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To solve $(2\sqrt{3} - 2i)^{-2}$ using De Moivre's Theorem and complex numbers, follow these steps:

1. Convert the complex number to polar form. For a complex number $z = x + yi$, the polar form is given by $z = r(\cos \theta + i \sin \theta)$, where:

   • $r = \sqrt{x^2 + y^2}$ (the modulus)
   • $\theta = \tan^{-1}(\frac{y}{x})$ (the argument)

   In this case:

   • $x = 2\sqrt{3}$
   • $y = -2$

   Calculate the modulus $r$:

   $$r = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4$$

   Now calculate the argument $\theta$:

   $$\theta = \tan^{-1}\left(\frac{-2}{2\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$

   Therefore, in polar form:

   $$2\sqrt{3} - 2i = 4\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$$

2. Use De Moivre's Theorem. According to De Moivre's theorem:

   $$(r (\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta))$$

   We need to find $(2\sqrt{3} - 2i)^{-2}$:

   $$(2\sqrt{3} - 2i)^{-2} = (4(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})))^{-2}$$

   This simplifies to:

   $$4^{-2} \left(\cos(-\frac{\pi}{6} \cdot -2) + i \sin(-\frac{\pi}{6} \cdot -2)\right) = \frac{1}{16} \left(\cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\right)$$

3. Calculate the trigonometric values:

   $$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$

   Thus:

   $$(2\sqrt{3} - 2i)^{-2} = \frac{1}{16} \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = \frac{1}{32} + i \frac{\sqrt{3}}{32}$$

4. Final Result:

   $$(2\sqrt{3} - 2i)^{-2} = \frac{1}{32} + i \frac{\sqrt{3}}{32}$$

This is the result in rectangular form after applying De Moivre's theorem to the complex number.