Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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??

share|cite|improve this question
It looks to me like your solution is correct! – Neal Dec 15 '11 at 11:44
You can also show that $f\,$ is bounded. If the function is entire and bounded, then it must be constant. See Liouville's theorem. – Henry Shearman Dec 15 '11 at 12:20
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!) – Georges Elencwajg Dec 15 '11 at 14:54
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. – Henry Shearman Dec 30 '11 at 13:17
up vote 13 down vote accepted

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.

share|cite|improve this answer
By any analytic function except for $w$ :). – Phira Dec 19 '11 at 16:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.