Actually, there is a trick for efficiently solving problem 43, which is similar to an exercise from L. Ahlfors complex analysis (I don't have the actually exercise number on hand, but if someone asks, I will look it up and post it):
Since z^5 = 1, we have that 1 + z + z^2 + z^3 + 5z^4 + 4z^5 + 4z^6 + 4z^7 + 4z^8 + 5z^9 = 5 + 5z + 5z^2 + 5z^3 + 10z^4 = 5(1 + z + z^2 + z^3 + z^4) + 5z^4.
The trick is that, z(1 + z + z^2 + z^3 + z^4) = z + z^2 + z^3 + z^4 + z^5 = 1 + z + z^2 + z^3 + z^4.
Since z != 1, we conclude that we must have that 1 + z + z^2 + z^3 + z^4 = 0.
So we have 5(1 + z + z^2 + z^3 + z^4) + 5z^4 = 5*0 + 5z^4 = 5e^(8ip/5) = 5e^(3ip/5 + ip) = -5e^(3ip/5). Where p = pie.
I dont know if that's what you had in mind, or if it is common enough to not consider it as a trick, but that's what I had in mind when I first came across the problem. However, by the construction of the problem, it seems apparent that they were looking for something like this.