Short (but cryptic) answer: complex numbers arise in quantum mechanics because we would like find solutions to the differential equation $$\frac{\partial}{\partial x}f(x) = cf(x)$$ which don't blow up as $x\to \pm\infty$.
Long answer:
Fundamentally, the shift from classical mechanics to quantum mechanics is replacing functions (observables) and numbers (states) by matrices (quantum observables) and vectors (quantum states). The first examples of this (say, in Griffiths) is where the vectors are complex valued functions and the "matrices" are differential operators which act on the space of such functions. As in classical mechanics, in order to make sense of measured quantities, we would like the results of measurements to be real, which is analogous to requiring that the eigenvalues of the matrices which correspond to our quantum observables be real (in other words, the matrices must be Hermitian). As long as the final, measured answer is real there is no reason to exclude the use of complex numbers.
However, having no reason to exclude something is very different from having a reason for it! So why do the complex numbers appear in quantum mechanics? This can probably be tied, at least in part, to the relationship between the position and momentum operators. The position operator is not hard to explain. If we take the wavefunction $\psi$ (really, the square of its norm, $|\psi|^2$) as a probability distribution telling us where a particle is likely to be, then the integral $$\langle x \rangle = \int x\,|\psi(x)|^2dx$$ measures the expected value of the position of the particle. However, the momentum operator $\hat{p} = i\frac{\partial}{\partial x}$ is more difficult to interpret. One possibility is that you know that the momentum of the particle (wave) should be related to the derivative of the wavefunction. But then you would check that, due to the minus sign acquired when doing integration by parts, the first derivative operator $\frac{\partial}{\partial x}$ is not Hermitian (this is the part where we secretly require that our wavefunctions not blow up as $x\to \pm\infty$). In fact, it is as far from having real eigenvalues as possible, all its eigenvalues are purely imaginary! Of course, we can fix this by multiplying by $i$, but we have had to introduce complex numbers into our theory in order to do so!
Furthermore, this definition of $\hat{p}$ seems almost inevitable. Once you know that the position operator and momentum operator must have commutator equal to a constant $$[\hat{x},\hat{p}] = c$$ you have pretty much forced $\hat{p}$ to be a first order differential operator. The Hermitian condition then forces us to introduce the $i$ for the reasons mentioned above. So we can trace the presence of complex numbers in our theory back to the requirement that $\hat{x}$ and $\hat{p}$ have a non-zero commutation relation. Why this is necessary also has an interesting answer, but I will end my ramblings here!
