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What are the solutions to x^3=−2−2i in polar form? Select all correct answers.

 

 Apr 29, 2020

Best Answer 

 #1
avatar+20981 
+2

To find the three cube roots of  -2 - 2i

find the radius:  sqrt( (-2)2 + (-2)2 )  =  sqrt( 4 + 4 )  =  sqrt(8)

and the angle:  tan-1( -2 / -2 )  =  5pi/4     [remember that it is in the third quadrant]

 

We now have:  81/2cis( 5pi / 4 )

 

To find the first cube root, we need to find the cube root of  81/2  which is  (81/2)1/3  =  81/6                            

and to divide the angle by 3, giving us  5pi /12.

So, for our first answer, we have  81/6 cis ( 5pi / 12 )

 

To find the next answer, we need to add 1/3 (or 2pi / 3 ) of the circle to our previous answer:   81/6 cis ( 13pi / 12 )

and to get our last answer we need to add another  2pi / 3 to this answer:   81/6 cis ( 21pi / 12 )  =  81/6 cis ( 7pi / 4 )

 Apr 30, 2020
 #1
avatar+20981 
+2
Best Answer

To find the three cube roots of  -2 - 2i

find the radius:  sqrt( (-2)2 + (-2)2 )  =  sqrt( 4 + 4 )  =  sqrt(8)

and the angle:  tan-1( -2 / -2 )  =  5pi/4     [remember that it is in the third quadrant]

 

We now have:  81/2cis( 5pi / 4 )

 

To find the first cube root, we need to find the cube root of  81/2  which is  (81/2)1/3  =  81/6                            

and to divide the angle by 3, giving us  5pi /12.

So, for our first answer, we have  81/6 cis ( 5pi / 12 )

 

To find the next answer, we need to add 1/3 (or 2pi / 3 ) of the circle to our previous answer:   81/6 cis ( 13pi / 12 )

and to get our last answer we need to add another  2pi / 3 to this answer:   81/6 cis ( 21pi / 12 )  =  81/6 cis ( 7pi / 4 )

geno3141 Apr 30, 2020

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