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What are the solutions to  x^3=−2−2i in polar form?

Select all correct answers.

8√6cis(5π/12)

8√6cis(13π/12)

8√6cis(7π/4)

8√6cis(2π/3)

22√3cis(2π/3)

22√3cis(7π/4)

22√3cis(13π/12)

22√3cis(5π/12)

8√6cis(5π/3)

 May 9, 2020
 #1
avatar+20906 
+1

-2 - 2i  in  r·cis(theta) form:

 

r  =  sqrt( (-2)2 + (-2)2 )  =  sqrt(8)

theta  =  tan-1( -2/-2 )  =  5pi/4

 

x3  =  sqrt(8) · cis ( 5pi/4 )     --->     x  =  [ sqrt(8) · cis ( 5pi/4 )  ]1/3

 

To find the principal cube root, take the cube root of  sqrt(8)  =  (81/2)1/3  =  (81/3)1/2  =  sqrt(2)

and divide 5pi/4 by 3  =  5p/12.

 

So, the principal cube root is:          sqrt(2)·cis( 5pi/12 )

Adding 2p/3, the second root is:     sqrt(2)·cis( 13pi/12 )

and another 2p/3, the third root is:  sqrt(2)·cis( 21pi/12 )

 

So, I don't get any of the answers ... 

 May 9, 2020

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