I've been playing with tan identities and I noticed the fact that $\tan(2\arctan(x))=\frac{2x}{1-x^2}$. This follows from the arctan sum identity of $\arctan(x)+\arctan(y)=\arctan(\frac{x+y}{1-xy})$. Using this identity, for any rational number $c$, we can find rational functions $u(x),v(x)$ such that $\tan(c\arctan(u(x))=v(x)$. I want to figure out if this is possible for any irrational numbers $c$. I feel like it isn't.
We can write arctan in log form and then take exponential on both sides to get the following $$\left(\dfrac{i+u(x)}{i-u(x)}\right)^c =\dfrac{i+v(x)}{i-v(x)}$$
I'm trying to prove that this is not possible for irrational $c$ We can phrase this in the following way. Let $F$ and $G$ be non-constant rational functions with complex coefficients such that $F(x)^c=G(x)$, where $c$ is real. Prove that $c$ must be a rational number.
Let $F$ and $G$ be non-constant rational functions with $F(z)^c = G(z)$. Since $F$ is non-constant, it must have at least one zero or pole, and thus $a\in \mathbb C$ such that as $z \to a$, $F(z) \sim k (z-a)^m$ where $k$ is a nonzero constant and $m$ a nonzero integer. Then as $z \to a$, $G(z) \sim k^c (z-a)^{mc}$. But since a zero or pole must have integer order, $mc$ must be an integer $n$, and so $c = n/m$ is rational.