Answer:
2^(1/6) (cos(-pi/12)+i sin(-pi/12))
2^(1/6) (cos(3pi/12)+i sin(3pi/12))
2^(1/6) (cos(7pi/12)+i sin(7pi/12))
2^(1/6) (cos(11pi/12)+i sin(11pi/12))
2^(1/6) (cos(5pi/4)+i sin(5pi/4))
2^(1/6) (cos(19pi/12)+i sin(19pi/12))
Step-by-step explanation:
Let's convert to polar form.
-2i=2(cos(A)+i sin(A) )
There is no real part so cos(A) has to be zero and since we want -2 and we already have 2 then we need sin(A)=-1 so let's choose A=-pi/2.
So z=2(cos(-pi/2)+i sin(-pi/2)).
There are actually infinitely many ways we can write this polar form which we will need.
z=2(cos(-pi/2+2pi k)+i sin(-pi/2+2pi k))
where k is an integer
Now let's find the 6 6th roots or z.
2^(1/6) (cos(-pi/12+2pi k/6)+i sin(-pi/12+2pi k/6))
Reducing
2^(1/6) (cos(-pi/12+pi k/3)+i sin(-pi/12+pi k/3))
Plug in k=0,1,2,3,4,5 to find the 6 6th roots.
k=0:
2^(1/6) (cos(-pi/12+pi (0)/3)+i sin(-pi/12+pi (0)/3))
=2^(1/6) (cos(-pi/12)+i sin(-pi/12))
k=1:
2^(1/6) (cos(-pi/12+pi/3)+i sin(-pi/12+pi/3))
2^(1/6) (cos(3pi/12)+i sin(3pi/12))
k=2:
2^(1/6) (cos(-pi/12+2pi/3)+i sin(-pi/12+2pi/3))
2^(1/6) (cos(7pi/12)+i sin(7pi/12))
k=3:
2^(1/6) (cos(-pi/12+3pi/3)+i sin(-pi/12+3pi/3))
2^(1/6) (cos(11pi/12)+i sin(11pi/12))
k=4:
2^(1/6) (cos(-pi/12+4pi/3)+i sin(-pi/12+4pi/3))
2^(1/6) (cos(15pi/12)+i sin(15pi/12))
2^(1/6) (cos(5pi/4)+i sin(5pi/4))
k=5:
2^(1/6) (cos(-pi/12+5pi/3)+i sin(-pi/12+5pi/3))
2^(1/6) (cos(19pi/12)+i sin(19pi/12))
