Case b.

Where $m_1 \not= m_2$ and both $m_1$ & $m_2$ are complex. We know that

$$y_1(x) = e^{m_1x} \quad\quad y_2(x) = e^{m_2x}$$

But now $y_1$ & $y_2$ are complex. This means we can rewrite the general solution using Euler's Formula.

$$\textrm{let}\quad m_1 = \mu + i\lambda$$

$$\textrm{let}\quad m_2 = \mu - i\lambda$$

So the general solution is

$$y(x) = Ae^{m_1x} + Be^{m_2x} = Ae^{(\mu + i\lambda)x} + Be^{(\mu - i\lambda)x}$$

$$= e^{\mu x}[Ae^{i\lambda x} + Be^{-i\lambda x}]$$

$$= e^{\mu x}[A(\cos(\lambda x) + i\sin(\lambda x)) + B(\cos(\lambda x) - i\sin(\lambda x))]$$

$$= e^{\mu x}[(A+B)(\cos(\lambda x) + (A-B)i\sin(\lambda x))$$

$$= e^{\mu x}[(C)(\cos(\lambda x) + (D)\sin(\lambda x))$$

The final general solution is

$$e^{\mu x}[(C)(\cos(\lambda x) + (D)\sin(\lambda x))$$

Where $C = A+B$ & $D=i(A-B)$ are both integration constants.