Expressing $A\sin x + B\cos x$ as a sine -- $A,B \in \mathbb{C}$ Suppose we have the expression $P=A\sin x + B\cos x$ , where $x$ is real and $A,B$ are complex numbers. Can we then write $P$ in the form $C \sin y$  , where obviously $C=C(A,B)$ and $y=y(A,B,x)$, with $y$ preferably being real ?
In the case where $A,B$ are real positive numbers, it's fairly easy to do. In fact it holds that $P=R\sin(x-a)$ where $R=\sqrt{A^2 +B^2}$ and $a=\arctan\frac{B}{A}$. We can find that by demanding $$A\sin x+B\cos x = R\sin(x+a) = R(\sin x \cos a + \cos x \sin a) $$ 
From which we can see that we need $R\cos a= A $ $(1)$ and $R\sin a=B$ $(2)$.
Thus, by doing the following two steps:
$$(2)/(1) \implies \tan a= \frac{B}{A} $$
$$ (1)^2 + (2)^2 \implies R^2 = A^2 + B^2$$
However when they are complex, dividing (1) and (2) gives us again $\tan a= \frac{B}{A} $ but now $a$ and thus $\tan a$ is real, but $ \frac{B}{A}$ is generally complex. Is there another (any?) way of doing this?

There is no way to work with $y$ real.
Indeed, when $x$ varies $A\sin x+B\cos x$ describes an ellipse in the complex plane, while $C\sin(y)$ is a line segment.
