I'm really iffy on complex numbers and I'd like to know if this is more or less correct. I first converted the numbers to trigonometric form:
$$-1+i=\sqrt{2}e^{\frac{3}{4}\pi i} \\ 1-\sqrt{3}i=2e^{\frac{5}{3}\pi i}$$
Hence,
$$\arg((-1+i)^{2n})= \arg((\sqrt{2}e^{\frac{3}{4}\pi i})^{2n})= n\arg(2e^{\frac{3}{2}\pi i}) - 2k\pi= \frac{3}{2}\pi n -2k\pi \\ \arg((1-\sqrt{3}i)^{n+1})= \arg((2e^{\frac{5}{3}\pi i})^{n+1}) = (n+1)\arg(2e^{\frac{5}{3}\pi i}) -2k'\pi=\frac{5}{3}\pi n - 2k'\pi$$
So we want to check that,
$$\begin{cases} \frac{3}{2}\pi n -2k\pi = \frac{1}{2} \pi \\ \frac{5}{3}\pi n - 2k'\pi = \frac{2}{3} \pi \end{cases} \iff \begin{cases} 3n \equiv 1 \pmod4 \\ 5n+5 \equiv 2 \pmod 6 \end{cases} \iff \begin{cases} n \equiv 3 \pmod4 \\ n \equiv 0 \pmod 3 \end{cases} $$
And the Chinese Remainder Theorem implies that this system of linear congruences has a unique solution (modulo $12$), namely,
$$n \equiv 3 \pmod {12}$$
Is this correct?
Your answer is correct, except I think you meant $\dfrac53\pi(n+1)$ where you wrote $\dfrac53\pi n.$