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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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    $\begingroup$ It looks to me like your solution is correct! $\endgroup$
    – Neal
    Dec 15, 2011 at 11:44
  • $\begingroup$ You can also show that $f\,$ is bounded. If the function is entire and bounded, then it must be constant. See Liouville's theorem. $\endgroup$
    – Henry Shearman
    Dec 15, 2011 at 12:20
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    $\begingroup$ Dear @Henry, I fail to see why $f$ should be bounded, since the sine function is definitely not bounded on $\mathbb C$ (because of Liouville's theorem , for example!) $\endgroup$
    – Georges Elencwajg
    Dec 15, 2011 at 14:54
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    $\begingroup$ 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. $\endgroup$
    – Henry Shearman
    Dec 30, 2011 at 13:17

1 Answer 1

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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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    $\begingroup$ By any analytic function except for $w$ :). $\endgroup$
    – Phira
    Dec 19, 2011 at 16:51

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