If a= ( 8 11 )+i ( 8 11 ) , then Re (a+a^2+a^3+a^4+a^5 )=

Step-by-step solution

a= ( 8 11 )+i ( 8 11 ) a=e^ i 8 11 a is 11 th root of unity and all roots are 1, a, a^2, , a¹⁰ Now, a¹⁰= a¹⁰ a a = a¹¹ a = 1 a = a Similarly, a^9= a^2 , a^8= a^3 , a^7= a^4 , a^6= a^5 , We know that, Sum of n roots of unity =0 gathered 1+a^1+a^2+a^3+ +a¹⁰=0 a+a^2+a^3+a^4+a^5+a^6+a^7+a^8 a^9+a¹⁰=-1 (a+ a )+ (a^2+ a^2 )+ (a^3+ a^3 ) + (a^4+ a^4 )+ (a^5+ a^5 )=-1 2 Re (a)+2 Re (a^2 )+2 Re (a^3 )+2 Re (a^4 )+2 Re (a^5 ) =-1[z+ z =2 Re (z)] 2 Re (a+a^2+a^3+a^4+a^5 )=-1 Re (a+a^2+a^3+a^4+a^5 )=- 1 2 gathered