Let's do it the hard way. We want to solve the equation
x5−1=(x−1)(x4+x3+x2+x+1)=0.
Then we are interested in solving x4+x3+x2+x+1=0. Note the symmetry: if r is a root, 1/r is also a root. Why? Thereafter, we have two ways out. First, we divide everything by x2 to get
x2+x+1+
1
x
+
1
x2
=(x+
1
x
)+(x+
1
x
)2−1=u2+u−1=0.
What did we do? We noted (x+x−1)2−2=x2+x−2 and substituted xu2+u=1+x−1=u. The solutions for u2+u=1 are −φ and φ−1, φ being the golden ratio. We now have two equations:
x+
1
x
=−φ⟹x2+φx+1=0⟹x=
±
√
φ−3
−φ
2
x+
1
x
=φ−1⟹x2+(1−φ)x+1=0⟹x=
±
√
−φ−2
+φ−1
2
.
You can now manipulate it to get a a+bi form. Alternatively, you can multiply (x−r)(x−r−1), divide x4+x3+x2+x+1 by it and discover which r makes remainder zero. This would be somewhat long, so I won't explictly do it, but here is the idea.
Maybe can help you
Or open this web
http://math.stackexchange.com/questions/603332/how-can-i-find-the-fifth-root-of-unity
