Just a minor detail, right?
Okay that means that x+1 and x- 3 must be factors: write the polynomial as [tex](ax+ b)(x+1)(x-3)= (ax+b)(x^2- 2x+ 1)= ax^3+ (b-2a)x^2+ (a-2b)x+ b[/quote]. Now choose a and b to be imaginary number such that neither b-2a nor a-2b is 0. a= b= i works nicely.