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144
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avatar+277 

What are the fourth roots of 6+6√(3i) ?

Enter your answer by filling in the boxes. Enter the roots in order of increasing angle measure.

 

(   )cis(   )

(   )cis(   )

(   )cis(   )

(   )cis(   )

 

hello, i posted this before, but i really do need help on this. i tried looking up videos on how to do this, but i'm really confused.. If someone could help me, I'd appreciate that a lot.

I can figure out the roots in order of increasing angle measure, so no worries about that, I know how to do that:)

Thank you so so much.

 Jun 5, 2020
 #1
avatar+21953 
+2

Because you have a sqrt(i) in the problem, let's start there ...

 

sqrt(i)  =  sqrt(2)/2 + (sqrt(2)/2)·i

 

Problem:  6 + 6·sqrt(3i)  =  6 + 6·sqrt(3)·sqrt(i)

  substituting:                      6 + 6·sqrt(3)·[ sqrt(2)/2 + (sqrt(2)/2)·i ]

  expanding:                        6 + 6·sqrt(6)/2 + 6·sqrt(6)/2 ·i

                                           ( 6 + 3·sqrt(6) ) + ( 3·sqrt(6) )·i

 

This is of the form  a + bi  where  a = ( 6 + 3·sqrt(6) )  and  b = ( 3·sqrt(6) )

 

We can now put this into  r·cis(theta)  form by using  r  =  sqrt( a2 + b2 )  and  theta  =  tan-1( b/a )

 

r  =  sqrt[ ( 6 + 3·sqrt(6) )2 + ( 3·sqrt(6) )2  =  sqrt( 144 + 36·sqrt(6) ]

 

theta  =  tan-1[ ( 3·sqrt(6) ) / ( 6 + 3·sqrt(6) ) ]

 

Now, find the fourth root of r and divide theta by 4, to get your primary root; then, you can get the other three roots. 

 Jun 5, 2020

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