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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??

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    Sorry, I didn't make that very clear. I was just suggesting a strategy (with little thought obviously), but in no way meant for this to be possible. Thanks for picking that up.2011-12-30

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In fact, we only need that $f$ is continuous with connected domain, while $\sin$ could be replaced by any analytic function. Since $\sin$ is analytic, the set of ponts $w$ satisfying $w=\sin w$ is discrete; hence the image of $f$ is discrete. But $f$ is continuous, and $\mathbb{C}$ is connected, and a continuous function from a connected space to a discrete space is constant.

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    By any analytic function except for $w$ :).2011-12-19