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've tried looking up videos on how to do this, but I'm very confused as there aren't any that look quite like the one I was given. If someone could help, I'd appreciate that so much!

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

Step-by-step explanation:

The genral form of a complex number in rectangular plane is expressed as z = x+iy

In polar coordinate, z =rcos ∅+irsin∅ where;

r is the modulus = √x²+y²

∅ is teh argument = arctan y/x

Given thr complex number z = 6+6√(3)i

r = √6²+(6√3)²

r = √36+108

r = √144

r = 12

∅ = arctan 6√3/6

∅ = arctan √3

∅ = 60°

In polar form, z = 12(cos60°+isin60°)

z = 12(cosπ/3+isinπ/3)

To get the fourth root of the equation, we will use the de moivres theorem; zⁿ = rⁿ(cosn∅+isinn∅)

z^1/4  = 12^1/4(cosπ/12+isinπ/12)

When n = 1;

z1 =  12^1/4(cosπ/3+isinn/3)

z1 = 12^1/4cis(π/3)

when n = 2;

z2 = 12^1/4(cos2π/3+isin2π/3)

z2 = 12^1/4cis(2π/3)

when n = 3;

z2 = 12^1/4(cosπ+isinπ)

z2 = 12^1/4cis(π)

when n = 4;

z2 = 12^1/4(cos4π/3+isin4π/3)

z2 = 12^1/4cis(4π/3)

answer: 298143

step-by-step explanation:

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12