I was trying to generate a method of rotating a function f(x) by angle a $y = f(x)$ is the Cartesian equation, which can be represented in parametric form in terms of $t$ as $$x = t \;\; \text{ and } \;\; y = f(t)$$
I rewrote it as a complex number $x + iy = t + i f(t)$, and multiplied it by $e^{i a} = (t + i f(t)) (\cos(a) + i \sin(a)) = t \cos(a) - \sin(a) f(t) + i(t \sin(a) + \cos(a) f(t))$, and re-expressed it as a parametric equation in terms of $t$ as:
$x = t \cos(a) - \sin(a) f(t)$ and $y = t \sin(a) + \cos(a) f(t)$.
Now I need to convert it back into Cartesian form, where I require a way to extract t from either equations.
Your relationships :
$$\begin{cases}x &=& \cos(a)t - \sin(a) f(t)\\y &=& \sin(a) t + \cos(a) f(t)\end{cases}\tag{1}$$
are equivalent to :
$$\begin{cases}t&=& \ \ \ \cos(a)x+\sin(a)y\\f(t)&=&-\sin(a)x+\cos(a)y \end{cases}\tag{2}$$
(replacement of multiplication by $e^{ia}$ by multiplication by $e^{-ia}$ ; see Remark at the bottom with a matrix-vector explanation).
"Plugging" the first equation of (2) into the second, we obtain an implicit equation :
$$f(\cos(a)x+\sin(a)y)=-\sin(a)x+\cos(a)y\tag{3}$$
In the general case, one cannot go further to obtain a cartesian equation $y=\varphi(x)$.
If we consider particular cases, we can, at least in certain cases, provide an explicit cartesian equation.
Take for example the case of a linear function $y=2x+1$ : (3) becomes,
$$2(\cos(a)x+\sin(a)y)+1=-\sin(a)x+\cos(a)y$$
from which you can extract $y$ as a function of $x$ and parameter $a$.
Remark : Matrix explanation of the equivalence above :
$$\pmatrix{x\\y}=\pmatrix{\cos(a) &- \sin(a)\\ \sin(a)& \ \ \ \cos(a)}\pmatrix{t\\f(t)}$$
is equivalent to :
$$\pmatrix{t\\f(t)}=\pmatrix{\ \ \ \cos(a) & \sin(a)\\ -\sin(a)&\cos(a)}\pmatrix{x\\y}$$