[Arkuski]

Yes yes, the imaginary roots translate into the series expansion of e^x. Of course, the solution for this type of ODE will be e^ax, but I'm guessing you need a better explanation as to why an imaginary root turns into something that has to do with sin and cos. Let's compute e^(a+bi). We can easily see that this is equal to (e^a)*(e^bi). So now, let's focus on e^bi. We plug this into the series expansion:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...

e^bi = 1 + (bi) + (bi)^2/2! + (bi)^3/3! + (bi)^4/4! + ...
= 1 + bi - b^2/2! - i(b^3) + b^4/4! + ...

We can separate even and odd terms in the series and find the following:

e^bi = (1 - b^2/2! + b^4/4! + ... ) + i(b - b^3/3! + b^5/5! + ... )

By recognizing these Taylor series, we can deduce the following:

e^bi = cos(b) + i*sin(b)

And, of course, we include our original e^a:

e^(a+bi) = (e^a) * [cos(b) + i*sin(b)]