Find \( z w \) and \( \frac{z}{w} \). Write each answer in polar form and in exponential form. \( \begin{array}{l}z=4+4 i \\ w=\sqrt{3}+i\end{array} \)

To find \( zw \) and \( \frac{z}{w} \), we'll first convert \( z \) and \( w \) into their polar forms. ### Step 1: Convert \( z \) to polar form For \( z = 4 + 4i \): 1. Modulus: \( |z| = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \) 2. Argument: \( \theta_z = \tan^{-1}\left(\frac{4}{4}\right) = \tan^{-1}(1) = \frac{\pi}{4} \) So in polar form, \( z = 4\sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \) or in exponential form: \[ z = 4\sqrt{2} e^{i\frac{\pi}{4}} \] ### Step 2: Convert \( w \) to polar form For \( w = \sqrt{3} + i \): 1. Modulus: \( |w| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \) 2. Argument: \( \theta_w = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \) So in polar form, \( w = 2 \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) \) or in exponential form: \[ w = 2 e^{i\frac{\pi}{6}} \] ### Step 3: Find \( zw \) Using polar forms: \[ zw = (4\sqrt{2})(2) \left(\cos\left(\frac{\pi}{4} + \frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right)\right) \] Calculating the modulus: \[ |zw| = 4\sqrt{2} \cdot 2 = 8\sqrt{2} \] Calculating the argument: \[ \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{5\pi}{12} \] Thus, in polar form: \[ zw = 8\sqrt{2} \left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right) \] In exponential form: \[ zw = 8\sqrt{2} e^{i\frac{5\pi}{12}} \] ### Step 4: Find \( \frac{z}{w} \) Using polar forms: \[ \frac{z}{w} = \frac{4\sqrt{2}}{2} \left(\cos\left(\frac{\pi}{4} - \frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{4} - \frac{\pi}{6}\right)\right) \] Calculating the modulus: \[ \left|\frac{z}{w}\right| = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \] Calculating the argument: \[ \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi}{12} - \frac{2\pi}{12} = \frac{\pi}{12} \] Thus, in polar form: \[ \frac{z}{w} = 2\sqrt{2} \left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right) \] In exponential form: \[ \frac{z}{w} = 2\sqrt{2} e^{i\frac{\pi}{12}} \] ### Final Answers 1. \( zw = 8\sqrt{2} e^{i\frac{5\pi}{12}} \) and \( 8\sqrt{2} \left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right) \) 2. \( \frac{z}{w} = 2\sqrt{2} e^{i\frac{\pi}{12}} \) and \( 2\sqrt{2}