Consider the function from a real value to a complex value:
$$f(x) = \cos(\sqrt{2} x ) + i \sin(\sqrt{3} x)$$
My contention is it never has the same complex value for two different real values of $x$.
i.e. $f(x)=f(y) \implies x=y$
Is this true? Is there a proof?
My second contention is that it has no definable inverse. $f^{-1}(z)$ from a complex number to a real number.
Edit:
I realised this must be wrong! But I think it is correct for this function to a quaternion:
$$g(x) = \cos(\sqrt{2} x ) + i \sin(\sqrt{2} x) + j\cos(\sqrt{3} x ) + k \sin(\sqrt{3} x)$$
Best Answer
The result is true for your $g(x)$.
For real numbers $u, v$, if $\cos(u) = \cos(v)$ and $\sin(u) = \sin (v)$, then we must have $u - v \in 2\pi \Bbb Z$.
Thus if $x, y$ are different real numbers such that $g(x) = g(y)$, then we must have $ax - ay \in 2\pi \Bbb Z$ and $bx - by \in 2\pi \Bbb Z$, where $a = \sqrt 2$ and $b = \sqrt 3$.
Taking quotient of the two, we get $a/b \in \Bbb Q$ which is false.
It is however not true for your $f(x)$. Take e.g. $x = -y = \frac \pi {\sqrt 3}$.