Question

How do I solve this problem? "One of the three cube roots of a complex number is 2sqrt3+2i. Determine the rectangular form of its other two cube roots. I'm not sure how to go about solving it.

Answer 1

\[x=2\sqrt{3}+2\iota ,x^3=\left( 2\sqrt{3}+2\iota \right)^3\]
\[=24\sqrt{3}+8\iota^3+12\sqrt{3}\iota \left( 2\sqrt{3}+2\iota \right)\]
\[=24\sqrt{3}+8 \iota^2\iota+72\iota+24\sqrt{3}\iota^2\]
\[=24\sqrt{3}-8\iota+72\iota-24\sqrt{3}=64\iota =64e ^{\left( 2n+\frac{ 1 }{2 } \right)\pi }\]
where n=0,1,2
taking cube roots
\[x=\left\{ 4^3\left( e ^{\left( 4n+1 \right)\frac{ \pi }{2 }\iota } \right) \right\}^{\frac{ 1 }{3 } }\]
\[x=4e ^{\left( 4n+1 \right)\frac{ \pi }{ 6 }\iota },n=0,1,2\]
\[x=4e ^{\iota \frac{ \pi }{ 6 }},4e ^{\iota \frac{ 5\pi }{6 }},4e ^{\iota \frac{ 9\pi }{6 }} \]
\[x=4e ^{\iota \frac{ \pi }{6 }}=4\left( \cos \frac{ \pi }{6 }+\iota \sin \frac{ \pi }{6 } \right)\]
\[=4\left( \frac{ \sqrt{3} }{ 2 }+\frac{1 }{ 2 }\iota \right)=2\sqrt{3}+2\iota \]
\[again~x=4e ^{\iota \frac{ 5\pi }{6 }}=4\left( \cos \frac{5\pi }{ 6 }+\iota \sin \frac{ 5\pi }{ 6 } \right)\]\[ =4\left( -\frac{ \sqrt{3} }{2 }+\frac{ 1 }{2}\iota \right)=-2\sqrt{3}+2\iota \]
\[and~ x=4e ^{\iota \frac{ 9\pi }{ 6 }}=4\left( \cos \frac{ 9 \pi }{6 } +\iota \sin \frac{ 9\pi }{6 }\right) \]
\[=4\left( \cos \frac{ 3\pi }{ 2 }+\iota \sin \frac{3 \pi }{2 } \right)=4\left( 0-\iota \right)=-4\iota \]

Answer 2

Thank you so much! I think I understand now

Answer 3

yw