To solve the given problem, we need to find the modulus and principal value of the argument for the complex number expression \(\frac{(1+i\sqrt{3})^{16}}{(\sqrt{3}-i)^{17}}\). We will first simplify the expression using polar form and then find the modulus and argument. #### Solution By Steps ***Step 1: Convert to Polar Form*** First, convert each complex number to its polar form. - For \(1+i\sqrt{3}\), the modulus \(r_1 = \sqrt{1^2 + (\sqrt{3})^2} = 2\), and the argument \(\theta_1 = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}\). - For \(\sqrt{3}-i\), the modulus \(r_2 = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2\), and the argument \(\theta_2 = \arctan\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}\). ***Step 2: Calculate Powers*** Calculate the powers using the polar form: - \((1+i\sqrt{3})^{16} = 2^{16} \operatorname{cis}(16\cdot\frac{\pi}{3})\) - \((\sqrt{3}-i)^{17} = 2^{17} \operatorname{cis}(17\cdot-\frac{\pi}{6})\) ***Step 3: Divide and Simplify*** Divide the two expressions: \[\frac{(1+i\sqrt{3})^{16}}{(\sqrt{3}-i)^{17}} = \frac{2^{16} \operatorname{cis}(16\cdot\frac{\pi}{3})}{2^{17} \operatorname{cis}(17\cdot-\frac{\pi}{6})}\] Simplify by dividing the moduli and subtracting the arguments: \[= \frac{2^{16}}{2^{17}} \operatorname{cis}\left(16\cdot\frac{\pi}{3} - 17\cdot-\frac{\pi}{6}\right)\] \[= \frac{1}{2} \operatorname{cis}\left(\frac{16\pi}{3} + \frac{17\pi}{6}\right)\] ***Step 4: Find Modulus and Argument*** - Modulus: \(r = \frac{1}{2}\) - Argument: \(\theta = \frac{16\pi}{3} + \frac{17\pi}{6} = \frac{49\pi}{6}\) Since the argument is greater than \(2\pi\), subtract \(2\pi\) to find the principal value: \[\theta = \frac{49\pi}{6} - 2\pi = \frac{49\pi}{6} - \frac{12\pi}{6} = \frac{37\pi}{6}\] Subtract another \(2\pi\) to get the principal value: \[\theta = \frac{37\pi}{6} - \frac{12\pi}{6} = \frac{25\pi}{6}\] Subtract \(2\pi\) again: \[\theta = \frac{25\pi}{6} - \frac{12\pi}{6} = \frac{13\pi}{6}\] Subtract \(2\pi\) one more time: \[\theta = \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6}\] #### Final Answer The modulus is \(\frac{1}{2}\) and the principal value of the argument is \(\frac{\pi}{6}\).

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