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}#
