For this problem, we will need to know how to find the n^"th" roots of a complex number. To do this, we will use the identity
e^(itheta) = cos(theta)+isin(theta)
Because of this identity, we can represent any complex number as
a+bi = Re^(itheta) where R = sqrt(a^2 + b^2) and theta = arctan(b/a)
Now we will go over the steps to find the 3^"rd" roots of a complex number a+bi. The steps for finding the n^"th" roots are similar.
Given a+bi = Re^(itheta) we are looking for all complex numbers z such that
z^3 = Re^(itheta)
As z is a complex number, there exist R_0 and theta_0 such that
z = R_0e^(itheta_0)
Then
z^3 = (R_0e^(itheta_0))^3 = R_0^3e^(3itheta_0) = Re^(itheta)
From this, we immediately have R_0 = R^(1/3). We also may equate the exponents of e, but noting that as sine and cosine are periodic with period 2pi, then from the original identity, e^(itheta) will be as well. Then we have
3itheta_0 = i(theta + 2pik) where k in ZZ
=> theta_0 = (theta+2pik)/3 where k in ZZ
However, as if we keep adding 2pi over and over, we will end up with the same values, we can ignore the redundant values by adding the restriction theta_0 in [0, 2pi), that is, k in {0, 1, 2}
Putting it all together, we get the solution set
z in {R^(1/3)e^(itheta/3), R^(1/3)e^(i((theta+2pi))/3), R^(1/3)e^(i(theta+4pi)/3)}
We may convert this back to a+bi form if desired using the identity
e^(itheta) = cos(theta) + isin(theta)
Applying the above to the problem at hand:
(z-1)^3 = 8i
=> z-1 = 2i^(1/3)
=> z = 2i^(1/3) + 1
Using the above process, we can find the 3^"rd" roots of i:
i = e^(ipi/2) => i^(1/3) in {e^(ipi/6), e^(i(5pi)/6), e^(i(3pi)/2)}
Applying e^(itheta) = cos(theta) + isin(theta) we have
i^(1/3) in {sqrt(3)/2 + i/2, -sqrt(3)/2 + i/2, -i}
Finally, we substitute in these values for z = 2i^(1/3)+1
z in {2(sqrt(3)/2 + i/2)+1, 2(-sqrt(3)/2 + i/2)+1, 2(-i)+1}
= {sqrt(3)+1+i, -sqrt(3)+1+i, 1-2i}
