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Let the roots of \(z^3 = 2 + 2i\) be \(a_1 + ib_1, a_2 + ib_2,\) and \(a_3 + ib_3.\) Compute \(a_1 a_2 a_3.\)

 Apr 16, 2019
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Let the roots of

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

be \(a_1 + ib_1\)\(a_2 + ib_2\) and \(a_3 + ib_3\).
Compute \(a_1 a_2 a_3\).

 

\(\begin{array}{|rcll|} \hline \mathbf{z^3} &\mathbf{=}& \mathbf{2 + 2i},\ \quad z^3=a+ib \\\\ && \text{convert 2+2i to polar form:}\\ && r = \sqrt{a^2+b^2} = \sqrt{2^2+2^2} = \sqrt{8}= \sqrt{2^3} \\ && \sqrt[3]{r} = \sqrt[3]{\sqrt{2^3} } = \sqrt[3]{ 2^\frac{3}{2} } = \left( 2^\frac{3}{2} \right)^{\frac{1}{3} } = 2^\frac{1}{2} = \sqrt{2} \\ && \theta = \arctan\left(\dfrac{b}{a}\right) = \arctan\left(\dfrac{2}{2}\right) = \dfrac{\pi}{4} \\ && \theta_i = \dfrac{\dfrac{\pi}{4}+2\pi k} {3} \\ && \theta_i = \dfrac{\pi}{12} + \dfrac{2}{3} \pi k ,\quad k= 0,\ 1, 2 \\\\ && \theta_1 = \dfrac{1}{12} \pi,\quad \theta_2 = \dfrac{3}{4} \pi, \quad \theta_3 = \dfrac{17}{12} \pi \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline z_0 &=& \sqrt{2} \left( \cos\left( \dfrac{1}{12} \pi\right) +i\sin\left(\dfrac{1}{12} \pi \right)\right) \quad | \quad \cos\left( \dfrac{1}{12} \pi\right) = \dfrac{\sqrt{2}(\sqrt{3}+1) }{4} \\ z_1 &=& \sqrt{2} \left( \cos\left( \dfrac{3}{4} \pi\right) +i\sin\left(\dfrac{3}{4} \pi \right)\right) \quad | \quad \cos\left( \dfrac{3}{4} \pi\right) = -\dfrac{\sqrt{2}}{2} \\ z_2 &=& \sqrt{2} \left( \cos\left( \dfrac{17}{12} \pi\right) +i\sin\left(\dfrac{17}{12} \pi \right)\right) \quad | \quad \cos\left( \dfrac{17}{12} \pi\right) = -\dfrac{\sqrt{2}(\sqrt{3}-1) }{4} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a_1 a_2 a_3} &=& \sqrt{2} \dfrac{\sqrt{2}(\sqrt{3}+1) }{4} \sqrt{2}\left(-\dfrac{\sqrt{2}}{2}\right) \sqrt{2}\left(-\dfrac{ \sqrt{2}(\sqrt{3}-1) }{4} \right) \\ &=& \dfrac{1}{4} (\sqrt{3}+1)(\sqrt{3}-1) \\\\ &=& \dfrac{3-1}{4} \\\\ &=& \dfrac{2}{4} \\\\ &\mathbf{=}& \mathbf{\dfrac{1}{2}} \\ \hline \end{array} \)

 

laugh

 Apr 17, 2019
edited by heureka  Apr 17, 2019

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