In a textbook I recently read that
$$ A_1e^{x}+A_2e^{-x}+A_3e^{ix}+A_4e^{-ix}$$
(where $A_n\in\mathbb{C}$, $A_i\neq A_j\;\forall\;i\neq j$) can be rewritten as
$$ A'_1\sin x+A'_2\cos x+A'_3\sinh x+A'_4\cosh x$$ (where $A'_n\in\mathbb{R}$).
Is that true? How can I prove that?
You know that $$ \sinh(x) = \frac{e^x - e^{-x}}{2} \\ \cosh(x) = \frac{e^x + e^{-x}}{2} \\ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \\ \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$ From that you can deduce $$ A_1e^x = A_1\cosh(x) + A_1\sinh(x) \\ A_2e^{-x} = A_2\cosh(x) - A_2\sinh(x) \\ A_3e^{ix} = A_3\cos(x) + A_3i\sin(x) \\ A_4e^{-ix} = A_4\cos(x) - A_4i\sin(x) $$ Summing everything gives us : $$ A_1e^x + A_2e^{-x} + A_3e^{ix} + A_4e^{-ix} = \\ (A_1+A_2)\cosh(x) + (A_3 + A_4)\cos(x) + (A_1 - A_2)\sinh(x) + i(A_3- A_4)\sin(x) $$ Therefore you get that $$ \begin{cases} A_1' = i(A_3 - A_4) \\ A_2' = A_3 + A_4 \\ A_3' = A_1 - A_2 \\ A_4' = A_1 + A_2 \end{cases} $$