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. $x = -y = \frac \pi {\sqrt 3}$.