If $$\mathcal{Z} = \frac{(2\sqrt3 + 2i)^8}{(1-i)^6} + \frac{(1-i)^6}{(2\sqrt3 - 2i)^8}$$
what is the argument of $\mathcal{Z}$?
considering the first term as ${Z}$ the other term becomes $\frac{1}{\overline{Z}}$. Is there any logic I could apply or any general method that may be there for these sort of problems , without actually expanding the whole thing ?
Hints (for a maybe simple evaluation): $$(2\sqrt 3 +2i) =2(\sqrt 3 + i) $$ $$= 4\left[\frac{\sqrt 3}2 +\frac{i} 2\right]=4e^{\frac{\pi i}6} \tag 1$$
Similarly, $$(2\sqrt 3 - 2i) =4\left[\frac{\sqrt 3}2-\frac{i}{2}\right] $$ $$= 4e^{-\frac{\pi i} 6} \tag 2$$
And, $$(1-i)=\sqrt 2\left[\frac1{\sqrt 2}-\frac{i}{\sqrt 2}\right]$$ $$= \sqrt 2e^{-\frac{\pi i} 4}\tag 3$$