Question

Find the fifth roots of 243(cos 300° + i sin 300°).

Answer 1

@jim_thompson5910

Answer 2

Given z = r*[ cos( theta ) + i*sin( theta ) ]


To find the five 5th roots of z


Use the formula the formula: r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ] where k = 0,1,2,3,4


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First 5th root:

k = 0


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(243)^(1/5)*[ cos( (300+360*k)/5 ) + i*sin( (300+360*k)/5 ) ]


(243)^(1/5)*[ cos( (300+360*0)/5 ) + i*sin( (300+360*0)/5 ) ]


3*[ cos( (300+360*0)/5 ) + i*sin( (300+360*0)/5 ) ]


3*[ cos( (300+0)/5 ) + i*sin( (300+0)/5 ) ]


3*[ cos( 300/5 ) + i*sin( 300/5 ) ]


3*[ cos( 60 ) + i*sin( 60 ) ]


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Second 5th root:

k = 1


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(243)^(1/5)*[ cos( (300+360*k)/5 ) + i*sin( (300+360*k)/5 ) ]


(243)^(1/5)*[ cos( (300+360*1)/5 ) + i*sin( (300+360*1)/5 ) ]


3*[ cos( (300+360*1)/5 ) + i*sin( (300+360*1)/5 ) ]


3*[ cos( (300+360)/5 ) + i*sin( (300+360)/5 ) ]


3*[ cos( 660/5 ) + i*sin( 660/5 ) ]


3*[ cos( 132 ) + i*sin( 132 ) ]


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Third 5th root:

k = 2


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(243)^(1/5)*[ cos( (300+360*k)/5 ) + i*sin( (300+360*k)/5 ) ]


(243)^(1/5)*[ cos( (300+360*2)/5 ) + i*sin( (300+360*2)/5 ) ]


3*[ cos( (300+360*2)/5 ) + i*sin( (300+360*2)/5 ) ]


3*[ cos( (300+720)/5 ) + i*sin( (300+720)/5 ) ]


3*[ cos( 1020/5 ) + i*sin( 1020/5 ) ]


3*[ cos( 204 ) + i*sin( 204 ) ]


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Fourth 5th root:

k = 3


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(243)^(1/5)*[ cos( (300+360*k)/5 ) + i*sin( (300+360*k)/5 ) ]


(243)^(1/5)*[ cos( (300+360*3)/5 ) + i*sin( (300+360*3)/5 ) ]


3*[ cos( (300+360*3)/5 ) + i*sin( (300+360*3)/5 ) ]


3*[ cos( (300+1080)/5 ) + i*sin( (300+1080)/5 ) ]


3*[ cos( 1380/5 ) + i*sin( 1380/5 ) ]


3*[ cos( 276 ) + i*sin( 276 ) ]


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Fifth 5th root:

k = 4


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(243)^(1/5)*[ cos( (300+360*k)/5 ) + i*sin( (300+360*k)/5 ) ]


(243)^(1/5)*[ cos( (300+360*4)/5 ) + i*sin( (300+360*4)/5 ) ]


3*[ cos( (300+360*4)/5 ) + i*sin( (300+360*4)/5 ) ]


3*[ cos( (300+1440)/5 ) + i*sin( (300+1440)/5 ) ]


3*[ cos( 1740/5 ) + i*sin( 1740/5 ) ]


3*[ cos( 348 ) + i*sin( 348 ) ]


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Given z = 243 * [ cos( 300 ) + i*sin( 300 ) ], the five 5th roots of z are:


3*[ cos( 60 ) + i*sin( 60 ) ]
3*[ cos( 132 ) + i*sin( 132 ) ]
3*[ cos( 204 ) + i*sin( 204 ) ]
3*[ cos( 276 ) + i*sin( 276 ) ]
3*[ cos( 348 ) + i*sin( 348 ) ]


Note: all angles are in degrees


I used the formulas found on this page: http://www.mathxtc.com/Downloads/NumberAlg/files/NthRootsComplex.pdf