I'm trying to show why an entire function with the property $f(z)= \sin(f(z))$ everywhere must be constant.
Is it sufficient to say that when taking the derivatives, we will get $f'(z)=f'(z) \cdot \cos(f(z))$, so either $f'$ is zero, so $f$ constant, or $\cos(f)=1$, so $f(z)=2 \pi k$ for all $z$, which means that by continuity, $f$ cannot be $2 \pi k_1$ at $z_1$ and $2 \pi k_2$ at $z_2$ for different $k$ (since in the image, along any path from $2\pi k_1$ to $2\pi k_2$, $f$ would not be $1$ anymore), so $f=2\pi k_0$ for some $k_0$, so again, $f$ constant.
Do we have the right to use normal chain rule here, since I first tried to use Cauchy-Riemann equations, and did not succeed with that. Or does this require some properties of sine, or is my solution even correct??