The main point in signature (1,3) is that the corresponding Laplacian is the wave operator, whereas in (0,4) you have the standard Laplacian (up to overall sign changes, but that doesn't matter). Hence your differential equations are of a different type, even though your functions may be complex-valued.
Hence there is a big difference between solutions of
$\Delta u=0$ for $u:M\rightarrow \mathbb{C}$ (elliptic) and
$\square u=0$ for $u:M\rightarrow \mathbb{C}$ (hyperbolic)
Now translate this to your Dirac equation (being some sort of square root of the above), where the functions you consider are the same (sections of isomorphic spinor bundles), but the Dirac operators $\gamma^i \nabla_i$ are different.
In order to answer the question why one uses the complexified Clifford algebra - that has to do with quantum mechanics, where it is more natural to work over the complex numbers!

The main point in signature (1,3) is that the corresponding Laplacian is the wave operator, whereas in (0,4) you have the standard Laplacian (up to overall sign changes, but that doesn't matter). Hence your differential equations are of a different type, even though your functions may be complex-valued.
Hence there is a big difference between solutions of
$\Delta u=0$ for $u:M\rightarrow \mathbb{C}$ (elliptic) and
$\square u=0$ for $u:M\rightarrow \mathbb{C}$ (hyperbolic)
Now translate this to your Dirac equation (being some sort of square root of the above), where the functions you consider are the same (sections of isomorphic spinor bundles), but the Dirac operators $\gamma^i \nabla_i$ are different.

The main point in signature (1,3) is that the corresponding Laplacian is the wave operator, whereas in (0,4) you have the standard Laplacian (up to overall sign changes, but that doesn't matter). Hence your differential equations are of a different type, even though your functions may be complex-valued.
Now translate this to your Dirac equation (being some sort of square root of the above)!

The main point in signature (1,3) is that the corresponding Laplacian is the wave operator, whereas in (0,4) you have the standard Laplacian (up to overall sign changes, but that doesn't matter). Hence your differential equations are of a different type, even though your functions may be complex-valued.
Hence there is a big difference between solutions of
$\Delta u=0$ for $u:M\rightarrow \mathbb{C}$ (elliptic) and
$\square u=0$ for $u:M\rightarrow \mathbb{C}$ (hyperbolic)
Now translate this to your Dirac equation (being some sort of square root of the above), where the functions you consider are the same (sections of isomorphic spinor bundles), but the Dirac operators $\gamma^i \nabla_i$ are different.
In order to answer the question why one uses the complexified Clifford algebra - that has to do with quantum mechanics, where it is more natural to work over the complex numbers!