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2 years ago

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

1 answer:

3 0

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

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 ) ]
-------------------------------------------------------------------
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 ) ]
-------------------------------------------------------------------
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 ) ]
-------------------------------------------------------------------
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 ) ]
-------------------------------------------------------------------
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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We can solve this problem by stating that the summation of momentums must be zero. That is, the momentum at one end of the pivot must be equal to the momentum at the other end.

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