+0

# so lost

0
48
1

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

#1
+5067
+2

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

.
Apr 23, 2019

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