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Find all the cube roots of \(8i \)

Guest Jun 22, 2017
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We have that

 

a0 =  ∛8 [ cos [ (pi/2)/3] +  i *sin [(pi/2)/3 ]  = ∛8 [ cos [ (pi/6)] +  i *sin [(pi/6) ] =

 2 [ √3/2 + 1/2 * i ] =    √3  + 1i

 

a1  = ∛8 [ cos ( (pi/2)/3 + 2pi/3 )  + i sin ( (pi/2 )/3 + 2pi/3 ) ]   =

 2 [ cos (5pi/6)  + i sin (5pi/6) ] =  2 [ - √3/2  +  (1/2)*i ]  =   - √3  +1i

 

a2  = ∛8 [ cos ( (pi/2)/3 + 4pi/3 )  + i sin ( (pi/2 )/3 + 4pi/3 ) ] =

2 [ cos(3pi/2) + i sin (3pi/2) ] =   2 [ -1 i ]  = -2i

 

 

 

 

cool cool cool

CPhill  Jun 22, 2017

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