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AfilCa [17]
2 years ago
11

Find the cube roots of 8i. Write the answer in a + bi form.

Mathematics
2 answers:
Murrr4er [49]2 years ago
8 0

Answer:

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

like this

AnnZ [28]2 years ago
3 0

Answer:

It's B

Step-by-step explanation:

Just took the quiz

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