Question

Find the fifth roots of 32(cos 280° + i sin 280°).

I have gotten to the point where I have 2(cos(128) + isin(128)) but I don't know where to go from there. Can anyone please help?

Answer 1

2(cos(128) + isin(128)) is just one of the 5 fifth roots

so you'll need to find 4 more

Answer 2

Here is the process used to find all 5 fifth roots of 32(cos 280° + i sin 280°)


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


Ex: when k = 3, we have found the fourth 5th root


First 5th root:

k = 0


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


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]


2*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]


2*[ cos( (280+0)/5 ) + i*sin( (280+0)/5 ) ]


2*[ cos( 280/5 ) + i*sin( 280/5 ) ]


2*[ cos( 56 ) + i*sin( 56 ) ]



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Second 5th root:

k = 1


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


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]


2*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]


2*[ cos( (280+360)/5 ) + i*sin( (280+360)/5 ) ]


2*[ cos( 640/5 ) + i*sin( 640/5 ) ]


2*[ cos( 128 ) + i*sin( 128 ) ]


This is the root you found

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Third 5th root:

k = 2


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


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]


2*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]


2*[ cos( (280+720)/5 ) + i*sin( (280+720)/5 ) ]


2*[ cos( 1000/5 ) + i*sin( 1000/5 ) ]


2*[ cos( 200 ) + i*sin( 200 ) ]


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Fourth 5th root:

k = 3


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


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]


2*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]


2*[ cos( (280+1080)/5 ) + i*sin( (280+1080)/5 ) ]


2*[ cos( 1360/5 ) + i*sin( 1360/5 ) ]


2*[ cos( 272 ) + i*sin( 272 ) ]


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Fifth 5th root:

k = 4


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


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]


2*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]


2*[ cos( (280+1440)/5 ) + i*sin( (280+1440)/5 ) ]


2*[ cos( 1720/5 ) + i*sin( 1720/5 ) ]


2*[ cos( 344 ) + i*sin( 344 ) ]


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Summary:



Given z = 32 * [ cos( 280 ) + i*sin( 280 ) ], the five 5th roots of z are:


2*[ cos( 56 ) + i*sin( 56 ) ]
2*[ cos( 128 ) + i*sin( 128 ) ]
2*[ cos( 200 ) + i*sin( 200 ) ]
2*[ cos( 272 ) + i*sin( 272 ) ]
2*[ cos( 344 ) + i*sin( 344 ) ]


Note: all angles are in degrees


You can check your answer by raising each fifth root to the fifth power to get 32 * [ cos( 280 ) + i*sin( 280 ) ] back again (use De Moivre's Formula)

Answer 3

I actually understand that. Thank you so much. I have been trying for so long to understand this nth root stuff. Thank you.

Answer 4

you're welcome