The problem stems from the fact that you have a linear polynomial with real coefficients and you want a complex (non-real) solution. If then must be a complex number in order for to be a root of .

That said, what you've written is good. You've shown that the only way for this complex number to be a root of this real polynomial is for both the coefficients to be 0. Now, I suspect you are actually supposed to have a quadratic polynomial.