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.

animebtsbiggestfan Aug 21, 2020

#2**+1 **

I think

\(z^3=-2(1+i) \)

just thinking about 1+i and where it lays on the complex plane.

I can see that the distance from the origin is sqrt2 and the angle with the positive x axis is pi/4

so this can be written as

\(z^3=-2*\sqrt2 * e^{\frac{\pi}{4}i}\\ z^3=-2^{\frac{3}{2}} * e^{\frac{\pi}{4}i}\\ z=-2^{\frac{3}{2}\cdot \frac{1}{3}} * e^{\frac{\pi}{4}i\cdot\frac{1}{3}}\\ z=-\sqrt{2} e^{\frac{\pi}{12}i}\\ \)

This is only one of the 3 cube roots, It is in the 3rd quadrant

the other two have angles of

\(\frac{\pi}{12}+\frac{2\pi}{3}=\frac{\pi}{12}+\frac{8\pi}{12}=\frac{9\pi}{12} =\frac{3\pi}{4} \\ and\\ \frac{\pi}{12}+\frac{4\pi}{3}=\frac{\pi}{12}+\frac{16\pi}{12}=\frac{17\pi}{12} \)

4th quad solution:

\(z=-\sqrt{2} e^{\frac{3\pi}{4}i}\\ z=-\sqrt{2} [cos(\frac{3\pi}{4})+isin(\frac{3\pi}{4})]\\ z=-\sqrt{2} [-\frac{1}{\sqrt2}+\frac{i}{\sqrt2}]\\ z=1-i \)

LaTex:

z^3=-2*\sqrt2 * e^{\frac{\pi}{4}i}\\

z^3=-2^{\frac{3}{2}} * e^{\frac{\pi}{4}i}\\

z=-2^{\frac{3}{2}\cdot \frac{1}{3}} * e^{\frac{\pi}{4}i\cdot\frac{1}{3}}\\

z=-\sqrt{2} e^{\frac{\pi}{12}i}\\

Melody Aug 22, 2020