By the polynomial remainder theorem, because x = 2i is a zero of f(x), we have no remainder upon dividing f(x) by x - 2i.
Computing the quotient yields
[tex]\dfrac{x^4+x^2+a}{x-2i} = x^3+2ix^2-3x-6i+\dfrac{12+a}{x-2i}[/tex]
Then if the remainder term is 0, follows that a = -12.
