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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.

 Aug 21, 2020
 #2
avatar+111628 
+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}\\

 Aug 22, 2020

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