The equation\(z^3 = -2 - 2i\)

has \(3\) solutions. What is the unique solution in the fourth quadrant? Enter in your answer in rectangular form.

Guest Jan 7, 2019

#1**+1 **

\(-2-2i = 2\sqrt{2}e^{i\frac{5\pi}{4}} = (\sqrt{2})^3e^{i\frac{5\pi}{4}} \\ (-2-2i)^{1/3} = \sqrt{2} e^{i\frac{\left(\frac{5\pi}{4}+2k\pi\right)}{3}},~k=0,1,2 = \\ \sqrt{2} e^{\frac{5\pi}{12}},~\sqrt{2}e^{i\frac{13\pi}{12}},~\sqrt{2}e^{i \frac{7\pi}{4}}\\ \text{of these only }\sqrt{2}e^{i \frac{7\pi}{4}} \text{ lies in the 4th quadrant}\)

\(\sqrt{2}e^{i \frac{7\pi}{4}} = \sqrt{2}\left(\dfrac{1}{\sqrt{2}} - i \dfrac{1}{\sqrt{2}}\right) = 1-i\)

.Rom Jan 7, 2019