Write the expression in rectangular form, \( x+y i \), and in exponential form, \( e^{i \theta} \). \( (-1-i)^{3} \)
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First, let's simplify \( (-1 - i)^3 \) in rectangular form. We can start by converting \( -1 - i \) to polar form. The modulus is \( |z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \) and the argument \( \theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \frac{3\pi}{4} \) (since both components are negative, it’s in the third quadrant). Therefore, in polar form, we have \( (-1 - i) = \sqrt{2} e^{i \frac{5\pi}{4}} \). Now, we raise it to the power of 3: \[ (-1 - i)^3 = \left(\sqrt{2} e^{i \frac{5\pi}{4}}\right)^3 = (\sqrt{2})^3 e^{i \frac{15\pi}{4}} = 2\sqrt{2} e^{i \frac{15\pi}{4}}. \] To convert \( \frac{15\pi}{4} \) into the standard range, subtract \( 2\pi \) to get \( \frac{15\pi}{4} - 2 \times 2\pi = \frac{15\pi}{4} - \frac{8\pi}{4} = \frac{7\pi}{4} \). Now rewriting in exponential form, we have \( (-1 - i)^3 = 2\sqrt{2} e^{i \frac{7\pi}{4}} \). To convert back to rectangular form, we use \( e^{i \frac{7\pi}{4}} = \cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right) \). Since \( \cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} \): \[ 2\sqrt{2} \left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = 2 - 2i. \] Thus, the expressions are \( 2 - 2i \) in rectangular form and \( 2\sqrt{2} e^{i \frac{7\pi}{4}} \) in exponential form.