Proof for the linear independence of exponentials generalised to complex powers Consider exponential functions $\exp(ax)$.  I've seen proofs online that shows that it is linearly independent when $a$ is real.  However, how do I generalise that $\exp(ax)$ are linearly independent even when $a$ are complex?
 A: The functions $x\mapsto\exp(ax)$ are eigenvectors of the linear map $f\mapsto f'$ with distinct eigenvalues. Therefore, they are linearly independent.
A: We can test for the linear independence of $n$ analytic functions, $f_1, \ldots, f_n$, by computing their Wronskian $W[f_1, \ldots, f_n]$, that is, the determinant of the matrix whose $(i, j)$ entry is $\partial_z^{i - 1} f_j$: $\{f_1, \ldots, f_n\}$ is linearly independent iff $W[f_1, \ldots, f_n] = 0$.
In our case, the functions are exponential maps $$f_j : z \mapsto \exp(a_j z) ,$$ and we have $(\partial_z^{i - 1} f_j)(z) = a_j^{i - 1} \exp(a_j z)$, so the Wronskian has a particularly simple form. For example, for $n = 3$ (and denoting the parameters in the exponential functions by $a, b, c$) we have $$W[f_1, f_2, f_3] = \det \pmatrix{e^{az}&e^{bz}&e^{cz}\\a e^{az}&b e^{bz}&c e^{cz}\\a^2 e^{az}&b^2 e^{bz}&c^2 e^{cz}&} = e^{(a + b + c) z} \det\pmatrix{1&1&1\\a&b&c\\a^2&b^2&c^2} .$$

Since $e^{(a + b + c) z}$ is never zero, it is enough to check whether $$\det\pmatrix{1&1&1\\a&b&c\\a^2&b^2&c^2}$$ vanishes. But this matrix is a square Vandermonde matrix (or perhaps the transpose of one, depending on convention), and its determinant is $(b - a)(c - a)(c - b)$. In particular, if $a, b, c$ are all pairwise distinct, this determinant is nonzero. The general case is similar.

