We will be assuming here that our roots are of the form, in this case,

r_1,2 = l+ mi

If we take the first root we'll find the following solution.

x^{l+ mi}

It is a problem as we don't want complex solutions, we only need real solutions. We can eliminate it by recalling,

x^r =e^{In xr =} e^{r In x}

Plugging the root in this provides,

x^{l+ mi} = e^{(l+ mi) In x}

= e^{l In x} e ^{mi In x}

= x^l cos(m In x) + i x^l sin(m In x)

Remember that we had to use Euler formula as well to find to the final step. This time, as we've done every other time we've noticed solutions as it we can take the real part and the imaginary part and utilize those for our two solutions.

Therefore, in the case of complex roots the general solution will be as,

 y(x) = c₁ x^l cos (m ln x ) + c₂ x^l sin (m ln x ) = x^l (c₁ cos (m ln x ) + c₂ sin (m ln x ))

Once again we can notice why we required x>0.