If z= ( 3 2 + i 2 )^5+ ( 3 2 - i 2 )^5 , then

Step-by-step solution

Let z₁ = 3 2 + i 2 = ( 6 ) + i ( 6 ) = e^ i /6 . Let z₂ = 3 2 - i 2 = (- 6 ) + i (- 6 ) = e^ -i /6 . The given expression is z = z₁^5 + z₂^5 . Using De Moivre's Theorem, z₁^5 = ( 5 6 ) + i ( 5 6 ) and z₂^5 = (- 5 6 ) + i (- 5 6 ) . Since (- ) = ( ) and (- ) = - ( ) , we have z₂^5 = ( 5 6 ) - i ( 5 6 ) . Adding these, z = 2 ( 5 6 ) + i ( ( 5 6 ) - ( 5 6 ) ) = 2 ( 5 6 ) . Since ( 5 6 ) = - 3 2 , we get z = 2 (- 3 2 ) = - 3 . Thus, Re (z) = - 3 and Im (z) = 0 . Answer: Im (z)=0