}-i\right)^m$.
I expanded out each side using De Moivre's, $\cos\left(30m\right)+i\sin\left(30m\right)=\cos\left(-30m\right)+i\sin\left(-30m\right)$.
I tried to compare when $\cos\left(30m\right)=\cos\left(-30m\right)$ and $\sin\left(30m\right)=\sin\left(-30m\right)$ but none of the quadrants work, the answer is $m=6$, which corresponds to Quadrant 3 and 4 (tan and cos).
I'm in year 10 just learning complex, so if there are harder methods to this, don't show me.
$$\left(\sqrt{3}+i\right)^m=\left(\sqrt{3}-i\right)^m$$ $$2^m\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^m=2^m\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^m$$
Using above formula we can write $$2^m(e^{i\pi/6})^m=2^m(e^{-i\pi/6})^m$$
$$e^{mi\pi/6}=e^{-mi\pi/6}$$
$$e^{2mi\pi/6}=e^{mi\pi/3}=1$$ $$e^{mi\pi/3}=e^{2ik\pi}$$
$$\frac{mi\pi}{3}=2ik\pi$$ $$\frac{m}{3}=2k$$ $$m=6k$$
where $k\in\mathbb Z$