z=1+ai
Cubing both sides,
z3=(1−a2)+2ai+(1−a2)ai−2a2
z3=1−3a2+(3a−a3)i
∵ z3 is a real number.
a(3−a2)=0⇒a=3 (a>0)
1+z+z2…..z11=z−1z12−1=1+3i−1(1+3i)12−1=3i(1+3i)12−1
(1+3i)12=212(21+23i)12=212(cos3π+isin3π)12=212(cos4π+isin4π)=212
=3i212−1=−13653i