If n is a positive integer, then (1 + i 3 )^ 2n + (1 - i 3 )^ 2n is equal to .........

Step-by-step solution

Let z = 1 + i 3 . Converting to polar form, we have r = 1^2 + ( 3 )^2 = 2 and = ⁻¹( 3 ) = 3 . Thus, 1 + i 3 = 2 ( 3 + i 3 ) . Similarly, 1 - i 3 = 2 ( 3 - i 3 ) . Using De Moivre's theorem, we get: (1 + i 3 )^ 2n = 2^ 2n ( 2n 3 + i 2n 3 ) (1 - i 3 )^ 2n = 2^ 2n ( 2n 3 - i 2n 3 ) Adding the two expressions: (1 + i 3 )^ 2n + (1 - i 3 )^ 2n = 2^ 2n (2 2n 3 ) = 2^ 2n+1 2n 3 . Answer: 2^ 2n+1 ( 2n 3 )