Let z = 1 + ai be a complex number, a > 0, such that z³ is a real number. Then the sum 1 + z + z² + .... + z¹¹ is equal to:

z=1+aiz2=1−a2+2aiz2.z=(1−a2)+2ai1+ai=(1−a2)+2ai+(1−a2)ai𕒶a2∵z3is real⇒2a+(1−a2)a=0⇒a(3−a2)=0⇒a=√3(a>0)1+z+z2.z11=z12𕒵z𕒵[Sum=a(rn𕒵)r𕒵whenr>1]=(1+√3i)12𕒵1+√3i𕒵=(1+√3i)12𕒵√3i(1+√3i)12=212(12+√32i)12=212(cosπ3+isinπ3)12=212(cos4π+isin4π)=212⇒212𕒵√3i=4095√3i=�√3i=𕒵365√3i