$$ \arg \frac { ( 2 + 2 i ) ( 1 + i \sqrt { 3 } ) } { 3 i ( \sqrt { 12 } - 2 i ) } $$

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To find the argument of the complex expression

$$
\arg \frac{(2 + 2i)(1 + i\sqrt{3})}{3i(\sqrt{12} - 2i)},
$$

we can use the property that the argument of a quotient is the difference of the arguments:

$$
\arg \frac{z_1}{z_2} = \arg z_1 - \arg z_2.
$$

Let $$ z_1 = (2 + 2i)(1 + i\sqrt{3}) $$ and $$ z_2 = 3i(\sqrt{12} - 2i) $$.

### Step 1: Calculate $$ \arg z_1 $$

First, we calculate $$ z_1 = (2 + 2i)(1 + i\sqrt{3}) $$.

Calculating $$ z_1 $$:

$$
z_1 = 2(1 + i\sqrt{3}) + 2i(1 + i\sqrt{3}) = 2 + 2i\sqrt{3} + 2i - 2\sqrt{3} = (2 - 2\sqrt{3}) + (2\sqrt{3} + 2)i.
$$

Now, we find the argument of $$ z_1 $$:

$$
\arg z_1 = \tan^{-1}\left(\frac{\text{Im}(z_1)}{\text{Re}(z_1)}\right) = \tan^{-1}\left(\frac{2\sqrt{3} + 2}{2 - 2\sqrt{3}}\right).
$$

Simplifying the fraction:

$$
\frac{2\sqrt{3} + 2}{2 - 2\sqrt{3}} = \frac{2(\sqrt{3} + 1)}{2(1 - \sqrt{3})} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}.
$$

To simplify further, we can multiply the numerator and denominator by the conjugate of the denominator:

$$
\frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{(\sqrt{3} + 1 + 3 + \sqrt{3})}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}.
$$

Thus,

$$
\arg z_1 = \tan^{-1}(-2 - \sqrt{3}).
$$

### Step 2: Calculate $$ \arg z_2 $$

Next, we calculate $$ z_2 = 3i(\sqrt{12} - 2i) $$.

Calculating $$ z_2 $$:

$$
z_2 = 3i\sqrt{12} - 3i(2i) = 3i\sqrt{12} + 6 = 6 + 3i\sqrt{12}.
$$

Now, we find the argument of $$ z_2 $$:

$$
\arg z_2 = \tan^{-1}\left(\frac{\text{Im}(z_2)}{\text{Re}(z_2)}\right) = \tan^{-1}\left(\frac{3\sqrt{12}}{6}\right) = \tan^{-1}\left(\frac{\sqrt{12}}{2}\right) = \tan^{-1}(2\sqrt{3}).
$$

### Step 3: Combine the arguments

Now we can find the argument of the original expression:

$$
\arg \frac{z_1}{z_2} = \arg z_1 - \arg z_2 = \tan^{-1}(-2 - \sqrt{3}) - \tan^{-1}(2\sqrt{3}).
$$

Using the tangent subtraction formula:

$$
\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b},
$$

where $$ a = \tan^{-1}(-2 - \sqrt{3}) $$ and $$ b = \tan^{-1}(2\sqrt{3}) $$.

Let $$ x = -2 - \sqrt{3} $$ and $$ y = 2\sqrt{3} $$:

$$
\tan(a - b) = \frac{x - y}{1 + xy} = \frac{(-2 - \sqrt{3}) - (2\sqrt{3})}{1 + (-2 - \sqrt{3})(2\sqrt{3})}.
$$

Calculating the numerator:

$$
-2 - \sqrt{3} - 2\sqrt{3} = -2 - 3\sqrt{3}.
$$

Calculating the denominator:

$$
1 + (-2)(2\sqrt{3}) - \sqrt{3}(2\sqrt{3}) = 1 - 4\sqrt{3} - 6 = -5 - 4\sqrt{3}.
$$

Thus,

$$
\tan(a - b) = \frac{-2 - 3\sqrt{3}}{-5 - 4\sqrt{3}}.
$$

The argument can be expressed as:

$$
\arg \frac{(2 + 2i)(1 + i\sqrt{3})}{3i(\sqrt{12} - 2i)} = \tan^{-1}\left(\frac{-2 - 3\sqrt{3}}{-5 - 4\sqrt{3}}\right).
$$

This is the final expression for the argument.

Thus, the final answer is:

$$
\arg \frac{(2 + 2i)(1 + i\sqrt{3})}{3i(\sqrt{12} - 2i)} = \tan^{-1}\left(\frac{-2 - 3\sqrt{3}}{-5 - 4\sqrt{3}}\right).
$$
The argument of the complex expression is $$ \tan^{-1}\left(\frac{-2 - 3\sqrt{3}}{-5 - 4\sqrt{3}}\right) $$.

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