Let
z=√2+√32+i⋅√2−√32
Find
1+z+z2+z3+⋯+z2016
z=√2+√32+i⋅√2−√32z=√(√2+√32)2+(√2−√32)2ei∗arctan(√2−√3√2+√3)z=1∗ei∗π12z=ei∗π12
Sum of a geometric sequence
1+z+z2+z3+⋯+z2016=z2017−1z−1=(ei∗π12)2017−1ei∗π12−1=ei∗201712π−1ei∗π12−1=ei∗30255∘−1ei∗15∘−1=cos(30255∘)+isin(30255∘)−1cos(15∘)+isin(15∘)−1=cos(15∘)+isin(15∘)−1cos(15∘)+isin(15∘)−11+z+z2+z3+⋯+z2016=1