Question

Find the cube roots of 27(cos 279° + i sin 279°).

Answer 1

Use the formulas found on this page


http://www.mathxtc.com/Downloads/NumberAlg/files/NthRootsComplex.pdf


to get the following


First cube root


\[\Large \sqrt[3]{27}\left(\cos\left(\frac{279}{3}\right)+i\sin\left(\frac{279}{3}\right)\right)\]


\[\Large 3\left(\cos(93)+i\sin(93)\right)\]


So the first cube root is \[\Large 3\left(\cos(93)+i\sin(93)\right)\]


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Second cube root


\[\Large \sqrt[3]{27}\left(\cos\left(\frac{279}{3}+\frac{360}{3}\right)+i\sin\left(\frac{279}{3}+\frac{360}{3}\right)\right)\]


\[\Large 3\left(\cos(93+120)+i\sin(93+120)\right)\]


\[\Large 3\left(\cos(213)+i\sin(213)\right)\]


The second cube root is \[\Large 3\left(\cos(213)+i\sin(213)\right)\]


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Third cube root


\[\Large \sqrt[3]{27}\left(\cos\left(\frac{279}{3}+\frac{2*360}{3}\right)+i\sin\left(\frac{279}{3}+\frac{360}{3}\right)\right)\]


\[\Large 3\left(\cos(93+240)+i\sin(93+240)\right)\]


\[\Large 3\left(\cos(333)+i\sin(333)\right)\]


and the third cube root is \[\Large 3\left(\cos(333)+i\sin(333)\right)\]


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