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.

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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.

auxiarc Jun 5, 2020

#1**+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( a^{2} + b^{2} ) 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.

geno3141 Jun 5, 2020