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}#