Write the expression in rectangular form, \( x+y i \), and in exponential form, \( \mathrm{r} e^{i \theta} \) \( (-1-i)^{3} \)
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To convert the expression \( (-1-i)^{3} \) to rectangular form, we first find the modulus and argument of the complex number \( -1 - i \). The modulus \( r \) is given by \( r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \). The argument \( \theta \) can be calculated using \( \tan \theta = \frac{-1}{-1} = 1 \), giving \( \theta = \frac{3\pi}{4} \) since the point lies in the third quadrant. Using the polar to exponential conversion, we have: \[ -1 - i = \sqrt{2} e^{i \frac{5\pi}{4}} \] Now we raise this to the power of 3: \[ (-1 - i)^{3} = \left(\sqrt{2} e^{i \frac{5\pi}{4}}\right)^{3} = (\sqrt{2})^{3} e^{i \frac{15\pi}{4}} = 2\sqrt{2} e^{i \frac{15\pi}{4}} \] Now, \( \frac{15\pi}{4} \) can be simplified since it is equivalent to \( \frac{15\pi}{4} - 2\pi = \frac{7\pi}{4} \). Now converting back to rectangular form: \[ (-1 - i)^{3} = 2\sqrt{2} \left(\cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}\right) = 2\sqrt{2} \left(\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = 2 - 2i \] So, in rectangular form, \( (-1 - i)^{3} = 2 - 2i \) and in exponential form, it can be expressed as \( 2\sqrt{2} e^{i \frac{7\pi}{4}} \).