f(1+i) = (1+i)^4 -2(1+i)^3 -(1+i)^2 +6(1+i) - 6;
But, (1+i)^2 = 1^2 + 2*1*i + i^2 = 1 + 2i - 1 = 2i; because i^2 = -1;
(1+i)^4 = [(1+i)^2]^2 = (2i)^2 = 4*i^2 = -4;
(1+i)^3 =(1+i)^2 * (1+i) = 2i * ( 1 + i ) = 2i + 2i^2 = 2i - 2 ;
Then, f(1+i)= -4 - 2( 2i - 2 ) - 2i + 6( 1 + i ) - 6 = - 4 + 4i + 4 + 6 + 6i - 6 = 10i which isn't 0 ;
Finally, 1+i isn't zero of f(x);
