When there are imaginary numbers, the synthetic division is essentially the same
process as when there are only real numbers.  There are basically just two
differences:

1. We just need to skip some extra space after the real part of each coefficient
to have room for the imaginary parts of the coefficients.  Put the imaginary
parts out to the right of the real numbers: 

2. Also we will have to do some scratch work to multiply complex numbers:

We start the same way as when there are only real numbers:

i | 1    1    -1    2   
  |       +i            ` 
    1    

Then we add the 1 and the +i and write the sum 1+i below the line:

i | 1    1    -1    2   
  |       +i            `   
    1    1+i

We have to do some scratch work here to multiply 1+i by i:
i(1+i) = i+i² = i+(-1) = -1+i

i | 1    1    -1    2   
  |       +i  -1+i      `
    1    1+i  -2+i

We do some more scratch work to multiply -2+i by i:
i(-2+i) = -2i+i² = -2i+(-1) = -1-2i

i | 1    1    -1     2   
  |       +i  -1+i  -1-2i`   
    1    1+i  -2+i   1-2i

So the quotient is a polynomial which has degree which is
1 less than the degree of the original polynomial.  The
original polynomial has degree 3, so the quotient will have
degree which is 1 less, so it will have degree 2.  Its
coefficients are on the bottom line of the synthetic 
division all but the last complex number, which is the 
remainder.

So the answers are:

Quotient = 1x² + (1+i)x + (-2+i)  [or x² + x + ix - 2 + i]

Remainder = 1-2i

Edwin