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The equation z^3 = -2 - 2i has 3 solutions. What is the unique solution in the fourth quadrant? Enter your answer in rectangular form.

 Apr 22, 2019

Best Answer 

 #1
avatar+6046 
+2

\(-2-2i = 2\sqrt{2} e^{i5\pi/4}\\ z^3 = 2\sqrt{2} e^{i5\pi/4}\\ z = \sqrt{2} \exp\left(i\dfrac{\frac{5\pi}{4}+2k\pi}{3}\right), ~k=0,1,2\\ z =\sqrt{2}\exp\left(i\dfrac{5\pi}{12}\right),~ \sqrt{2}\exp\left(i\dfrac{13\pi}{12}\right),~\sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right)\\ \text{and of these }\\ z = \sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right) \text{ lies in the fourth quadrant}\)

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 Apr 23, 2019
 #1
avatar+6046 
+2
Best Answer

\(-2-2i = 2\sqrt{2} e^{i5\pi/4}\\ z^3 = 2\sqrt{2} e^{i5\pi/4}\\ z = \sqrt{2} \exp\left(i\dfrac{\frac{5\pi}{4}+2k\pi}{3}\right), ~k=0,1,2\\ z =\sqrt{2}\exp\left(i\dfrac{5\pi}{12}\right),~ \sqrt{2}\exp\left(i\dfrac{13\pi}{12}\right),~\sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right)\\ \text{and of these }\\ z = \sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right) \text{ lies in the fourth quadrant}\)

Rom Apr 23, 2019

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