\(4 \sqrt{(-8+8 \sqrt{3} i)}=\) (a) \(\pm(1+\sqrt{3 i})\) (b) \(\pm(2+2 \sqrt{3 i})\) (c) \(\pm(\sqrt{3}+\mathrm{i})\) (d) \(\pm(2-2 \sqrt{3} \mathrm{i})\)

First, we need to rewrite the given complex number in polar form. To do this, we can first express the number inside the square root in rectangular form using the real part and the imaginary part. So we have: \(-8+8\sqrt{3}i\) Now, we need to find the magnitude (r) and argument (θ) of the complex number. We can use the following formulas to find r and θ: \(r = \sqrt{a^2+b^2}\) \(\theta = \arctan{\frac{b}{a}}\) where a and b are the real and imaginary parts of the complex number. In our case, we have a = -8 and b = 8√3. So we have: \(r = \sqrt{(-8)^2+(8\sqrt{3})^2} = \sqrt{64+192} = \sqrt{256} = 16\) \(\theta = \arctan{\frac{8\sqrt{3}}{-8}} = \arctan{-\sqrt{3}} = \frac{4\pi}{3}+2k\pi \rm{, k\in Z}\) since arguments are only unique up to integer multiples of \(2\pi\) Hence, the polar form of the given complex number is: \(16\left(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\right)\). Now, we can rewrite the given expression as: \(4\sqrt{16\left(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\right)}\)

Applying De Moivre's theorem, we obtain: \(4\left(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}}\right)\), where \(\theta =\frac{4\pi}{3}+2k\pi \rm{, k\in Z}\) So, substituting for \(\theta\): \(4\left(\cos{\frac{2\pi}{3}+k\pi}+i\sin{\frac{2\pi}{3}+k\pi}\right)\).

Now, the expression is in trigonometric form. Converting the polar form to rectangular form, we use the standard equations: \(a=r\cos\theta\) \(b=r\sin\theta\) So the rectangular form is: \(4\left(2\cos\left(\frac{2\pi}{3}+k\pi\right)+2i\sin\left(\frac{2\pi}{3}+k\pi\right) \right)\) Considering integer values of k, for even numbers (k = 0, 2, -2, ...), the expression becomes: \(4\left(2\cos\left(\frac{2\pi}{3}\right)+2i\sin\left(\frac{2\pi}{3}\right) \right)=4\left(-1 + \sqrt{3} i\right) = -4 + 4\sqrt{3}i\) And for odd values of k (k = 1, -1, 3, ...), the expression becomes: \(4\left(2\cos\left(\frac{5\pi}{3}\right)+2i\sin\left(\frac{5\pi}{3}\right) \right)=4\left(-1 - \sqrt{3} i\right) = -4 - 4\sqrt{3}i\) From this, we can summarize that the result is \(\pm(2-2\sqrt{3}i)\).

Our final answer is \(\pm(2-2\sqrt{3}i)\). Comparing this with the options, we can see that it matches option (d). Hence, the correct choice is (d).