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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.

 May 17, 2019
 #1
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The equation \(z^3 = -2 - 2i\) has 3 solutions.
What is the unique solution in the fourth quadrant?.

 

\(\begin{array}{|rclrcl|} \hline z^3 &=& -2-2i \\ &=& \sqrt{(-2)^2+(-2)^2}e^{\left(\pi+\arctan(\frac{-2}{-2}) \right)i} \\ &=& 2^{\frac{3}{2}}e^{ \frac{5\pi}{4} i}\qquad \Rightarrow & z_0&=& 2^{\frac{3}{2\cdot 3}}\cdot e^{ \frac{5\pi}{4\cdot 3} i} \\ & & & &=& \sqrt{2} \cdot e^{ \frac{5\pi}{12} i} \\ & & & &=& \sqrt{2}\left(\cos(\frac{5\pi}{12} ) +i\cdot \sin(\frac{5\pi}{12}) \right) \\ & & & \mathbf{z_0} &=& \mathbf{0.3660+1.3660i} \quad \text{first quadrant} \\\\ & & &z_1&=&\sqrt{2} \cdot e^{\left( \frac{5\pi}{12}+\frac{2\pi}{3} \right) i} \\ & & & &=&\sqrt{2} \cdot e^{ \frac{13\pi}{12} i} \\ & & & &=& \sqrt{2}\left(\cos(\frac{13\pi}{12} )+i\cdot \sin(\frac{13\pi}{12}) \right) \\ & & & \mathbf{z_1} &=& \mathbf{-1.3660-0.3660i} \quad \text{third quadrant} \\\\ & & &z_2&=&\sqrt{2} \cdot e^{\left( \frac{5\pi}{12}+2\cdot \frac{2\pi}{3} \right) i} \\ & & & &=&\sqrt{2} \cdot e^{ \frac{21\pi}{12} i} \\ & & & &=& \sqrt{2}\left(\cos(\frac{21\pi}{12} )+i\cdot \sin(\frac{21\pi}{12}) \right) \\ & & & \mathbf{z_2} &=& \mathbf{1-i} \quad \text{fourth quadrant} \\ \hline \end{array} \)

 

The unique solution in the fourth quadrant is \(\mathbf{1-i}\).

 

laugh

 May 17, 2019

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