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

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