For any integer \(\mathrm{n}, \arg \left[(\sqrt{3}+\mathrm{i})^{4 n+1} /(1-\mathrm{i} \sqrt{3})^{4 \mathrm{n}}\right]=\) (a) \((\pi / 3)\) (b) \((\pi / 6)\) (c) \((2 \pi / 3)\) (d) \((5 \pi / 6)\)

We first simplify the powers of complex numbers in the numerator and denominator using De Moivre's theorem: \((\cos\varphi + i\sin\varphi)^n = \cos(n\varphi)+ i\sin(n\varphi)\) So, let's first find the polar form of the given complex numbers: \(\sqrt{3} + \mathrm{i}=\) 2\((\frac{\sqrt{3}}{2} + \frac{1}{2} \mathrm{i}) = 2(\cos(\pi/3) + \mathrm{i}\sin(\pi/3))\) \(1-\mathrm{i}\sqrt{3} =\) 2\((\frac{1}{2} - \mathrm{i}\frac{\sqrt{3}}{2}) = 2(\cos(5\pi/3) + \mathrm{i}\sin(5\pi/3))\) Now, we can use De Moivre's theorem to simplify the powers: \((\sqrt{3}+\mathrm{i})^{4 n+1} = 2^{4n+1} (\cos((4n+1)\pi/3) + \mathrm{i}\sin((4n+1)\pi/3))\) \((1-\mathrm{i} \sqrt{3})^{4 \mathrm{n}} = 2^{4n} (\cos(20n\pi/3) + \mathrm{i}\sin(20n\pi/3))\)

Now we need to simplify the fraction inside the argument function: \(\frac{(\sqrt{3}+\mathrm{i})^{4 n+1}}{(1-\mathrm{i} \sqrt{3})^{4 \mathrm{n}}}= \frac{2^{4n+1}(\cos((4n+1)\pi/3) + \mathrm{i}\sin((4n+1)\pi/3))}{2^{4n}(\cos(20n\pi/3) + \mathrm{i}\sin(20n\pi/3))}\) The coefficients (powers of 2) will cancel out, and we get: \(\frac{\cos((4n+1)\pi/3) + \mathrm{i}\sin((4n+1)\pi/3)}{\cos(20n\pi/3) + \mathrm{i}\sin(20n\pi/3)}\)

To find the principal argument of the given expression, we can simply subtract the argument of the denominator from that of the numerator: \((4n+1)\pi/3 - 20n\pi/3 = (4n - 20n+1)\pi/3 = -16n\pi/3 + \pi/3\) Since n is an integer, -16nπ/3 represents a multiple of 2π, and thus it won't contribute to the principal argument. So, the principal argument of the expression is: \(\arg\left[\frac{(\sqrt{3}+\mathrm{i})^{4 n+1}}{(1-\mathrm{i} \sqrt{3})^{4 \mathrm{n}}\right] = \boxed{\pi/3}\) Hence, the correct choice is (a) \((\pi / 3)\).